Class 10 – Page 21 – AP Board Solutions

AP SSC 10th Class English Solutions Chapter 1B Every Success Story is Also a Story of Great Failures

March 29, 2023February 5, 2023 by Mahesh

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 1B Every Success Story is Also a Story of Great Failures Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 1B Every Success Story is Also a Story of Great Failures

10th Class English Chapter 1B Every Success Story is Also a Story of Great Failures Textbook Questions and Answers

Comprehension

I. Answer the following questions.

Question 1.
‘To Lincoln, defeat was a detour and not a dead end.’ Justify this statement from the life of Lincoln.
Answer:
Abraham Lincoln was elected President of the United States at age 52. Before he became President, he faced many setbacks. He failed in business, lost in legislative, senatorial and congressional races. He lost his sweetheart when he was 22. Thus, Lincoln had failed many times before he got the big success. He proved that defeat was a detour and not a dead end.

Question 2.
What did Socrates suggest as the secret to success? Do you agree or disagree with it? Give reasons for your answer.
Answer:
Socrates suggested that burning desire is the only secret to one’s success. I agree with this statement. If we don’t have a desire, we can’t achieve anything. A burning desire is the starting point of all accomplishments. A weak desire can’t produce great results. The motivation to get the success comes from the burning desire to achieve a purpose.

Question 3.
Complete the following table about the failures and successes of the great personalities mentioned in the lesson you have just read. The first one is done for you.

Answer:

Person	Failures/Rejections he faced	Final success
1. Abraham Lincoln	Failed in business, defeated in legislative elections, had nervous breakdown, lost congressional race and senatorial race	Elected President of the United States of America
2. Lee de Forest	Charged by the district attorney	Invented the triodes tube
3. Wright Brothers	A New York Times editorial questioned the wisdom of the Wright Brothers	Took their famous flight at Kitty Hawk
4. Colonel Sanders	Knocked many doors before he got the first order	Became famous as a businessman and restaurateur who founded the Kentucky Fried Chicken chain restaurants.
5. Walt Disney	Newspaper editors rejected him by saying that he had no talent.	Created some of the world’s most well-known fictional characters including Mickey Mouse. He was well known as an ani mator, film producer, director, screenwriter, voice actor, entrepreneur, entertainer, international icon and philanthropist.
6. Thomas Edison	As a child, he was considered to be a stupid, lost his factory to fire.	Grew up to be a great scientist.
7. Henry Ford	Forgot to put the reverse gear in his first car, broke at the age of 40.	Established cars company
8. Lee lacocca	Fired by Henry Ford 11	Well known for engineering the Ford Mustang and Ford Pinto cars and became one of the most famous business people in the world.
9. Ludwig van Beethoven	He was told that he had no talent for music.	Gave some of the best music to the world.

Study skills

Read the following story.

An elderly carpenter was ready to retire. He told his employer-contractor of his plans to leave the house-building business to live a more leisurely life with his wife and enjoy his extended family. He would miss the paycheck each week, but he wanted to retire. They could get by.

The contractor was sorry to see his good worker go and asked if he could build just one more house as a personal favour. The carpenter agreed. But over time it was easy to see that his heart was not in his work. He resorted to shoddy workmanship and used inferior materials. It was an unfortunate way to end a dedicated career.

When the carpenter finished his work, his employer came to inspect the house. Then he handed the front-door key to the carpenter and said, “This is your house… my gift to you.”
The carpenter was shocked!

What a shame! If he had only known that he was building his own house, he would have done it all so differently.
So it is with us. We build our lives, a day at a time, often putting less than our best into the building. Then, with a shock, we realize we have to live in the house we have built. If we could do it over, we would do it much differently.

But, you cannot go back. You are the carpenter, and every day you hammer a nail, place a board, or erect a wall. Someone once said, “Life is a do-it-yourself project.” Your attitude, and the choices you make today, help build the “house” you will live in tomorrow. Therefore, build wisely!

I. Complete the table given below.

Answer:

Point to be observed	Observation/Point of view
1. The setting/location in which the story took place	(A building) Retirement of a carpenter
2. The main characters	A contractor and a carpenter
3. The main events	a. The carpenter decided to retire. b. He told his employer (contractor) about his decision. c. The contractor requested the carpenter to build one more house as a personal favour. d. The carpenter agreed and finished his work. e. The contractor gave the very house to the carpenter as a gift.
4. The turning point in the story	The contra.ctor inspected the house, handed the front-door key to the carpenter and told him to take the house as a gift.
5. The irony in the story	The carpenter thought that the house was for the contractor. As he had already decided to retire, he didn’t do the work with good workmanship. But he was shocked to know that the house was for him.
6. The message in the story	“Build your life with good attitude.”
7. Possible title	“Building One’s Life.”

II. Extended Writing

Rewrite the above story by giving another ending.

Answer:
An elderly carpenter was ready to retire. He told his employer-contractor of his plans to leave the house-building business to live a more leisurely life with his wife and enjoy his extended family. He would miss the paycheck each week, but he wanted to retire. They could get by.

The contractor was sorry to see his good worker go and asked if he could build just one more house as a personal favour. The carpenter agreed. He wanted to build the house using all his skill. He knew that his employer had confidence and faith in him. He began his work and completed with determination. He gave the final touch to the house and invited his employer. The employer came to inspect the house.

The contractor was amazed to see the house with all its fine work. He praised the carpenter for building such a beautiful house. He thanked the carpenter and gave him a big amount of money. The carpenter felt very happy thanked his employer and went away.

Listening

Listen to the story about a farmer.

Just now you have listened to a story about a farmer. Here are some sentences that describe the events in the story. However, they are not in the order in which they occur in the story. Listen to the story and arrange them in the order in which they occur. Put 1 after the event that occurs first, 2 against the next and so on.

He threw himself into the Barcelona River and committed suicide.
The wise man said, “Because that is a diamond.”
He picked up the stone and put it in the living room.
He looked all through Europe and couldn’t find any.
They found that the farm was indeed covered with acres and acres of diamonds.
That night the farmer couldn’t sleep.
The person who had bought his farm was watering the camels at a stream.
He made arrangements to sell off his farm and went in search of diamonds.
He was unhappy and he was discontent.
A wise man told him about the glory of diamonds.
Answer:
A wise man told him about the glory of diamonds.
That night the farmer couldn’t sleep.
He was unhappy and he was discontent.
He made arrangements to sell off his farm and went in search of diamonds.
He looked all through Europe and couldn’t find any.
He threw himself into the Barcelona River and committed suicide.
The person who had bought his farm was watering the camels at a stream.
He picked up the stone and put it in the living room.
The wise man said, “Because that is a diamond.”
They found that the farm was indeed covered with acres and acres of diamonds.

Oral Activity

After listening to the story, work in pairs and discuss the advantages and disadvantages of travelling abroad. You may use the hints given in the box for your understanding.

Answer:

Advantages and Disadvantages of Travelling Abroad
Nowadays we see many students, educated people and some others going abroad everywhere due to several reasons. Students go for studies. Educated people go for jobs. Other people go for touring. Going abroad has not only advantages but also disadvantages.
Advantages:

As far as merits are concerned while travelling abroad you will come to know about the other cultures and rituals. It will definitely add to one’s knowledge.
Students go abroad for their studies. Educated people go abroad for getting better opportunities.
By going abroad they can get good and quality education.
They can study and get jobs abroad.
They can settle in their lives by getting better earning. Thus they can improve their social status.
They can get more opportunities and more income.
They can learn new languages.
They can make friends with people from different countries.
They can know about recent technological developments.
The new environment can allow us to see the world and create memories that will last forever.

Disadvantages:

People who go abroad would cut off themselves from their roots from the native country.
They should forget their own environment which is friendly and lively.
They should live away from their culture, rituals and religions.
They should face difficulties in adapting other cultures.
They would miss the native food and tastes.
They would miss the nearness of their parents.
They should live far away from their parents. They miss the sacred love their parents.
Their parents would suffer from old age problems as there is nobody to take care of them.

Every Success Story is Also a Story of Great Failures Summary in English

All stories of success are also stories of great failures. Failures are the stepping stones to success. We all know that Abraham Lincoln was the President of the United States, It was not an easy thing for him to reach such a great height in his life. He faced major setbacks in business as well as politics before he was elected President of the United States at age 52. He didn’t lose hope and give up his efforts.

Lee De Forest was the inventor of the triodes tube. In 1913, he was charged for using dishonest means to mislead the public. Though he was publicly humiliated, he didn’t stop his efforts and made so many inventions.

The Wright Brothers successfully took their flight after their wisdom was questioned by a New York Times editorial.

Colonel Sanders was the founder of KFC chain restaurants. He, at age 65* worked hard and succeeded in selling food items. He had knocked on more than a thousand doors before he got his first order.

Walt Disney, the famous cartoonist faced many rejections before his getting the success. He created Mickey Mouse when he was working out of a small mouse infested shed near the church.

Thomas Edison, who was partially deaf, failed 10,000 times while he was working on the light bulb. He didn’t give up and got the success finally. He was the man with great attitude. Though he lost his factory to fire, he went on inventing the phonograph. He didn’t lose hope at any stage of his life.

Henry Ford forgot to put the reverse gear in the first car he made and went broke at the age of 40. He succeeded later and reached top position in selling of cars.

Ludwig van Beethoven was a German composer and pianist. Though he was told that he had no talent for music, he gave some of the best music.

All these persons succeeded in spite of their problems. A setback can act as a driving force and teach us humility. We have to find courage and faith to overcome the setback. We must get rid of fear and doubt. We need to try to become victors. We need to have burning desire to achieve our task. A burning desire is the key to one’s success. We can’t get great results with a weak desire.

IF YOU THINK

If you think you are beaten, you are.
If you think you dare not, you don’t!
If you like to win, but think you can’t,
It’s almost a cinch you won’t.
If you think you’ll lose, you’re lost;
For out in the world we find
Success begins with a fellow’s will;
It’s all in the state of mind.
If you think you are outclassed, you are,
You’ve got to think high to rise,
You’ve got to be sure of yourself before
You can ever win a prize.
Life’s battles don’t always go
To the stronger and faster man,
But sooner or later the man who wins
Is the man who thinks he can.

IF YOU THINK Summary
Success, failure, win, defeat – these ail depend upon the man’s thinking. Success begins with a fellow’s will. If one thinks he is beaten or lost, it certainly happens. If one thinks one is outclassed, one is outclassed. So, one needs to be positive in one’s thoughts. One needs to think high to get success. The person who thinks can easily win in the end. ‘ Victory is not always on the side of the stronger and faster man.

Every Success Story is Also a Story of Great Failures About the Author

Shiv Khera is an Indian author of self-help books and activist.
While working in the United States, he was inspired by a lecture delivered by Norman Vincent Peale and followed his motivational teachings. Khera has written several books including You Can Win.
His mission is “to ensure freedom through education and justice.”

Shiv Khera was born in a business family who used to have coal mines Kendua in Dhanbad, India. Soon after the nationalisation of coal mines by the Indian government, he had to search for his own living. In his early years he worked as a car washer, a life insurance agent, and a franchise operator before he became a motivational speaker.

Every Success Story is Also a Story of Great Failures Glossary

overcome(v): to defeat or succeed in controlling or dealing with something

detour(n): a road or route that is used when the usual one is closed

triodes tube (n): A triode is an electronic amplification tube having three active electrodes. Invented in 1906, the triode vacuum tube was the first electronic amplification device. This invention founded the electronic age, making possible amplified radio technology and long-distance telephony.

Lee de Forest: an American inventor with over 180 patents to his credit. He invented the Audion, a vacuum tube that takes relatively weak electrical signals and amplifies them. He is one of the fathers of the “electronic age,” as the Audion helped to usher in the widespread use of electronics. He is also credited with one of the principal inventions that brought sound to motion pictures.

fraudulent (adj): intended to cheat someone, usually in order to make money illegally

attorney (n): a lawyer

humiliate (v): to make someone feel ashamed or stupid

Colonel Sanders: Colonel Harland David Sanders was an American business man and restaurateur who founded the Kentucky Fried Chicken (KFC) chain restaurants

beat-up (adj): old and damaged

quit (v): to leave ; to stop doing something

recipe (n): set of instructions for cooking a particular type of food

Walt Disney: an American animator, film producer, director, screenwriter, voice actor, entrepreneur entertainer, international icon and philanthropist, well known for his influence in the field of entertainment during the 20th century. He created some of the world’s most well-known fictional characters including Mickey Mouse.

rejection (n): the act of not accepting

Infest(v): to exist in large numbers in a particular place, often causing damage or disease (especially of insects or animals such as rats)

Henry Ford: an American industrialist, the founder of the Ford Motor company, and sponsor of the development of the assembly line technique of mass production. He did not Invent the automobile but he developed and manufactured the first automobile that many middle class Americans could afford to buy.

disaster (n): a sudden event such as a flood, storm or accident

Lee Iacocca: an American Business man known for engineering the Ford Mustang and Ford Pinto cars. He was the former chairman of the Chrysler Corporation. He was one of the most famous business people in the world. He is the author (or coauthor) of several books, including Iacocca.

Ludwig van Beethoven: a German composer and pianist. He was a crucial figure in the transition between the Classical and Romantic eras in Western art music. He remains one of the most famous and influential of all composers. His best known compositions include 9 symphonies. 5 concertos, 2 piano sonatas and 16 string quartets. He also composed other chamber music, choral works (Including the celebrated Missa Solemnis and songs).

inevitable (adj): certain to happen : that you cannot avoid or prevent

setback (n): a problem that delays or prevents progress

humility (n): the quality of not being too proud about oneself.

victor (n): the winner

stumbling block (n): a problem or difficulty that stops oneself from achieving something

Napoleon Hill: an American author in the area of the new thought movement who was one of the earliest producers of the modern genre of personal-success literature. He is widely considered to be one of the great writers on success.

conceive(n): to form an idea, a plan, etc. In your mind; to imagine something

Socrates (469 BC – 399 BC): a classical Greek Athenian philosopher credited as one of the founders of Western Philosophy; He is an enigmatic figure known chiefly through the accounts of later classical writers, especially the writings of his students Plato and Xenophon and the plays of his contemporary Aristopanes. Many would claim that Plato’s dialogues are the most comprehensive accounts of Socrates to survive from antiquity. Socrates has become renowned for his contribution to the field of ethics.

ducked(v): pushed somebody under water and held them there for a short time

gasp (v): to breathe quickly in a way that can be heard because one is having difficulty in breathing .

accomplishment (n): something successful or impressive that is achieved after a lot of effort and hard work

cinch(n): a thing that is certain to happen

outclassed (v): overcame

will (n): determination to do something that one has decided to do

AP SSC 10th Class English Solutions Chapter 1A Attitude is Altitude

March 29, 2023February 5, 2023 by Mahesh

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 1A Attitude is Altitude Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 1A Attitude is Altitude

10th Class English Chapter 1A Attitude is Altitude Textbook Questions and Answers

Look at the pictures and answer the questions.

Question 1.
What qualities have made these persons great?
Answer:
The persons in the above pictures are Sudha Chandran, Stephen Hawking and Helen Keller.
Sudha Chandran is an accomplished Bharatanatyam dancer, Indian film and television actress, who showed great courage even though she had lost her leg in an accident.

Stephen Hawking is one of the famous scientists. Despite his debilitating illness, he has done ground-breaking work in physics.

Helen Keller overcame the adversity of being blind and deaf to become one of the 20th century’s leading humanitarians.
These three great persons showed that anything can be done, if one tries hard enough. From their lives, it is understood that one need not lose hope. Their lives show us that disability need not prevent one from reaching great heights. The qualities such as grit, determination, confidence, will power, dedication, discipline, etc. have made these persons great.

Question 2.
What qualities do you think a person requires to become successful ?
Answer:
The qualities a person requires to become successful are determination, dedication, discipline, grit, courage, faith, positive thinking, responsibility, hard work, commitment, desire, persistence, good character, etc. These qualities make a person highly successful.

Comprehension

I. Answer the following questions.

Question 1.
‘I call it my chicken drumstick,’ joked Nick. What does ‘it’ refer to? How does it help him ?
Answer:
‘It’ refers to the small foot that Nick has on his left hip. It helps him balance and enables him to kick. Nick uses his one foot to type, write with a pen and pick things up between his toes.

Question 2.
‘His parents decided not to send him to a special school.’ Was Nick happy about his parents’ decision? Why or why not? What do you think about this decision?
Answer:
Nick’s was not happy with his parents’ decision because there might be the risk of bullying. His companions at mainstream school might tease him.
But later, he realised that his parents’ decision was the best one. I think that this decision was the best one his parents could have made for him as he was able to achieve a degree in Financial Planning and Real Estate. The decision made him independent.

Question 3.
How did Nick’s parents help him to become independent?
Answer:
Right from the start Nick’s parents did their best to make him independent. His father put Nick in the water at 18 months and gave him the courage to learn how to swim. He encouraged Nick to get into football and skating. He taught Nick how to type with his toe at just six years old. Nick’s mother invented a special plastic device that meant Nick could hold a pen and pencil. Despite the risk of bullying, Nick’s parents insisted on his attending mainstream school.

Question 4.
Pick out the symptoms of Nick’s depression. Do you think his depression was normal or something unique about him? Give reasons for your opinion.
Answer:
Nick was deeply depressed when he was eight years old. When he was in depression, he went to his mother crying and told her that he wanted to kill himself. He felt cold and bitter. He hated God for doing this to him. He was terrified about his future without his parents. At age ten Nick tried to drown himself in the bath. These were the symptoms of Nick’s depression. I think his depression was normal. Any human being born like Nick might have the same depression.

Question 5.
Which incident in this text is funny? What makes it funny?
Answer:
Once Nick was in a car and a girl at traffic lights was looking at him interestingly. She could only see Nick’s head so he decided to do a 360 in the car seat to freak her out. When she saw the torso, she sped off really quickly. This is the funny incident in the text. The idea of Nick showing the girl his torso makes the scene funny.

Question 6.
What made Nick choose Bethany Hamilton as his teacher to learn surfing?
Answer:
Nick went to Hawaii in 2008 and met surfing master Bethany Hamilton. She had her arm bitten off by a shark when she was 12. Nick chose Bethany Hamilton as she too was a disabled person and still she performed surfing very well. Though she had no arm she did well. As a disabled person, she could know how a disabled person could learn how to surf. She could know the psychological condition of a disabled person.

Question 7.
Which aspect of Nick’s physical condition helps him to do a 360 degree spin?
Answer:
Nick has a very low centre of gravity so he has got pretty good balance. This condition helps Nick to do 360 degree spin.

Question 8.
What are the distinct features of this text? List them.
Answer:

It is an autobiography.
Varied sentences are used.
Ideas are arranged in a logical sequence.
Emotive and persuasive language is used.
Ideas are exemplified very well.
Cohesion is found in the text. The sentences are connected with proper linkers.
Unity of ideas is maintained for coherence. The sentences are connected in a clear and reasonable way.
Certain quotations and examples are given.
Proper introduction, development and conclusion are ensured.
The ideas are given formally and precisely.
As far as the content is concerned, it touches the emotions of a reader.
The content is message-oriented.
Physical description of the main character is given properly.
The main character’s attitudes, likes, dislikes, strengths and weaknesses are given effectively.

Question 9.
What do you learn from the life of Nick Vujicic?
Answer:
Nick Vujicic uses his life story to inspire people of all around the world. From the life of Nick, we learn that if we fail, we should try again and again until we succeed. Nick shows us that anything can be done, if we try hard enough. Undoubtedly, I feel that his life is a source of inspiration to everyone of us. He has been able to do all these because of his faith, hope, perseverance and strong will power. Nick proves that disability need not prevent one from reaching great heights. Hence I learn that we shouldn’t lose our faith and hope under any circumstances.

II. Given below are some of the statements made by Nick Vujicic. Which of these are substantiated by Nick’s life ?

I realised why God had made us like this – to give hope to others.
I wanted to concentrate on something good that / had.
I tell people to keep on getting up when they fall and to always love themselves.
If I can encourage just one person, then my job in this life is done.

Answer:
I think all the above statements are substantiated by Nick’s life.
i) I realised why God had made us like this – to give hope to others:
When Nick was 13, he read a newspaper article about a disabled man who had managed to achieve great things and help others. Nick was inspired very much by the disabled man mentioned in the article. He got a hope from him. Then he decided to use his life to encourage others and give them hope.

ii) I wanted to concentrate on something good that I had:
Nick was not discouraged with his disability. He knew that he had no arms and legs but he thought that they would never take away the beauty of his eyes. With his positive approach, he thought that his eyes were very good ones for him. He wanted to love his positive qualities such as faith, will power and perseverance. Further, he was quite satisfied with what he had, i.e. a small foot on his left hip. He said he would be lost with it.

iii) I tell people to keep on getting up when they fall and to always love themselves: If Nick fails, he tries again and again until he succeeds. When Nick was learning how to surf, he was terrified at first but later he did very well. Though he has no limbs, he always loves himself. I know this by his statement, “I decided to be thankful for what I do have, not get angry about what I don’t.”

iv) If I can encourage just one person, then my job in this life is done:
This statement was substantiated by Nick aptly. He encouraged many people in the world. He used his life story to inspire others. He made many people realise their inner beauty and personality. He was able to rise many people from the state of despair to the state of great hope.

Vocabulary

I. Look at the list of adjectives in the box. Which of them describe Nick Vujicic. Tick (✓) them.

Answer:

Explanation:

positive: hopeful and confident
negative: harmful, unpleasant or not wanted
optimistic: believing that good things will happen In the future
pessimistic: expecting that bad things will happen in the future
generous: willing to give money, spend time, etc. in order to help people or give them pleasure
nervous: worried or frightened about something
confident: sure that something will happen in the way that you want or expect
smart: fashionable, clever
sociable: friendly and enjoys being with other people
cowardly: not at all brave
courageous: brave
reliable: dependable
Independent: confident and able to do things by oneself in ones own way
dependent: needing someone or something in order to exist, be successful, be healthy, etc.
obedient: always doing what one is told to do
trusting: willing to believe that other people are good and honest
valiant: very brave in difficult situation
self-centred: paying so much attention to oneself that one doesn’t notice what is happening
arrogant: behaving in an unpleasant or rude way
sarcastic: saying things that are the opposite of what one means, in order to make an unkind joke
quarrelsome: one who quarrels a lot with people
rude: not polite and annoys others
creative: good at using imagination to make things
stubborn: determined not to change one’s mind when people think one is being unreasonable
lazy: not making any effort to do anything
pompous: trying to seem magnificent or very important; fond of display, acting too proudly, self-important.
level-headed: sensible in making judgements

Pick out from the text the examples that support these attributes.

Attribute	Example from the text
1. Positive	In spite of his disability he has become great.
2. Optimistic	He always thinks, “If I fail, I try again, and again, and again. If I can encourage just one person, then my job is done.”
3. Generous	He spends most of his time delivering motivational speeches all over the world.
4. Confident	Though he has no limbs, he gains faith and hope and tries to give them to others.
5. Smart	He appears to be smart and shows his cleverness in learning things.
6. Sociable	He enjoys being with others by delivering speeches, playing football, golf, surfing, etc.
7. Courageous	Despite the absence of his limbs, he didn’t get discouraged. With great courage, he tries again and again and becomes a football player, a swimmer and a surfer.
8. Reliable	He is the most reliable person. As an Evangelical Christian he has faith and gives the same to others with his motivational speeches. The people believe his reliability and are attracted towards him.
9. Independent	He independently has done the things what he has wanted to do. e.g.: Playing football and golf, swimming, surfing, delivering speeches, etc.
10. Obedient	He seems to be obedient in accepting the unlucky birth.
11. Valiant	With great determination, he has reached a great position in his life. He becomes an international symbol of triumph over adversity.
12. Creative	He also plays golf with a club tucked under his chin.
13. Level-headed	He judges well when he says, “The challenges in our lives are there to strengthen our convictions. They are not there to run us over.”

II. Given below are words that describe the positive or negative attitude of a person. Discuss them with your partner and put a tick (✓) against the positive qualities and a cross (✕) against the negative qualities.

Answer:

Explanation:

sullen: angry and silent (negative quality)
sneaky: doing things in a secret and dishonest way (negative quality)
exuberant: happy and full of energy and excitement (positive quality)
bossy: always telling other people what to do in a way that is annoying (negative quality)
cautious: careful to avoid danger or risk (positive quality)
aggressive: behaving in an angry threatening way (negative quality)
competitive: determined or trying very hard to he more successful than other people (positive quality)
finicky: too concerned with unimportant details and small things that you like or dislike (negative quality)
imaginative: good at thinking of new and interesting ideas (positive quality)
observant: good or quick at noticing things (positive quality)
enthusiastic: showing a lot of interest and excitement about something (positive quality)
outgoing: someone who Is outgoing likes to meet and talk to new people (positive quality)
haughty: behaving in a proud. unfriendly way (negative quality)
intrepid: willing to do dangerous things; fearless, brave (positive quality)
malicious: very unkind and cruel and is trying to do harm to others (negative quality)
meticulous: very careful about small details and always makes sure that everything is done correctly (positive quality)
slovenly: lazy untidy and careless (negative quality)
obnoxious: very offensive, unpleasant or rude (negative quality)
stingy: not generous. especially with money (negative quality)
Idealistic: believing that one should live according to high standards and principles, even when they are very difficult to achieve (positive quality)
affable: friendly and easy to talk to (positive quality)
logical: reasonable and sensible (positive quality)
mean: cruel (negative quality)
cultured: intelligent, polite and interested in art, literature, music etc. (positive quality)

III. Tick (✓) the word/expression which is nearest in meaning to the word underlined in each sentence given below.

1. Water sports aren’t Nick’s only thing – he also plays golf with a dub tucked under his chin.
a) kept
b) pulled
c) covered
d) tied
Answer: a

2. His distraught mother couldn’t bring herself to hold him until he was four months old.
a) angry
b) careless
c) shy
d) anxious
Answer: d

3. Despite the risk of bullying, his parents insisted on Nick’s attending mainstream school.
a) discrimination
b) humiliation
c) negligence
d) failure
Answer: b

4. Nick, who was teased and bullied, had an electric wheelchair for mobility, and a team of carers to help him.
a) socialising
b) movement
c) travel
d) safety
Answer: b

5. Nick managed to pull through to become an international symbol of triumph over adversity.
a) disability
b) poverty
c) difficulties
d) opposition
Answer: c

6. The challenges in our lives are there to strengthen our convictions.
a) feelings
b) practices
c) beliefs
d) characters
Answer: c

IV. Read the following descriptions about persons and fill in the blanks with the words that match with these descriptions.

1. Ashok is a very unpleasant and rude fellow. I don’t talk to him. He is ———–.
2. Srinu is very confident, elegant and polite. But sometimes he does not appear to be sincere. He is ———–.
3. Sujatha is quiet and calm. She doesn’t like to spend time with others. She is interested in her thoughts and feelings. So she is an ———–.
4. Ramesh is very sensitive. He always shows sympathy for people who are suffering. He is ———–.
5. My neighbour is such a person that he has a feeling of hatred for others. He desires to hurt others’ feelings. He is ———–.
6. My father pays careful attention to every detail. He makes thorough planning for everything in our home. He is ———–.
7. Prem is an extremely unpleasant person. He always tries to offend people. He is ———–.
8. My mother is very calm and peaceful. She doesn’t get excited or irritated easily. She is ———–.
9. Rahul is a very lively and confident person. He enjoys being with his friends. He is an ———–.
10. Sridhar is always afraid of speaking in the class. He is ———–.
Answers:
1. boorish
2. suave
3. introvert
4. compassionate
5. malicious
6. meticulous
7. obnoxious
8. placid
9. extrovert
10. timid

GRAMMAR

I. Read the text again and identify some more Defining and Non-defining Relative Clauses.

1. —————————————————————
2. —————————————————————
3. —————————————————————
4. —————————————————————
5. —————————————————————
Answer:
1. Nick has a small foot, which helps him balance. (Non-defining Relative Clause)
2. Nick, who was born in Melbourne, now lives in Los Angeles. (Non-defining Relative Clause)
3. I read about a disabled man who had managed to achieve great things and help others. (Defining Relative Clause)
4. I decided to give people the courage that the article had given me. (Defining Relative Clause)
5. Nick, who later achieved a degree, says this. (Non-defining Relative Clause)

II. Read the following sentences and circle the Non-defining Relative Clauses and put a comma/a hyphen wherever necessary.

1. The meeting which was held in the town hall was a great success.
2. The people who live in glass houses must not throw stones.
3. The teachers who have taught Abhiram say that he is an exceptionally creative student.
4. Napoleon who won the French honour died at St. Helena.
5. The silk sarees which are made in Banares are popular all over the world.
Answer:
1. The meeting which was held in the town hall was a great success. (Defining Relative Clause)
2. The people who live in glass houses must not throw stones. (Defining Relative Clause)
3. The teachers who have taught Abhiram say that he is an exceptionally creative student. (Defining Relative Clause)
4. Napoleon, who won the French honour, died at St. Helena. (Non-defining Relative Clause)
5. The silk sarees which are made in Benares are popular all over the world. (Defining Relative Clause)

III. Complete the sentences with Defining or Non-defining Relative Clauses.

1. This is the peaceful place __________________________ .
2. The Mahaprasthanam’ is a popular book __________________________ .
3. Suresh ____________________________ has left the school just now.
4. Is this the street _______________________________________________ ?
5. The express __________________________________ has just arrived.
6. The place ___________________________ has been closed to traffic.
7. This is a famous engineering college _____________________________ .
8. My sister_______________ has secured ajob as an engineer in BHEL.
9. The theatre _____________________________ was designed by Mr Rao.
10. Viswanath is one of the outstanding directors in Telugu film industry______________________ .
Answer:
1. This is the peaceful place that is liked by every naturalist.
2. The Mahaprasthanam’ is a popular book which I have bought in Hyderabad.
3. Suresh, who is mv cousin, has left the school just now.
4. Is this the street where meat is sold?
5. The express which runs between Delhi and Chennai has just arrived.
6. The place which is behind our house has been closed to traffic.
7. This is a famous engineering college that is founded by Mr. Yashwanth.
8. My sister, who has completed her education in Bengaluru, has secured a job as an engineer in BHEL.
9. The theatre which was built in 2012 was designed by Mr. Rao.
10. Viswanath is one of the outstanding directors in Telugu film industry which produces a number of creative directors.

WRITING

I. Writing a biographical sketch

A biographical sketch is an account of the life and activities of an individual. It includes information about the person’s name, place of residence, education, occupation, life and activities and other important details. It is written by another person. If a person writes an account of his/her own life, it is called an autobiography.

Given below in the table is the biographical information about Stephen Hawking. Based on this information write a biographical sketch.

Points to be covered	Details of the person
Date and place of birth	January 8,1942, Oxford, England
Information about the family	Eldest one of four sons Mother: Isobel Hawking Father: Frank Hawking, a medical researcher with a speciality in tropical diseases
Important events in life	Early academic life: recognized as a bright student High school years: third from the bottom of his class Pursuits outside of school: loved board games, constructed a computer out of recycled parts at the age of 16, solved rudimentary mathematical equations Marriage: married Jane Wilde Health: Suffered from Amyotrophic Lateral Sclerosis Research: on black holes Discovery: Hawking Radiation
Awards, prizes, honours	Fellow of the Royal Society at the age of 32 Albert Einstein Award. The Pius XI- Gold Medal for Science from Pope Paul VI in 1975
Contribution to his field and society	Contributed to the advancement of science and research

Answer:

Biographical Sketch of Stephen Hawking

Stephen William Hawking is one of the greatest scientists of the world. He was born at Oxford in England on 8 January, 1942 to Isobel Hawking and Frank Hawking. He was the eldest of four sons. His father, Frank Hawking, was a medical researcher with a speciality in tropical diseases.

Hawking was recognised as a bright student. But he was not initially successful academically. During his high school years, Stephen stood the third from the bottom of his class.

Hawking used to remain with a close group of friends with whom he enjoyed board games, the manufacture of fireworks, model aeroplanes and boats. With the help of the mathematics teacher, Dickran Tahta, Hawking and his companions built a computer from clock parts, an old telephone switchboard and other recycled components when he was 16. Hawking solved rudimentary mathematical equations.

He married Jane Wilde on 14 July, 1965. The first years after their marriage were hectic. Hawking has achieved many things in the field of science in spite of his prolonged illness. He has been suffering from an incurable disease, Amyotrophic Lateral Sclerosis (ALS) which has weakened his muscles since he was 17 years old.

Stephen Hawking is known for his work regarding black holes and for authoring several popular science books. In 1974, Hawking’s research turned him into a celebrity within the scientific world when he showed that black holes are not the information vacuums that scientists had thought they were. In simple terms, Hawking demonstrated that matter, in the form of radiation, can escape the gravitational force of a collapsed star. Thus Hawking Radiation was invented.

As a mark of his eminence, Professor Hawking obtained twelve honorary degrees which he deserved. He was named a Fellow of the Royal Society at the age of 32, and later earned the prestigious Albert Einstein Award. In 1975, Hawking journeyed to Rome, where he was honoured with the Pius XI – Gold Medal for science from Pope Paul VI in 1975.

Attitude is Altitude Summary in English

This is the story of a great person, Nicholas James Nick Vujicic, who stands as an inspiration to other people with his great qualities-faith, will power and perseverance- even though he is a disabled person. His disability has not stopped him from doing what he has wanted to do. Nick Vujicic was born on 4 December 1982 in Melbourne, Australia with a rare disability called Phocomelia characterised by the absence of all four limbs. Despite the absence of his limbs, he plays football and golf, swims and surfs. He is an Australian Christian evangelist and motivational speaker. As a child, he struggled mentally and emotionally as well as physically. It was a shock to Nick’s parents that he was born without limbs. His mother was a nurse and his father was a computer programmer and accountant. His parents did their very best when they decided to send him to mainstream school despite the risk of humiliation.

Nick has a small foot on his left hip. It gives him balance and enables him to kick. With his foot, he writes with a pen and picks things up between his toes. He also plays golf with a heavy stick tucked under his chin. He is a huge fan of the English Premier League. His parents did their best to make him independent from the start. His father gave him the courage to learn how to swim by putting him in water at 18 months. His mother invented a special plastic device to enable him to hold a pen and pencil.

At age eight, Nick was deeply depressed and went to his mother crying. He told her that he wanted to kill himself. When he was ten years old, he decided to end his life by drowning himself in a bathtub but luckily he was unsuccessful in his attempt. When Nick was 13, he was inspired when he read about a disabled man who had managed to achieve great things and helped others. He understood that God had given him the life to give hope to others and so he decided to use his life to encourage other people and give them the courage. He believes that challenges strengthen our opinions. With the help of his religion, friends and family, Nick becomes an international symbol of triumph over adversity. Nick won the Australian Young Citizen of the Year award in 1990 for his bravery and perseverance. His persistence made him achieve a degree in Financial Planning and Real Estate. He went to Hawaii in 2008 and was trained in surfing by Bethany Hamilton. Nick was on the cover of Surfer magazine with his feat of doing the 360 degree spins. He is able to do this feat because of his low centre of gravity.
Nick and Kanae Miyahara, whose mother was a Mexican and father was a Japanese, got married on February 12, 2012 and on February 13, 2013 they were blessed with the birth of a healthy baby boy with full body.

Nick has travelled to over 24 countries speaking to groups of up to 110,000 people. He keeps telling the people to keep on getting up when they fall and to always love themselves. He spreads his messages of faith and hope around the world.

Nick uses his story to inspire and inform people all around the world. Nick presents values, principles, attitudes and perspectives that challenge and equip people from all walks of life to overcome their adversities.

Attitude is Altitude Glossary

stretch (v): to straighten one’s arms, legs or body

wrap (v): put ones arms around something

skateboard (n): a short narrow board with small wheels at each end, which you stand on and ride as a sport

torso (n): the main part of the body, not including the head, arms or legs

surf: to take part in the sport of riding on waves on a surfboard

despite (pre): In spite of

limb (n): an arm or leg

hip (n): one of the two parts on each side of one’s body between the top of one’s leg and one’s waist

chicken drumstick (n): the lower part of the leg of a chicken or other bird that is cooked and eaten as food

faith (n): belief or trust in God

virgin (n): someone who has never had sex

modest (adj): shy about attracting sexual interest

publicist (n): someone whose job is to make sure that people know about a new product, film, book, etc.

club (n): a heavy stick with one end thicker than the other

tuck (v): to put something into a space

distraught (adj): so upset and worried

bully (v): to threaten to hurt someone or frighten them

bullying (n): humiliation

carer (n): someone who looks alter an old or ill person at home

wall-mounted (adj): attached to a wall

pump-action (adj): a pump-action piece of equipment is operated by pulling or pressing a part in or out. so that the contents come out in short bursts

drown (v): to die from being under water for too long time

lack: to not have something that one needs

triumph (n): an important success or victory

adversity (n): a difficulty or unpleasant situation

Inspirational (adj): providing encouragement or new ideas for what one should do

conviction (n): a very strong belief or opinion

run over (phr.v.): to hit someone or something with something (a vehicle, challenge, etc.) and drive over them

perseverance (n): the quality of continuing to try to achieve a particular aim despite difficulties

give somebody the eye (n.phrase): to look at someone in a way that shows one thinks they are sexually attractive

freak out (phr.v.): make someone very anxious. upset or afraid

amazing (adj): (here) very good

spin (n): an act of turning around quickly

centre of gravity (n.phrase): the point in any object on which it can balance

AP SSC 10th Class Chemistry Solutions Chapter 13 Principles of Metallurgy

March 29, 2023February 3, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 13 Principles of Metallurgy Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 13th Lesson Principles of Metallurgy

10th Class Chemistry 13th Lesson Principles of Metallurgy Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Can you mention some articles that are made up of metals?
Answer:
Jewellery, conducting wires, utensils, etc.

Question 2.
Do metals exist in nature in the form same as that we use in our daily life?
Answer:
No, they exist as ores and minerals.

Question 3.
Have you ever heard the words like ore, mineral and metallurgy?
Answer:
Yes, these words are related to extraction of metals.

Question 4.
Do you know how these metals are obtained?
Answer:
These metals are generally extracted from their ores.

Improve Your Learning

Question 1.
List three metals that are found in nature as oxide ores.
Answer:
The three metals that are found in nature as oxide ores are

Bauxite (Al₂O₃ 2H₂O)
Haematite (Fe₂O₃)
Zincite (ZnO).

Question 2.
List three metals that are found in nature in uncombined form.
Answer:
The three metals that are found in nature in uncombined form are

Gold
Silver
Platinum.

Question 3.
Write a note on dressing of ore in metallurgy.
(OR)
What is concentration of Ore? List various physical methods that are used to enrich the ore.
Answer:

Ores that are mined from the earth are usually contaminated with large amount of impurities such as soil and sand, etc.
Concentration or dressing means, simply getting rid of as much of the unwanted rocky material as possible from the ore. The impurities like sand and clay are called gangue.

The physical methods adopted in dressing of the ore or enriching the ore depends upon the difference between physical properties of ore and gangue.

Methods of dressing or concentration of the ore:
1. Hand picking :
If the ore particles and the impurities are different in one of the properties like colour, size, etc. using that property, the ore particles are handpicked separating them from other impurities.

2. Washing:

We use washing method with water to separate dust from rice, dal and vegetable fruits, etc.
Ore particles are crushed and kept on a slopy surface. They are washed with controlled flow of water. Less densive impurities are carried away by water flow, leaving the more densive ore particles behind.

3. Froth flotation:
This method is mainly useful for sulphide ores. The ore with impurities is finely powdered and kept in water taken in a flotation cell. Air under pressure is blown to produce froth in water. Froth so produced, takes the ore particles to the surface whereas impurities settle at the bottom. Froth is separated and washed to get ore particles.

4. Magnetic separation:
If the ore or impurity, one of them is magnetic substance and the other is non-magnetic substance, they are separated using electromagnets.

Question 4.
What is an ore? On what basis a mineral is chosen as an ore?
Answer:
Ore :
A mineral from which a metal can be extracted economically and conveniently is called ore.
A mineral is chosen as an ore if the mineral is economical and profitable to extract.

Example:
Aluminium is the common metal in the Earth’s crust in all sorts of minerals. It is economically feasible and profitable to extract from bauxite which contains 50-70% of aluminium oxide.

Question 5.
Write the names of any two ores of iron.
Answer:
The names of two ores of iron :

Haematite (Fe₂O3)
Magnetite (Fe₃O₄).

Question 6.
How do metals occur in nature ? Give examples to any two types of minerals.
Answer:

The earth’s crust is the major source of metals.
Sea water also contains some soluble salts such as sodium chloride and magnesium chloride etc.
Some metals like gold (Au), silver (Ag) and copper (Cu) are available in nature in free state as they are least reactive.
Other metals are found in nature in the combined form due to their more reactivity.
The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals.

Examples :
Minerals in oxide form :
Bauxite, Zincite, Magnetite, etc.

Minerals in sulphide form :
Copper iron pyretes, Galena, etc.

Question 7.
Write short notes on froth flotation process.
(OR)
Which method is useful for concentration of sulphide ore? Explain the method.
Answer:
Froth Floatation process :

This method is mainly useful for sulphide ores which have no wetting property whereas impurities get wetted.
The ore with impurities is finely powdered and kept in water taken in a floatation cell.

Froth floatation process for the concentration of sulphide ores

Air under pressure is blown to produce froth in water.
Froth so produced takes the ore particles to the surface whereas impurities settle at the bottom.
Froth is separated and washed to get ore particles.

Question 8.
When do we use magnetic separation method for concentration of an ore? Explain with an example.
(OR)
Write the name of the method we use to separate the ore or impurity in which one of them is magnetic substance. Draw a neat diagram indicating the method.
Answer:
If the ore or impurity, one of them is magnetic substance and the other non-magnetic substance they are separated using electromagnets.
Ex :
Iron from iron ore (Fe₃O₄) is separated from its impurity by passing through a magnetic field. The field attracts magnetic ore (Iron) and repels the non-magnetic impurities.

Question 9.
Write short notes on each of the following :
i) Roasting
ii) Calcination
iii) Smelting
Answer:
i) Roasting :
Roasting is a pyrochemical process in which the ore is heated in the presence of oxygen or air, below its melting point. Generally, reverberatory furnace is used for roasting.
Ex:
Zinc blende on heating with oxygen in reverberatory furnace forms zinc oxide as solid and liberating sulphur dioxide as gas.
2Zns_(s) + 3O_2(g) → 2ZnO_(s) + 2SO_2(g)

ii) Calcination :
Calcination is a pyrochemical process in which the ore is heated in the absence of air. The ore gets generally decomposed in this process.
Ex: MgCO_3(s) → MgO_(s) + CO_2(g)

iii)Smelting:
Smelting is a pyrochemical process, in which the ore is mixed with flux and fuel, then is strongly heated.

Question 10.
What Is the difference between roasting and calcination? Give one example for each.
(OR)
Roasting and Calcination are the methods to extract crude metals from ores. What is the difference between Roasting and Calcination?
Answer:

Roasting	Calcination
1. Roasting is a pyrochemical process in which the ore is heated in the presence of air below its melting point.	1. Calcination is a pyrochemical process in which the ore is heated in the absence of air.
2. It is an oxidation reaction.	2. It is a decomposition reaction.
3. It requires oxygen.	3. It doesn’t require oxygen.
4. It is suitable to sulphide ores.	4. It is suitable to carbonate ores.
5. Ex : 2ZnS + 3O₂ → 2ZnO + 2SO₂	5. Ex : CaCO₃ → CaO + CO₂

Question 11.
Define the terms:
i) gangue
ii) slag.
Answer:
i) Gangue:
The impurity present in the ore is called gangue.
(or)
Unwanted impurity associated with ore.

ii) Slag:
The impurities found from molten metal during poling process of refining are called slag.

Question 12.
Magnesium is an active metal if it occurs as a chloride in nature, which method of reduction is suitable for its extraction?
Answer:

The method of reduction which is useful for chloride of magnesium which is active is electrolytic reduction.
Fused MgCl₂ is electrolysed with steel cathode (-) and graphite anode (+). The metal (Mg) will be deposited at cathode and chlorine gas liberates at the anode.
At cathode : Mg²⁺ + 2 e^– → Mg
At anode : 2 Cl^– → Cl₂ + 2e^–

Question 13.
Mention two methods which produce very pure metals.
Answer:
Methods which produce very pure metals are :

Electrolytic Reduction
Smelting.

(OR)

Distillation
Poling.

Question 14.
Which method do you suggest for extraction of high reactivity metals? Why?
Answer:
1.The only method which is suitable for extraction of high reactivity metals is electrolysis of their fused compounds.

2. Other methods are not suitable due to following reasons :
a) Simple reduction methods like heating with C, CO, etc. to reduce the ores of these metals are not suitable because the temperature required for the reduction is too high and more expensive.

b) Electrolysis of their aqueous solutions are also not preferable because water in the solution would be discharged at the cathode in preference to the metal ions.

Question 15.
Suggest an experiment to prove that the presence of air and water is essential for corrosion. Explain the procedure.
(OR)
Write the experimental procedure to prove that water and air are essential for rusting of iron articles.
(OR)
How can you prove that the presence of air and humid are essential for corrosion?
(OR)
Explain in brief, an experiment to prove that the presence of air and water are essential for corrosion.
Write the precautions to be taken in the experiment to show air and water are essential for rusting iron articles and also write the experimental procedure.
Answer:
Aim :
To prove that the presence of air and water is essential for corrosion or for rusting of iron articles.

Apparatus :
3 boiled test tubes, 3 corks, boiled distilled water, anhydrous calcium chloride, clean iron nails.

Procedure :

Take three test tubes and place clean iron nails in each of them.
Label these test tubes A, B and C. Pour some water in test tube A and cork it.
Pour boiled distilled water in test tube B, add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture.
Leave these test tubes for a few days and then observe.
We will observe that iron nails rust in test tube A, but they do not rust in test tubes B and C.

Observation :

In test tube A, the nails are exposed to air and water. Hence, the nails rusted.
In test tube B, the nails are exposed only to water, but not to air, because the oil float on water and prevent the air from dissolving in the water. Hence, the nails are not rusted.
In test tube C, the nails are exposed to dry air, because anhydrous calcium chloride will absorb the moisture, if any, from the air. Hence, the nails are not rusted.

Conclusion :
From the above experiment we can prove that air and water are essential for corrosion.

Question 16.
Collect information about extraction of metals of low reactivity silver, platinum and gold and prepare a report.
(OR)
Prepare a report with the collected information about extraction of metals of low reactivity silver, platinum and gold.
Answer:
Extraction of Silver:

Silver can be extracted from Ag2S by using displacement from aqueous solution.
Ag₂S is dissolved in KCN solution to get dicyanoargentate (I) ions.
From these ions Ag is precipitated by treating with Zinc dust powder.
Ag₂S + 4 CN^– → 2 [Ag(CN)₂]^– + S^2-
2[Ag(CN)₂]^–_(aq) + Zn_(s) → [Zn(CN)₄]^2-_(aq) + 2Ag_(s)

Extraction of Gold :

Gold is extracted from gold ore like electrum. Impurities are separated from the gold by treating gold ore with a weak cyanide solution.
Zinc is added and a chemical reaction takes place which separates the gold from ore.
Pure gold is removed from the solution with a filter press.

Extraction of Platinum:

The extraction of platinum from ore is a complex process and includes milling the ore, a froth flotation process, and smelting at high temperatures.
This removes the base metals, notably iron and sulphur and concentrate platinum.

Question 17.
Draw the diagram showing
i) Froth flotation
ii) Magnetic separation.
(OR)
Draw a neat diagram and label the parts that shows froth floatation process for the concentration of sulphide ore.
Answer:
i)

ii)

Question 18.
Draw a neat diagram of reverberatory furnace and label it neatly.
(OR)
Draw a neat labelled diagram of a reverberatory furnace.
Answer:

Question 19.
What is activity series? How does it help in extraction of metals?
Answer:
Activity Series :
We can arrange metals in descending order of their reactivity. This series of writing metals is called activity series.

Uses of activity series the extraction of metals :

Activity series is extremely useful in extraction of metals because we can judge the nature of metal and how it exists.
High reactive metals like K, Na, Ca, Mg and Al are so reactive that they are never found in nature in free state.
The moderate reactive metals like Zn, Fe, Pb, etc. are found in the earth’s crust mainly as oxides, sulphides and carbonates.
The least reactive metals like Au, Ag, Pt are found even in free state in nature.

Question 20.
What is thermite process? Mention its applications in daily life.
Answer:
Thermite Process :

Thermite process is the chemical reaction which takes place between metal oxides and aluminium.
When highly reactive metals such as sodium, calcium, aluminium, etc. are used as reducing agents, they displace metals of lower reactivity from the compound.
This reaction is highly exothermic. The amount of heat evolved is so high that the metals can he directly converted into molten state.

Applications in daily life :

The reaction if Iron (III) oxide (Fe203) with aluminium is used to join railing of railway tracks or cracked machine parts.
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe + Heat.
And also used for joining of cracked metal utensils in the house.

Question 21.
Where do we use hand picking and washing methods in our daily life ? Give examples.
How do you correlate these examples with enrichment of ore?
Answer:
Daily life examples for hand picking:

Separating mud particles from rice is an example for hand picking because the colour and size of these two are different.
Similarly, the ore particles and the impurities are different in one of the properties like colour, size, etc. are separated by hand picking.

Daily life examples for washing :

We can clean some vegetables like potatoes by controlled flow of water. Less densive impurities are carried away by the flow leaving the more densive potatoes.
Similarly, ores are washed with controlled flow of water. Less densive impurities i are carried away by water flow, leaving the more densive ore particles behind.

Fill In The Blanks

1. The method is suitable to enrich the sulphide ores.
2. Arranging metals in the decreasing order of their reactivity is called
3. The method suitable for purification of low boiling metals.
4. Corrosion of iron occurs in the presence of and
5. The chemical process in which the ore is heated in the absence of air is called
Answer:

Froth flotation
activity series
distillation
air, water
calcination

Multiple Choice Questions

1. The impurity present in the ore is called ………………….
A) Gangue
B) Flux
C) Slag
D) Mineral
Answer:
A) Gangue

2. Which of the following is a carbonate ore?
A) Magnesite
B) Bauxite
C) Gypsum
D) Galena
Answer:
A) Magnesite

3. Which of the following is the correct formula of Gypsum?
A) CuSO₄ • 2 H₂O
B) CaSO₄ • ½ H₂O
C) CuSO₄ • 5 H₂O
D) CaSO₄ • 2 H₂O
Answer:
D) CaSO₄ • 2 H₂O

4. The oil used in the froth flotation process is
A) kerosene oil
B) pine oil
C) coconut oil
D) olive oil
Answer:
B) pine oil

5. Froth flotation is method used for the purification of ………………. ore.
A) sulphide
B) oxide
C) carbonate
D) nitrate
Answer:
A) sulphide

6. Galena is an ore of ………………..
A) Zn
B) Pb
C) Hg
D) Al
Answer:
B) Pb

7. The metal that occurs in the native form is ………………
A) Pb
B) Au
C) Fe
D) Hg
Answer:
B) Au

8. The most abundant metal in the earth’s crust is …………………
A) silver
B) aluminium
C) zinc
D) iron
Answer:
B) aluminium

9. The reducing agent in thermite process is ………………….
A) Al
B) Mg
C) Fe
D) Si
Answer:
A) Oxidise

10. The purpose of smelting an ore is to ……………….. it.
A) Oxidise
B) Reduce
C) Neutralise
D) None of these
Answer:
B) Reduce

10th Class Chemistry 13th Lesson Principles of Metallurgy InText Questions and Answers

10th Class Chemistry Textbook Page No. 238

Question 1.
How are the metals present in nature?
Answer:
Some metals like gold (Au), silver (Ag) and copper (Cu) are available in nature in free sjate. Other metals mostly are found in nature in the combined form.

10th Class Chemistry Textbook Page No. 240

Question 2.
What metals can we get from the ores mentioned in the Table – 1?
Answer:
The metals are Aluminium (Al), Copper (Cu), Magnesium (Mg), Silver (Ag), Manganese (Mn), Iron (Fe), Zinc (Zn), Sodium (Na), Mercury (Hg), Lead (Pb), Calcium (Ca).

Question 3.
Can you arrange metals in the order of their reactivity?
Answer:
The order of reactivity is like this : Ag < Cu < Pb < Mn < Fe < Zn < Al < Mg < Ca < Na.

Question 4.
What do you notice in Table – 2?
Answer:
We notice that ores of many metals are oxides and sulphides. ______

Question 5.
Can you think how we get these metals from their ores?
Answer:
We can get metals from their ores by using various extracting techniques.

Question 6.
Does the reactivity of a metal and form of its ore (oxides, sulphides, chlorides, carbonates, sulphates) has any relation with process of extraction?
Answer:
Yes, they have relation. Metals like K, Na, Ca, Mg and Al are so reactive. They exist in all forms whereas moderate reactive metals like Zn, Fe, Pb, etc. exist as oxides, sulphides and carbonates. The least reactive metals are found even in free state.

Question 7.
How are metals extracted from mineral ores?
Answer:
The extraction of a metal from its ores involves mainly in three states. They are :

Concentration or Dressing
Extraction of crude metal
Refining or purification of the metal.

10th Class Chemistry Textbook Page No. 248

Question 8.
Do you know why corrosion occurs?
Answer:
Corrosion occurs due to reaction of metal with both air and water.

Question 9.
What are the conditions under which iron articles rust?
Answer:
Iron articles get rust due to both air and water.

10th Class Chemistry Textbook Page No. 251

Question 10.
What is the role of furnace in metallurgy?
Answer:
Furnace is used to carry out pyrochemical process in metallurgy.

Question 11.
How do furnaces bear large amounts of heat?
Answer:
Furnaces have metallic lining. So they bear large heats.

Question 12.
Do all furnaces have same structure?
Answer:
No, they have different structures.

10th Class Chemistry Textbook Page No. 239

Question 13.
Do you agree with the statement “All ores are minerals but all minerals need not be ores.? Why?
Answer:

Yes, I agree with the statement. The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals whereas ore is a mineral from which the metal is profitably extracted.
For example, aluminium exists in two mineral forms that is clay and bauxite. But aluminium is mainly extracted from bauxite which contains 70% aluminium oxide.
So Bauxite is an ore of aluminium whereas clay is not ore.
So all ores are minerals but all minerals need not be ores.

10th Class Chemistry 10th Lesson Principles of Metallurgy Activities

Activity – 1
1. How do you classify ores based on their formula?
Answer:
1) Look at the following ores.

2) Identify the metal present in each ore.
3) Classification of ores as oxides, sulphides, chlorides, carbonates, and sulphates as follows :

AP SSC 10th Class Physics Solutions Chapter 12 Electromagnetism

March 29, 2023February 2, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 12 Electromagnetism Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 12th Lesson Electromagnetism

10th Class Physics 12th Lesson Electromagnetism Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
How do electric appliances work?
Answer:
Electrical appliances work due to the electric force. Electrical force works in displacing the charges. Electric force is independent of the state of rest or the motion of the charged particle. Electric motor, washing machine are some of the examples of electric appliances.

Question 2.
How do electromagnets work?
Answer:
An electromagnet acquires the magnetic properties only when electric current is passed through the solenoid. Once the current is switched off, it almost loses its magnetic properties as retentivity of soft iron is very low. The strength of the electromagnet depends upon number of turns per unit length of the solenoid and the current through the solenoid.

Question 3.
Is there any relation between electricity and magnetism?
Answer:
The first evidence that there exists such a relationship between electricity and magnetism was observed by Oersted. When current carrying conductor was parallel to the axis of the needle, and the needle was deflected. This was much against his expectations. On reversing the direction of the current the needle moved in opposite direction.

Question 4.
Can we produce magnetism from electricity?
Answer:
We can produce magnetism from electronic current. Ampere with his Ampere’s swimming rule explained the direction of electric current and the deflection of magnetic needle.

Improve Your Learning

Question 1.
Are the magnetic field lines closed? Explain. (AS1)
Answer:

Magnetic field lines are closed.
If we observe the field lines formed by a current carrying straight wire, circular field lines are formed. They are closed circles.
If we observe the field lines by a current carrying solenoid the field lines out side the solenoid are continuous with those inside.
Thus the magnetic field lines are closed loops.

Question 2.
See figure, magnetic lines are shown. What is the direction of the current flowing through the wire? (AS1)

Answer:
If field lines are in anti-clockwise direction as shown in the diagram, the direction of current is vertically upwards. This can be demonstrated with right hand thumb rule.

Question 3.
A bar magnet with north pole facing towards a coil moves as shown in figure. What happens to the magnetic flux passing through the coil? (AS1)

(OR)
Why would induced current be generated in the coil when a north pole of a bar magnet pushed into it ?
Answer:
If north pole of the magnet moves towards the coil, there is a continuous change of magnetic flux linked with closed coil, then current is generated in the coil.

Question 4.
A coil is kept perpendicular to page. At P, current flows into the page and at Q it comes out of the page as shown in figure. What is the direction of magnetic field due to the coil? (AS1)

Answer:
At the top, anti-clockwise direction.
At the bottom, clockwise direction.
Try This:

Take a test tube and wound minimum 50 turns of 24 guage insulated copper wire with 2cms length at the centre of test tube as shown in figure, ‘l Now solenoid is ready. Take 3cms length of iron nail and make it floats on water with appropriate foam (thermocol) on the water. Now connect the j two ends of solenoid two 3-6 volts battery eliminator and switch on the eliminator. You can observe the motion of the nail towards the solenoid. (If not move decrease the water level or increase the potential).
Try to explain motion of the nail into the water using solenoid concept.

Question 5.
The direction of current flowing in a coil is shown in the figure. What type of magnetic pole is formed at the face that has flow of current as shown in the figure? (AS1)

Answer:
North. Since the current in the coil flows in anti-clockwise direction, north pole is formed at the face we are watching.

Question 6.
Why does the picture appear distorted when a bar magnet is brought close to the screen of a television? Explain. (AS1)
(OR)
Explain magnetic force on moving charge and current carrying wire.
(OR)
What happens when you bring a bar magnet near a picture of TV screen ? What inference do you conclude from this activity?
Answer:
This is due to the fact that magnetic field exerts a force on the moving charge.

TV screen Activity :

Take a bar magnet and bring it near the TV screen.
Then the picture on the screen is distorted.
Here the distortion is due to the motion of the electrons reaching the screen are affected by the magnetic field.
Now move the bar magnet away from the screen.
Then the picture on the screen stabilizes.
This must be due to the fact that the magnetic field exerts a force on the moving charges. This force is called magnetic force.
The magnitude of the force is F = Bqv where B is magnetic induction, ‘q’ is the charge and v is the velocity of the charged particle.

Question 7.
Symbol ‘X’ indicates the direction of a magnetic held into the page. A straight long wire carrying current along its length is kept perpendicular to the magnetic field. What is the magnitude of force experienced by the wire? In what direction does it act? (AS1)

Answer:
1) Magnetic force (F) experienced by the wire with the magnitude of ILB :
Here I = Current, L = Length of the wire B = Magnetic field

2) Direction of the magnetic force (F) :

1) The direction of force can be find by using Right hand rule.
2) Fore finger → i (North)
Middle finger → B (into the page)
Thumb → F (Towards west parallel to the paper)

Question 8.
Explain the working of electric motor with a neat diagram. (AS1)
(OR)
Which device converts electrical energy into mechanical energy? Explain the working of that device with a neat diagram.
Answer:
Electric motor converts electrical energy into mechanical energy.
Electric motor:
It is a device which converts electrical energy into mechanical energy.

Principle :
It is based on the principle that a current carrying conductor placed perpendicular to the magnetic field experiences a force.

Construction :
a) Armature coil:
It contains a single loop of an insulated copper wire in the form of a rectangle.

b) Strong magnetic field :
Armature coil is placed between two permanent poles (N & S) of a strong magnet.

c) Slip-ring Commutator:
It consists of two halves (C₁ and C₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring. Commutator reverses the direction of current in the armature coil.

d) Brushes:
Two carbon brushes B₁ and B₂ press against the commutator. These brushes act as the contact between the commutator and terminals of the battery.

e) Battery :
A battery is connected across the carbon brushes. The battery supplies the current to the armature coil.

Working and Theory :

When current flows through the coil, AB and CD experience magnetic force.
In the arm, AB of the coil experiences a force in one direction, similarly, in CD it experiences in opposite direction.
These two equal and opposite forces constitute a couple; which rotates the coil.
At this position, the supply of current to the coil is cut off because contacts of commutator and brushes break.
Hence no force acts on the arms of the coil.
The coil will not come to rest because of rotational inertia of motion, till the commutator again comes in contact with the brushes B₁ and B₂.
Now the direction of the current in the arms AB and CD is reversed.
Then the couple again rotates in opposite direction.
The coil of DC motor continues to rotate in the same direction. Hence electrical energy is converted into mechanical energy.
The speed of rotation of the motor depends on
a) current through the armature
b) number of turns of the coil
c) area of the coil
d) magnetic induction.

Question 9.
Derive Faraday’slaw of induction from law of conservation of energy. (AS1)
Answer:
Faraday’s law :
Whenever there is a continuous change in magnetic flux linked with coil closed the current is generated in the coil.

Consider a pair of parallel bare conductors which are separated by 7′ meters.
They are placed in uniform magnetic field of induction ‘B’ supplied by ‘N’ and ‘S’ poles of the magnet.
A galvanometer is connected to the ends of the parallel conductors.
We can close the circuit by touching the parallel conductor with another bare conductor which is taken in our hand.
If we move our hand to the left, the galvanometer needle will deflect in one direction.
If we move our hand to the right, the needle in the galvanometer moves in opposite direction.
A current will be set up in the circuit only when there is an EMF in the circuit. Let EMF be ‘ε’.
The principle of conservation energy tells us that this electric energy must come from the work that we have done in moving the cross wire.

Question 10.
The value of magnetic field induction which is uniform is 2T. What is the flux passing through a surface of area 1.5 m2 perpendicular to the field? (AS1)
Answer:
B = 2T ; Φ == ? ; A = 1.5 m²
We know B = $\frac{\phi}{\mathrm{A}}$
or Φ = BA = 2 × 1.5 =3 Webers

Question 11.
An 8N force acts on a rectangular conductor 20 cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40 A. (AS1)
Answer:
F = 8N ; l = 20 cm or 20 × 10^-2 m ; B = ? ; i = 40 Amp

Question 12.
Explain with the help of two activities that current carrying wire produces magne tic field. (AS1)
(OR)
How can you verify that a current carrying wire produces a magnetic field with the help of experiment?
Answer:
Activity – 1

Take a thermocole sheet and fix two thin sticks ol height 1cm.
Join the two sticks with the help of copper wire.
Take a battery, tap key and connect them in series with the copper wire which is thin.
Keep a marine compass needle beneath the wire.
If you press the tap key, current flows in the copper wire.
Immediately the magnetic needle gets deflected.
This indicates that the magnetic field is increased when current flows through the conductor.

Activity – 2

Take a wooden plank and make a hole.
Place the plank on the table.
Place the retort stand on it.
Pass copper wire through the hole.
Connect the two ends of the wire with battery through switch.
Place some compass needle around the hole.
When the current flows the magnetic needle deflects.
We can verify this by changing the direction of current.
So we can conclude the magnetic field surrounds a current carrying conductor.

Question 13.
How do you verify experimentally that the current carrying conductor experiences a force when it is kept in magnetic field? (AS1)
Answer:

A copper wire is passed through splits of wooden sticks.
Connect the wire to 3 volts battery.
Close the switch of the battery and pass the current.
Bring the horse-shoe magnet near the wire.
Then a force is experienced on the wire.
Reverse the polarities of the magnet, then the direction of the force is also reversed.
The right hand rule helps the direction of flow of current and the direction of current.

Question 14.
Explain Faraday’s law of induction with the help of an activity. (AS1)
(OR)
Write an activity which proves changing magnetic flux produces induced current in the circuit.
Answer:

Connect the terminals of a coil to a sensitive ammeter.
Push a bar magnet towards the coil, with its north pole facing the coil, the needle in the galvanometer deflects.
It shows that a current is set up in the coil.
The galvanometer does not deflect if the magnet is at rest.
If the magnet is moved away from the coil, the needle in the galvanometer again deflects in opposite direction.
Further this experiment enables us to understand that the relative motion of the magnet and coil set up a current in the coil. It makes no difference whether the magnet is moved towards the coil. This is one form of Faraday’s law.

Question 15.
Explain the working of AC electric generator with a neat diagram. (AS1)
(OR)
Which device converts mechanical energy into electrical energy? Explain the working of that device with a neat diagram.
Answer:
Generator converts mechanical energy into electrical energy.

As armature is rotated about an axis, the magnetic flux linked with armature changes. Therefore, an induced current is produced in the armature.
If the armature rotates in anti-clockwise direction, from Flemming’s right hand rule the direction of current and deflection of the coil are noted.
Alter armature has turned through 180°, it occupies another position.
By applying Flemming’s right hand rule we can find the direction of current and deflection of the needle.
Hence we can conclude the induced current is alternating in nature.

Question 16.
Explain the working of DC generator with a neat diagram. (AS1)
Answer:

The principle and working of D.C generator is same as that of AC generator except that in place of slip – rings as sliding contacts, we have a slip-ring or a commutator.
In a slip ring, there are two half rings.
The ends of armature coil are connected to these rings and these rings rotate the armature.
By using slip-ring, the direction of induced current does not change in the external circuit throughout the complete rotation of the armature. In other words, the current in the external circuit always flows in the same direction. Hence the induced current is unidirectional.

Question 17.
Rajkumar said to you that the magnetic field lines are open and they start at north pole of bar magnet and end at south pole. What questions do you ask Rajkumar to correct him by saying “field lines are closed”? (AS2)
Answer:

If the magnetic field lines start at north pole and end at south pole, where do the lines go from south pole?
What is happening within the bar magnet?
Are the magnetic field lines passing through bar magnet?
What is the direction of magnetic field lines inside the bar magnet? (Recall the solenoid activity).
Can you say now, that the magnetic field lines are open?

Question 18.
As shown in figure, both coil and bar magnet move in the same direction. Your friend is arguing that there is no change in flux. Do you agree with his statement? If not, what doubts do you have? Frame questions about the doubts you have regarding change in flux. (AS2)

Answer:

What happens if both magnet and coil move in same direction?
What happens if both magnet and coil move in opposite direction?
What is the direction of the current in the coil?
If both move in same direction, is there any linkage of flux with the coil?
When ‘N’ pole is moved towards the coil what is the direction of current?
If magnet is reversed, what is the direction of current in the coil?

Question 19.
What experiment do you suggest to understand Faraday’s law? What items are required? What suggestions do you give to get good results of the experiment? Give precautions also. (AS3)
Answer:
Aim :
To understand Faraday’s law of induction.

Materials required :
A coil of copper wire, a bar magnet, Galvanometer, etc.

Procedure :

Connect the terminal of a coil to a sensitive galvanometer as shown in the figure.
Normally, we would not expect any deflections of needle in the galvanometer because there is to be no electromotive force in this circuit.
Now if we push a bar magnet towards the coil, with its north pole facing the coil, we observe the needle in the galvanometer deflects, showing that a current is set up in the coil.
The galvanometer does not deflect if the magnet is at rest.
If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction.
If we use the end of south pole of a magnet instead of north pole in the above activity, the deflections are exactly reversed.
This experiment proves “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Precautions :

The coil should be kept on an insulating surface.
Bar magnet should be of good magnetic moment.
The centre of the galvanometer scale must be zero.
The deflections in the galvanometer must be observed while introducing the bar magnet into the coil and also while withdrawing it.

Question 20.
How can you verify that a current carrying wire produces a magnetic field with the help of an experiment? (AS3)
Answer:
Experiment:

Take a thermocole sheet and fix two thin wooden sticks of height 1cm.
These sticks are joined with the help of a copper wire.
Connect battery and tap key to this copper wire.
Place a magnetic compass beneath the wire.
Now press the tap key and allow the current through the wire. It is observed that magnetic needle deflects.
If you change the direction of the current, the direction of deflection of needle also changes.
So we can say current carrying conductor produces magnetic field.

Question 21.
Collect information about generation of current by using Faraday’s law. (AS4)
Answer:
Faraday’s law is useful in generation of current.

According to this law, the change in magnetic flux induces EMF in the coil.
Fie also proposed electromagnetic induction.
Electromagnetic induction is a base for generator, which produces electric current.
Transformer also works on the principle of electromagnetic induction, which is helpful in transmission of electricity.
Hence Faraday’s law is used in the generation and transmission of current.

Question 22.
Collect information about material required and procedure of making a simple electric motor from internet and make a simple ntotor on your own. (AS4)
Answer:
Aim :
Preparation of a simple electric motor.

Material requried :
A wire of nearly 15 cm, 1.5v Battery, Iron nail, strong magnet, paper clip.

Procedure:

Attach the magnet to the head of the iron nail.
Attach a paper clip to the open end of the magnet.
Now attach the other end of the nail (Free end) to the cap (positive terminal) of the battery.
Now connect the negative terminal of the battery and the head of the iron nail through a wire.
We observe that the paper clip rotates.

Another model:
Materials required :
1.5 m enamelled copper wire (about 25 gauge), 2 safety pins,
1.5 v battery, magnets, rubber bands or bands cut from cycle tube.

Procedure :

Wind copper wire on the battery nearly 10 – 15 turns to make a coil.
Remove the coil and fix the ends as shown in the figure.
Scrape the insulation com¬pletely on one end of the coil.
Scrape the insulation on top, left and right of the other end. The bottom should be insulated.
Now complete the electric mo¬tor as shown in the figure. “5

Question 23.
Collect information of experiments done by Faraday. (AS4)
Answer:
Experiment – 1

Connect the terminals of a coil to a sensitive galvanometer as shown in the figure.
Normally, we would not expect any deflection of needle in the galvanometer because there is no EMF in the circuit.
Now, if we push a bar magnet towards the coil, with its north pole facing the coil, the needle in the galvanometer deflects, showing that a current has been set up in the coil, the galvanometer does not deflect if the magnet is at rest.
If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction.
If we use the end of south pole of a magnet instead of north pole, the results i.e., the deflections in galvanometer are exactly opposite to the previous one.
This activity proves that the change in magnetic flux linked with a closed coil, produces current.
From this Faraday’s law of induction can be stated as “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”. This induced EMF is equal to the rate of change of magnetic flux passing through it.

Experiment – 2

Prepare a coil of copper wire C₁ and connect the two ends of the coil to a galvanometer.
Prepare another coil of copper wire similar to C₂ and connect the two ends of the coil to a battery via switch.
Now arrange the two coils C₁ and C₂ nearby as shown in the figure.
Now switch on the coil C₂. We observe a deflection in the galvanometer connected to the coil C₁.
The steady current in C₂ produces steady magnetic field. As coil C₂ is moved towards the coil C₁ the galvanometer shows a deflection.
This indicates that electric current is induced in coil C₁.
When C₂ is moved away, the galvanometer shows a deflection again, but this time in the opposite direction.
The deflection lasts as long as coil C₂ is in motion.
When C₂ is fixed and C₁ is moved, the same effects are observed.
This shows the induced EMF due to relative motion between two coils.

Question 24.
Draw a neat diagram of electric motor. Name the parts. (AS5)
Answer:

Question 25.
Draw a neat diagram of an AC generator. (AS5)
(OR)
Draw the diagram of electric generator and label its parts. A.
Answer:

Question 26.
How do you appreciate the Faraday’s law, which is the consequence of conservation of energy? (AS6)
Answer:

Law of conservation of energy says energy neither be created nor be destroyed, but can be converted from one form to another.
Faraday’s law says whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil. This induced EMF is equal to the rate of change of magnetic flux passing through it.
We have to do some work to move the magnet through a coil. This work produces energy.
This energy is converted into electrical energy in the coil.
In this way conservation of energy takes place in electromagnetic induction.

Question 27.
How do you appreciate the relation between magnetic field and electricity that changed the lifestyle of mankind? (AS6)
Answer:

Changing life style of mankind is a result of many inventions, utilising a lot of scientific principles.
Scientists all ways going on searching for new principles and new applications to make our life more comfortable.
If you consider electricity, right from amber stone to nuclear power, so many changes have been incorporated.
The idea of Oersted and Faraday that current carrying wire produces electricity and electromagnetic induction, enable us to use electric motors, generators, fans, mixers, grinders, induction stoves, etc.
All these appliances makes our life more comfortable. Hence Faraday and Oersted rendered a lot of servies in this field.
Hence, I appreciate the relation between magnetic field and electricity that changed the life style of mankind.
So if current is more, induction is also more.

Question 28.
Give a few applications of Faraday’s law of induction in daily life. (AS7)
Answer:
Applications:
The daily life applications of Faraday’s law of induction are

Generation of electricity
Transmission of electricity
Metal detectors in security checking
The tape recorder
Use of ATM cards
Induction stoves
Transformers
Induction coils (spark plugs in automobiles)
Break system in railway wheels
AC and DC generators
Windmills, etc.

Question 29.
Which of the various methods of current generation protects the nature well? Give examples to support your answer. (AS7)
Answer:
Windmill :

Electricity is produced when an armature of a generator rotates between two poles of a strong magnet.
Whereas when wind falls on the wheel of a windmill, it rotates. So the armature of the generator rotates between two poles of a magnet along with the rotation of the wheel of the windmill.
Thus electric current is produced.
This is how, KE of the wind is converted into electric energy.

Advantages :
Wind energy produces no smoke and no harmful gases. So this form of energy is pollution free or environment-friendly.

Fill in The Blanks

1. The SI unit of magnetic field induction is ………………….
2. Magnetic flux is the product of magnetic field induction and …………………
3. The charge is moving along the direction of magnetic field. Then force acting on it is ………………..
4. A current carrying wire of length L is placed perpendicular to a uniform magnetic field B. Then the force acting on the wire with current I is ……………..
5. Faraday’s law of induction is the consequence of …………………
Answer:

weber/m² (or) Tesla
area
zero
ILB
Law of conservation of energy

Multiple Choice Questions

1. Which converts electrical energy into mechanical energy?
A) motor
B) battery
C) generator
D) switch
Answer:
A) motor

2. Waich converts mechanical energy into electrical energy?
(OR)
The device used to convert mechanical energy into electrical energy among the following is
A) motor
B) battery
C) generator
D) switch
Answer:
C) generator

3. The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field, is
A) 0
B) ILB
C) 2ILB
D) ILB/2
Answer:
B) ILB

10th Class Physics 12th Lesson Electromagnetism InText Questions and Answers

10th Class Physics Textbook Page No. 221

Question 1.
Why does the needle get deflected by the magnet?
Answer:
Because of strength of the magnetic field of the magnet, the needle gets deflected since it is in the field.

10th Class Physics Textbook Page No. 213

Question 2.
How can we find the strength of the field and direction of the field?
Answer:
We can find the strength of the field with magnetic flux and the direction of the field from the tangent drawn to the line of force.

10th Class Physics Textbook Page No. 214

Question 3.
Can we give certain values to magnitude of the field at every point in the magnetic field?
Answer:
In uniform magnetic field it is same whereas in non-uniform magnetic field it is different.

10th Class Physics Textbook Page No. 215

Question 4.
What is the flux through unit area perpendicular to the field?
Answer:
Flux density or magnetic induction.

Question 5.
Can we generalize the formula of flux for any orientation of the plane taken in the field?
Answer:
Yes, Φ = BA cos θ

Question 6.
What is the flux through the plane taken parallel to the field?
Answer:
Magnetic flux (or) Magnetic field.

Question 7.
What is the use of introducing the ideas of magnetic flux and magnetic flux density?
Answer:
Magnetic flux and flux density help in understanding the concept of electromagnetic induction and relation between electricity and magnetism.

Question 8.
Are there any sources of magnetic field other than magnets?
Answer:
Current carrying straight wires and loops act as sources of magnetic filed.

Question 9.
Do you know how old electric calling bells work?
Answer:
Yes. They work on the principle of magnetic effect of electric currents.

10th Class Physics Textbook Page No. 218

Question 10.
What happens when a current carrying wire is kept in a magnetic field?
Answer:

Magnetic field applies force on current carrying wire.
So it gets deflected and the direction of deflection is given by right hand rule.
Or there will be no force acting on the wire when wire is in the direction of the field.

10th Class Physics Textbook Page No. 219

Question 11.
Do you feel any sensation on your skin?
Answer:
Yes. The hair on my skin rises up when I stand near TV screen.

Question 12.
What could be the reason for that?
Answer:
It is due to the magnetic field produced by electric charges in motion.

Question 13.
Why does the picture get distorted?
Answer:
Due to motion of electrons that form the picture is affected by the magnetic field of bar magnet.

Question 14.
Is the motion of electrons reaching the screen affected by the magnetic field of the bar magnet?
Answer:
Yes. The motion of electrons reaching the screen is affected by the magnetic field of the bar magnet.

Question 15.
Can we calculate the force experienced by a charge moving in a magnetic field?
Answer:
Yes, If the force is F, it is given by the expression F = qvB.

Question 16.
Can we generalize the equation for magnetic force on charge when there is an angle ‘0’ between the directions of field “B” and velocity “v”?
Answer:
No, Then force F is given by the formula F = qvB sin θ.

Question 17.
What is the magnetic force on the charge moving parallel to a magnetic field?
Answer:
When the charge moves parallel to the magnetic field the value of “θ” becomes zero. In the equation F = qvB sin θ, since θ = θ, the value of force F also becomes zero.

Question 18.
What is the direction of magnetic force acting on a moving charge?
Answer:
By applying right hand rule we can guess the direction of magnetic force acting on a moving charge is the “thumb” direction.

10th Class Physics Textbook Page No. 221

Question 19.
Can you determine the magnetic force on a current carrying wire which is placed along a magnetic field?
Answer:
F = BIl sin θ. If the current carrying wire is placed along direction field θ = 0.
∴ F = 0

Question 20.
What is the force on the wire if its length makes an angle ‘θ’ with the magnetic field?
Answer:
F = Bqv sin θ or F = Bil sin θ, where ‘i’ is current. WorhA
Here B = magnetic induction, q = charge, v = velocity of the charge and ‘θ’ is the angle between direction of field and velocity.

10th Class Physics Textbook Page No. 222

Question 21.
How could you find its (current carrying wire) direction?
Answer:
We can find by using right hand rule.

Question 22.
Is the direction of deflection observed experimentally same as that of the theoretically expected one?
Answer:
Yes. But it depends on polarities of the horse shoe magnet.

Question 23.
Does the right hand rule give the explanation for the direction of magnetic force exerted by magnetic field on the wire?
Answer:
The right hand rule does not help us to explain the reason for deflection of wire.

Question 24.
Can you give a reason for it (deflection of wire)?
Answer:
There exists only magnetic field due to external source. When there is a current in the wire, it also produces a magnetic field. These fields overlap and give non-uniform field. This is the reason for it.

10th Class Physics Textbook Page No. 223

Question 25.
Does this deflection fit with the direction of magnetic force found by right hand rule?
Answer:
Yes. This deflection fits with the direction of magnetic force found by right hand rule.

Question 26.
What happens when a current carrying coil is placed in a uniform magnetic field?
Answer:
It gets deflected since magnetic lines of force are perpendicular to the length of the coil.

Question 27.
Can we use this knowledge to construct an electric motor?
Answer:
Yes. This is the principle of electric motor.

Question 28.
What is the angle made by AB and CD with magnetic field?
Answer:

AB and CD are at right angles to the magnetic field.

10th Class Physics Textbook Page No. 224

Question 29.
Can you draw the direction of magnetic force on sides AB and CD?
Answer:
Yes. The direction of magnetic force on sides AB and CD can be determined by applying right hand rule.

Question 30.
What are the directions of forces on BC and DA?
Answer:

ADBC, magnetic force pulls the coil up and at DA magnetic force pulls it down.

Question 31.
What is the net force on the rectangular coil?
Answer:
Net force on the rectangular coil is zero.

Question 32.
Why does the coil rotate?
Answer:

The rectangular coil rotates in clockwise direction because of equal and opposite pan’ of forces acting on the two sides of the coil.

Question 33.
What happens to the rotation of the coil if the direction of current in the coil remains unchanged?
Answer:
The coil comes to halt and rotates in anti-clockwise direction.

Question 34.
How could you make the coil rotate continuously?
Answer:
If the direction of current in coil, after the first half rotation, is reversed, the coil will continue to rotate in the same direction.

10th Class Physics Textbook Page No. 225

Question 35.
How can we achieve this (convertion of electrical energy to mechanical energy)?
Answer:
Brushes B₁ and B₂ are used to achieve this.

Question 36.
What happens when a coil without current is made to rotate in magnetic field?
Answer:
When the coil rotated due to the change in magnetic flux electricity is generated.

Question 37.
How is current produced?
Answer:
The current is produced from the battery to the coil.

Question 38.
Why is there a difference in behaviour in these two cases?
Answer:
The A.C. supply changes its direction a number of times in a second. But D.C. is unidirectional current. So there is a difference in the behaviour of the metal ring in these two cases.

Question 39.
What force supports the ring against gravity when it is being levitated?
Answer:
The magnetic force developed in the coil of copper wire supports the ring against gravity when it is being levitated.

10th Class Physics Textbook Page No. 226

Question 40.
Could the ring be levitated if DC is used?
Answer:
The metal ring is levitated because the net force on it should be zero according to Newton’s second law.

Question 41.
What is this unknown force acting on the metal ring?
Answer:
The change in polarities at certain intervals at the ends of the solenoid causes the unknown force acting on the metal ring.

Question 42.
What is responsible for the current in the metal ring?
Answer:
The field through the metal ring changes so that flux linked with the metal ring changes and this is responsible for the current in metal ring.

Question 43.
If DC is used, the metal ring lifts up and falls down immediately. Why?
Answer:
The flux linked with metal ring is zero. When the switch is on, at that instant there should be a change in the flux linked with ring. So the ring rises up and falls down. If the switch is off, the metal ring again raises up and falls down. There is no change in flux linked with ring when the switch is off.

10th Class Physics Textbook Page No. 227

Question 44.
What could you conclude from the above analysis (metal ring lifts up and falls down)?
Answer:
The relative motion of the magnet and coil sets up a current in the coil.

10th Class Physics Textbook Page No. 228

Question 45.
What is the direction of induced current?
Answer:
The direction of the induced current is such that it opposes the charge that produced it.

Question 46.
Can you apply conservation of energy for electromagnetic induction?
Answer:
Yes, we can apply. The mechanical energy is converted into electrical energy.

10th Class Physics Textbook Page No. 229

Question 47.
Can you guess what could be the direction of induced current in the coil in such case?
Answer:
The direction of the induced current in the coil must be in anti-clockwise direction.

Question 48.
Could we get Faraday’s law of induction from conservation of energy?
Answer:
Yes, we can get. Here we have to ignore the friction everywhere.

10th Class Physics Textbook Page No. 230

Question 49.
Can you derive an expression for the force applied on crosswire by the field “B”?
Answer:
Yes. The force applied F = BIl.

10th Class Physics Textbook Page No. 232

Question 50.
How could we use the principle of electromagnetic induction in the case of using ATM card when its magnetic strip is swiped through a scanner? Discuss with your friend or teacher.
Answer:
If the card is moved through a card reader, then a change in magnetic flux is produced in one direction, which induced potential or EMF. The current received by the pickup coil goes through signal amplification and translated into binary code, so that it can be read by computer.

Question 51.
What happens when a coil is continuously rotated in a uniform magnetic field?
Answer:
An induced current is generated in the coil.

Question 52.
Does it (continuous rotation of coil) help us to generate electric current?
Answer:
Yes. Continuous rotation of coil helps us to generate electric current.

10th Class Physics Textbook Page No. 233

Question 53.
Is the direction of current induced in the coil constant? Does it change?
Answer:
Yes, it changes. When the coil is at rest in vertical position, with side (A) of coil at top position side (B) at bottom position, no current will be induced in it.

Question 54.
Can you guess the reason for variation of current from zero to maximum and vice-versa during the rotation of coil?
Answer:
The reason for variation of current from zero to maximum and vice-versa during the rotation of coil current generated follows the same pattern so that in first half except that the direction of current is reversed.

Question 55.
Can we make use of this current? If so, how?
Answer:
Two carbon brushes are arranged in such a way that they press the slip rings to obtain current from the coil. When these brushes are connected to external devices like TV, Radio we can make them work with current supplied from ends of carbon brushes.

10th Class Physics Textbook Page No. 234

Question 56.
How can we get DC current using a generator?
Answer:
By connecting two half-slip rings instead of a slip ring commutator on either side to the ends of the coil we can get D.C. current.

Question 57.
What changes do we need to make in an AC generator to be converted into a DC generator?
Answer:
Instead of two slip rings, we have to use a slip ring commutator to change A.C. generator into a D.C. generator.

10th Class Physics 12th Lesson Electromagnetism Activities

Activity – 2

Question 1.
Show that the magnetic field around a bar magnet is three dimensional and its strength and direction varies from place to place.
Answer:

Take a sheet of white paper and place it on the horizontal table.
Place a bar magnet in the middle of the sheet.
Place a magnetic compass near the magnet it settles to a certain direction.
Use a pencil and put dots on the sheet on either side of the needle. Remove the compass. Draw a small line segment connecting the two dots. Draw an arrow on it from south pole of the needle to north pole of the needle.
Repeat the same by placing the compass needle at various positions on the paper. The compass needle settles in different directions at different positions.
This shows that the direction of magnetic field due to a bar magnet varies from place to place.
Now take the compass needle to places far away from magnet, on the sheet and observe the orientation of the compass needle in each case.
The compass needle shows almost the same direction along north and soiath at places far from the magnet.
This shows that the strength of the field varies with distance from the bar magnet.
Now hold the compass a little above the table and at the top of the bar magnet.
We observe the deflection in compass needle. Hence we can say that the mag¬netic field is three dimensional i.e., magnetic field surrounds its source.
From the above activities we can generalize that a magnetic field exists in the region surrounding a bar magnet and is characterized by strength and direction.

Activity – 3

Question 2.
Explain how you draw magnetic lines of force in the magnetic field.
(OR)
What is the name given to the imaginary lines joining from north pole to south pole of a bar magnet called? Explain how you can draw those lines around a bar magnet.
Answer;
These lines are called magnetic lines of forces.

Procedure:

Take a white drawing sheet.
Place a marine compass at the centre of the sheet.
Draw a line which shows north and south of the earth on the drawing sheet.
Now remove the compass needle and place a bar magnet at the centre of the sheet showing north of the bar magnet pointing north of the earth.
Place the magnetic compass near the bar magnet without contact. The needle comes to rest after oscillations.
Locate the end of the pointer with pencil. Now place the compass needle at this point and once again notice the end of the pointer.
We can repeat the same around the magnet, and draw all the points with the help of the pencil.
We can draw the lines taking the needle too far to the magnet and we can observe the orientation of needle of compass.
So we can conclude that the strength of field varies with distance from the bar magnet.
These lines of force are from north of the bar magnet to south of the bar magnet.

Activity – 4

Question 3.
Explain the direction of magnetic field around the straight conductor carrying current.
(OR)
What field would be formed around straight conductor carrying current? How do you find the direction of that field experimentally?
Answer:
Magnetic field would be formed around current carrying conductor.

Procedure:

Take a wooden plank and make a hole and place il on the table.
Place a stand on the plants, and suspend a c opper wire from the stand and see that it passes through the hole made to the plank.
Connect a battery and switch to this wire in series. Place some magnetic needle at the hole.
If the current is passed through the wire, the magnetic needle deflects and it is directed as the tangent to the circle
If the current flows in downward direction, the field lines are in anti-clockwise direction and if the current flows in upward direction, the field lines are in clockwise direction.
The direction ol the current and magnetic fines of force can be easily explained with the help of right hand thu mb rule. If you hold the current carrying conductor with your right hand grip stretching the thumb, the direction of the I humb shows the direction of the current, the direction of the other four fingers shows direction of magnetic lines of force.

Activity 5

Question 4.
Explain the direction of magnetic field due to circular coil.
Answer:

Procedure :

Take a thin wooden plank and cover it with whitepaper.
Make two holes to the plank and pass insulated copper wire through the holes and wind the wire 4 to 5 times through the holes such that it looks like a coil.
The ends of the wire are connected to the battery terminals.
Now place a compass needle at the centre of the coil.
Put dots on either side of the compass. Repeat this by keeping at the dots. We can observe that field lines are circular.
Here the direction of the field is perpendicular to the plane of the coil.
The direction of the magnetic field due to coil points towards you when the current in the coil is in anti-clockwise direction.
When you curl your right hand fingers in the direction of current, thumb gives the direction of magnetic field.

Activity – 6

Question 5.
Explain the magnetic field due to solenoid.
(OR)
What is the name given to the device which is a long wire wound in a close pack helix? Find the direction of magnetic field around that device.
Answer:
It is called solenoid.

Procedure :

Take a wooden plank covered with white paper.
Make holes on its surface.
Pass copper wire through the holes.
Join the ends of the coil to a battery through a switch.
Current passes through the coil, when we switch on the circuit.
Now sprinkle iron filings on the surface of the plank, around the coil. Then orderly pattern of iron filings is seen on the paper.
The iron filings arrange themselves in orderly way and look like lines of force.
The long coil is known as solenoid. The direction of the field due to solenoid is determined by using right hand rule.
One end of the solenoid behaves like a north pole and the other behaves like south pole.
Outside the solenoid the direction of the field lines of force is from north to south while inside the direction is from south to north. Thus the magnetic field lines are closed loops.
Hence electric charges in motion produce magnetic field.

Activity – 8

Question 6.
Explain the field lines due to horse-shoe magnet between its poles.
(OR)
Which field is set up between poles of a horse-shoe magnet? Explain the field lines due to horse magnet between its poles.
Answer:
Non-uniform magnet is set up between poles of a horse shoe magnet.

Procedure :

The field in between north and south pole of horse-shoe magnet are straight and parallel.
If the wire is passing perpendicular to the paper, the magnetic lines of force are concentric circles, when the current is passed.
The direction of field lines due to the wire in upper part coincides with the direction of field lines of horse-shoe magnet.
The direction of field lines by the wire in lower part is opposite to the direction of field lines of horse-shoe magnet.
Hence the net field in upper part is strong and in the lower part is weak.
Hence a non-uniform field is created around the wire.

Activity – 9

Question 7.
Explain electromagnetic induction.
(OR)
Which current will levitate the ring in the following figure? Explain the experimental activity.
Answer:
AC will levitate the ring.

Procedure :

Fix a soft iron cylinder on the wooden base vertically.
Wind copper wme around the soft iron.
Take a metal ring which is slightly greater in radius than the radius of soft iron cylinder and insert it through the soft iron cylinder.
Connect the ends of the coil to an AC source and switch on the current.
Here metal ring levitates on the coil (appears to rise and floats in the air).
In this experiment we can conclude that if AC current is used, the magnetic induction changes in both magnitude and direction in the solenoid and in the ring. The field through the metal ring changes, so that flux linked with the metal ring changes.
If DC current is used, the metal ring lifts up and falls down immediately.

AP SSC 10th Class Physics Solutions Chapter 11 Electric Current

March 29, 2023February 2, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 11 Electric Current Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 11th Lesson Electric Current

10th Class Physics 11th Lesson Electric Current Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What do you mean by electric current?
(OR)
Define electric current.
Answer:
Electric current is defined as the amount of charge crossing any cross-section of the conductor in one second.

Question 2.
Which type of charge (positive or negative) flows through an electric wire when it is connected in an electric circuit?
Answer:
Negative type of charge flows through an electric wire when it is connected in an electric circuit.

Question 3.
Is there any evidence for the motion of charge in daily life situations?
Answer:
Yes, lightning is a live example.

Improve Your Learning

Question 1.
Explain how electron flow causes electric current with Lorentz – Drude theory of electrons. (AS1)
(OR)
How does electron flow cattle elfectric current with Lorentz – Drude theory of electrons? Explain.
Answer:
Lorentz – Drude theory :

Lorentz – Drude proposed that conductors like metals contain a large number of free electrons.
The positive ions are fixed in their locations. The arrangement of the positive ions is called lattice.
The negative ions (electrons) move randomly in lattice in an open circuit.
When the lattice is closed the electrons are arranged in ordered motion.
When the electrons are in order motion, there will be a net charge (crossing through any cross section.
This order motion of electrons is called electric current.

Question 2.
How does a battery work? Explain. (AS1)
(OR)
How does a battery maintain a constant potential difference between its terminals?
Answer:
Working of a battery :

A battery consists of two metal plates (positive electrode = anode and negative electrode = cathode) and a chemical (electrolyte).
The electrolyte between the two metal plates consists of positive and negative ions which move in opposite directions.
The electrolyte exerts a chemical force on these ions and makes them move in a specified direction.
Depending upon the nature of the chemical, positive ions move towards one of the plates and accumulate on that plate.
As a result of this accumulation of charges on this plate it becomes anode.
Negative ions move in a direction opposite to the motion of positive ions and accumulate on the other plate.
As a result of this the plate becomes negatively charged called cathode.
This accumulation of different charges on respective plates continues till both plates are sufficiently charged.
But the ions in motion experience electric force when sufficient number of charges are accumulated on the plates.
The motion of ions continues towards their respective plates till the chemical force is equal to electric force.
Thus the battery works.

Question 3.
Write the difference between potential difference and emf. (AS1)
Answer:
Potential Difference:
Work done by the electric force on unit charge is called potential difference.
$\mathbf{V}=\frac{\mathbf{W}}{q}=\frac{\mathbf{F} l}{\mathbf{q}}$

Electromotive force (emf):
The work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.
$\varepsilon=\frac{W}{q}=\frac{F d}{q}$

Question 4.
How can you verify that the resistance of a conductor is temperature dependent? (AS1)
(OR)
How do you prove increase in temperature affects the resistance with an activity?
Answer:
Resistance :
The resistance of a conductor is the obstruction offered to the flow of electrons in a conductor.

Resistance is temperature dependent:
Aim:
To show that the value of resistance of a conductor depends on temperature for constant voltage between the ends of the conductor.

Materials required :

A bulb
A battery
Key
Insulated wire
Multimeter

Procedure :

Take a bulb and measure the resistance when it is in open circuit using a multimeter.
Note the value of resistance in your notebook.
Connect a circuit with components as shown in figure.
Switch on the circuit. After few minutes, measure the resistance of the bulb again.
Note this value in your notebook.

Observation :

The value of resistance of the bulb in second instance is more than the resistance of the bulb in open circuit.
The bulb gets heated.

Result:
The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.

Question 5.
What do you mean by electric shock? Explain how it takes place. (AS1)
Answer:
Electric shock:
The electric shock is combined effect of potential difference, electric current, and resistance of the human body.

An electric shock can be experienced when there exists a potential difference between one part of the body and another part.
When current flows through human body, it chooses the path which offers low resistance.
The resistance of a body is not uniform throughout it.
As long as current flow continues inside the body, the current and resistance of human body go on changing inversely.
This is called the electric shock.

Question 6.
Derive $\mathbf{R}=\frac{\rho l}{\mathbf{A}}$. (AS1)
(OR)
What are laws of resistance and derive a formula for resistance.
Answer:
Resistance of a conductor is directly proportional to the length of the conductor,
i.e., R ∝ l ………………….. (1)
Resistance of a conductor is inversely proportional to the cross-section area of the conductor.
i.e., R ∝ $\frac{1}{\mathrm{~A}}$ ………………….. (2)
From (1) and (2) R ∝ $R \propto \frac{l}{A} \Rightarrow R=\frac{\rho l}{A}$
where ρ is a constant,
ρ is called specific resistance or resistivity.

Question 7.
How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross-section area and temperature? (AS1)
Answer:

Collect manganin wires of different lengths with the same cross-sectional areas.
Make a circuit as shown in figure.
Connect one of the manganin wires between the ends P and Q.
Measure the value of the current using the ammeter.
Repeat the same for other lengths of the wires.
Note the values of currents.
We notice that the current decreases with increase in the length of the wire.
∴ R ∝ l (at constant temperature and cross-section area) …………… (1)
Do the same with manganin wires with equal lengths but different cross-section area.
We notice that the resistance was more when the cross-section area was less.
∴ R ∝ $\frac{1}{\mathrm{~A}}$ ………………. (2)
∴ R ∝ $[latex]\frac{1}{\mathrm{~A}}$[/latex]
Thus we verify l and A.

Question 8.
Explain Kirchhoff’s laws with examples. (AS1)
(OR)
Write two examples of Kirchhoffs laws and explain it.
Answer:
Kirchhoff’s laws :
Two simple rules called Kirchhoff’s rules are applicable to any DC circuit containing batteries and resistors connected in any way.
The two laws are (i) Junction law and (ii) Loop law.

i) Junction law :

Here P is called junction point where conducting wires meet. The junction law states that, at any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents leaving the junction.
i.e., I₁ + I₄ + I₆ = I₂ + I₃ + I₅
This law is based on the conservation of charge.

ii) Loop law:

Loop law states that, the algebraic sum of the increases and decreases in potential difference (voltage) across various components of the circuit in a closed circuit loop must be zero.

This law is based on the conservation of energy.

Question 9.
What is the value of 1 KWH in Joules? (AS1)
Answer:
1 KWH = 1 KW x 1h
= 1000 W × 60 min = 1000 W × 60 × 60 s = 3.6 × 10₆ Ws = 3.6 × 10₆ J.
∴ 1 KWH = 3.6 × 10₆ J.

Question 10.
Explain overloading of household circuit. (AS1)
Answer:

Electricity enters our homes through two wires called lines. These lines have low resistance and the potential difference between the wires is usually about 240 V.
All electrical devices are connected in parallel in our home. Hence, the potential drop across each device is 240 V.
Based on the resistance of each electric device, it draws some current from the supply. Total current drawn from the mains is equal to the sum of the currents passing through each device.
If we add more devices to the household circuit the current drawn from the mains also increases.
This leads to overheating and may cause a fire. This is called “overloading”.

Question 11.
Why do we use fuses in household circuits? (AS1)
(OR)
What is the use of fuses?
Answer:

The fuse consists of a thin wire of low melting point.
When the current in the fuse exceeds 20 A, the wire will heat up and melt.
The circuit then becomes open and prevents the flow of current into the household circuit.
Hence all the electric devices are saved from damage that could be caused by overload.
Thus we can save the household wiring and devices by using fuses.

Question 12.
Deduce the expression for the equivalent resistance of three resistors connected in series. (AS1)
(OR)
Derive R = R₁ + R₂ + R₃
(OR)
The second end of a first resistor is connected to first end of second resistor. Then how are the resistors connected? Derive the expression for the resultant resistance of this connection.
Answer:
Series connection:
In series connection of resistors, there is only one path for the flow of current in the circuit. Hence, the current in the circuit is equal to I.
According to Ohm’s law,
∴ V₁ = IR₁ ; V₂ = IR₂ and V₃ = IR₃.
⇒ Let R be the equivalent resistance of the combination of resistors in series.

Also V = I R_eq
V = V₁ + V₂ + V₃
I R_eq = IR₁ + IR₂ + IR₃
⇒ I R_eq = I (R₁ + R₂ + R₃)
⇒ R_eq = R₁ + R₂ + R₃
∴ The sum of individual resistances is equal to their equivalent resistance when the resistors are connected in series.

Question 13.
Deduce the expression for the equivalent resistance of three resistors connected in parallel. (AS1)
(OR)
Derive : $\frac{1}{\mathbf{R}}=\frac{1}{\mathbf{R}_{1}}+\frac{1}{\mathbf{R}_{2}}+\frac{1}{\mathbf{R}_{3}}$
(OR)
Explain the expression for the equivalent resistance of three resistors which are connected in parallel.
(OR)
If all the first ends of resistors are connected to and second ends are connected to another point, then what type of connection is this? Derive the resultant resistance for this connection.
Answer:
Parallel Connection :
In parallel connection of resistors, there is same potential difference at the ends of the resistors. Hence the voltage in the circuit is equal to V.
Let Ip I2 and I3 be the currents flowing through R₁, R_2, and R₃ resistors respectively.
Hence, we can write I = I₁ + I₂ + I₃.
According to the Ohm’s law,

∴ The equivalent resistance of a parallel combination is less than the resistance of each of the resistors.

Question 14.
Silver is a better conductor of electricity than copper. Why do we use copper wire for conduction of electricity? (AS1)
Answer:
Silver is costlier than copper. So, we use copper wire for conduction of electricity even though silver is a better conductor of electricity.

Question 15.
Two bulbs have ratings 100 W, 220 V and 60 W, 220 V. Which one has the greater resistance? (AS1)
Answer:

∴ The second bulb possessing 60 W, 220 V has the greater resistance.

Question 16.
Why don’t we use series arrangement of electrical appliances like bulb, television, fan, and others in domestic circuits? (AS1)
Answer:

If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases. To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series. They are connected in parallel.
In series combination same current passes through all resistors. This is not suggestable for household appliances. Hence, they are connected in parallel.

Question 17.
A wire of length 1 m and radius 0.1 mm has a resistance of 100 Ω. Find the resistivity of the material. (AS1)
Answer:
1) Given l = 1 m, r = 0.1 mm = 10^-4 m, R = 100 Ω

Question 18.
Why do we consider tungsten as a suitable material for making the filament of a bulb? (AS2)
(OR)
What is the reason for using Tungsten as a filament in electric bulb?
Answer:
Tungsten has higher resistivity values and melting point. So, we consider tungsten as a suitable material for making the filament of a bulb.

Question 19.
Are the head lights of a car connected in series or parallel? Why? (AS2)
Answer:
The headlights of a car are connected in parallel.
Reason :

When they are connected in parallel, same voltage (RD) will be maintained in the two lights.
If one of the light damaged, the other will work without any disturbance.

Question 20.
Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?
Answer:

The electric appliances are connected in parallel in a household circuit. Because in parallel wiring if any electric appliance is switched off, other appliances don’t get off.
If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases.
To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series.

Question 21.
Suppose that you have three resistors each of value 30Ω. How many resistors can you obtain by various combinations of these three resistors? Draw diagrams in support of your predictions. (AS2)
Answer:
Let R₁ = 30Ω, R₂ = 30Ω, R₃ = 30Ω
We get different resistors by different combinations as shown below.

Question 22.
State Ohm’s law. Suggest an experiment to verify it and explain the procedure. (AS3)
How do you prove experimentally the ratio V/l is a constant for a given conductor?
Answer:
Ohm’s law :
The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.

Verification :
Aim :
To verify Ohm’s law or to show that $\frac{\mathrm{V}}{\mathrm{I}}$ = constant for a conductor.

Materials required :
6V Battery eliminator, 0 to 1A Ammeter, 0 – 6V volt meter, copper wires, 50 cm manganin coil, Rheostat, switch and 3V LED, etc.

Procedure :

Complete the circuit as shown in figure. Knob should be adjusted to 4.5V at battery eliminator.
Using Rheostat change the potential difference between two ends of manganin wire from 0V to 4.5V (maximum).
By using Rheostat adjust the potential difference 1V between two ends of manganin wire.
Now observe the electric current through Ammeter in the circuit and note down in the following table.

Using Rheostat change the potential difference with different values upto 4.5V and note down the current value (I) in the table.
Take atleast five values of V and I and note down in the table.
Find $\frac{\mathrm{V}}{\mathrm{I}}$ for each set of values.
We notice that $\frac{\mathrm{V}}{\mathrm{I}}$ is a constant.
V ∝ I ⇒ $\frac{\mathrm{V}}{\mathrm{I}}$ = constant
This constant is known as resistance of the conductor, denoted by R.
⇒ $\frac{\mathrm{V}}{\mathrm{I}}$ = R
∴ Ohm’s law is verified.

How to Make Rheostat:

Make two holes at the two ends of 30cm Wooden scale. Through these holes fix two bolts with the help of nuts.Then take iron box filament i. e., nichrome wire and tie one end of thewire to the first bolt and wound wire with equal distance on the wooden scale to other end of the second bolt. Place this scale on the other scale perpendicularly as shown in the figure and stick them with glue. Now Rheostat is ready. Take support of your teacher to know the connection and functioning of Rheostat.

Question 23.
a) Take a battery and measure the potential difference. Make a circuit and measure the potential difference when the battery is connected in the circuit. Is there any difference in potential difference of battery? (AS4)
b) Measure the resistance of a bulb (filament) in open circuit with a multi-meter. Make a circuit with elements such as bulb, battery of 12 V and key in series. Close the key. Then again measure the resistance of the same bulb (filament:) for every 30 seconds. Record the observations in a proper table. What can you conclude from the above results? (AS4)
Answer:

a) When the battery is connected in a circuit, the voltage slowly decreases due to consumption of it. So, there is difference in voltage before using and after connecting.

b) After connecting battery (12 V), key in ammeter and bulb as shown in figure, we measure current (I) using the ammeter and voltage using multi-meter or voltmeter.

Note these values in the following table. Measure the resistance of the same bulb for every 30 seconds.

We conclude that the resistance is constant.

Question 24.
Draw a circuit diagram for a circuit in which two resistors A and B are connected in series with a battery and a voltmeter is connected to measure the potential difference across the resistor A. (AS5)
Answer:

V : Volt meter
A and B : Resistors
B : Battery
K: Key

Question 25.
How can you appreciate the role of a small fuse in house wiring circuit in preventing damage to various electrical appliances connected in the circuit? (AS7)
(OR)
We can save the household wiring and devices by using fuses. Write any four points by appreciating the role of fuse.
Answer:

The fuse consists of a thin wire of low melting point. When the current in the fuse exceeds 20 A, the wire will heat up and melt.
The circuit then becomes open and prevents the flow of current into the household circuit. So all the electric devices are saved from damage that could be caused by overload.
Thus we can save the household wiring and devices by using fuses.
In this way a small fuse prevents a great damage to costly electrical appliances in the circuit.

Question 26.
In the figure, the potential at A is………….. when the potential at B is zero. (AS7)

Answer:
Potential difference at A = V
Potential difference atB = V + 5 × 1 + 2 = 0 ⇒ V + 7V = 0
∴ V = – 7V

Question 27.
Observe the circuit and answer the questions given below. (AS7)

i) Are resistors C and D in series?
ii) Are resistors A and B in series?
iii) Is the battery in series with.any resistor?
iv) What is the potential drop across the resistor C?
v) What is the total emf in the circuit if the potential drop across resistor A is 6 V?
Answer:
The given circuit is written / drawn as
i) Yes, resistors ‘C’ and ‘D’ are connected in series. (Because, passing of the current is same to those resistors)
ii) No, resistors A’ and ‘B’ are not in series. (Because, different currents are passing through A and B. i.e., I₁ and I₂)
iii) The battery is in series with the resistor ‘A’. (Because, same current is passing through battery and resistor ‘A’, i.e., I)

iv) Potential drop across the resistor ‘C’
V₂ = V₃ + V₄
14V = V₃ + 8V
V₃ = 6V
Potential drop = 6V

v) Total emf
emf of combination of V₃ and V₄ = 14V ……………….. (1)
emf of combination of (1) and V₂ = 14 V ………………. (2)
emf of combination of (2) and V₁ = 6V + 14V = 20V
(Given, emf of ‘A’ = 6V)
Total emf = 20V

Question 28.
If the resistance of your body is 100000 Cl, what would be the current that flows in your body when you touch the terminals of a 12 V battery? (AS7)
Answer:
We know that, $I=\frac{V}{R}$; here V = 12 V, R = 1,00,000Ω.
∴ The current passing through our body $I=\frac{12 \mathrm{~V}}{100000 \Omega}$ = 0.00012 Ampere.

Question 29.
A uniform wire of resistance 100 Ω is melted and recast into wire of length double that of the original. What would be the resistance of the new wire formed? (AS7)
Answer:
Given R = 100 Ω
When ‘l = l’, R = 100 Ω.
When’l = 2l’, A’ = A / 2.

∴ Resistance is increased by four times.
∴ R = 4 × 1ooΩ = 400Ω.

Question 30.
A house has 3 tube lights, two fans and a Television. Each tube light draws 40 W. The fan draws 80 W and the Television draws 60 W. On the average, all the tube lights are kept on for five hours, two fans for 12 hours and the television for five hours every day. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWh. (AS7)
Answer:
Given 3 tube lights, two fans and a television.
Power consumed by 1 tube light = 40 W
∴ Power consumed by 3 tube lights = 3 × 40W = 120W
3 tube lights are kept on for five hours. So, consumption of power by 3 tube lights
= 5 × 120 W = 600 W ……………. (1)
Power consumed by 1 fan = 80 W
∴ Power consumed by 2 fans = 2x80W=160W
2 fans are kept on for 12 hours. So, consumption of power by 2 fans
= 12 × 160 W = 1920 W ……………. (2)
Power drawn by TV = 60 W
TV is kept on for 5 hours = 5 x 60 W = 300 W ………………. (3)
∴ Consumption of power in one day = (1) + (2) + (3)
= 600W+ 1920 W + 300 W = 2820 W = 2.820 KW
∴ Total consumption of power in 30 days at Rs. 3 per KW
= 2.820 × 30 × 3 = Rs. 253.80/-

Fill in The Blanks

1. The kilowatt hour is the unit of …………………..
2. A thick wire has ………………….. resistance than a thin wire.
3. An unknown circuit draws a current of 2 A from a 12 V battery. Its equivalent resistance is …………………..
4. The SI unit of potential difference is …………………..
5. The SI unit of current is …………………..
6. Three resistors of values 2Ω, 4Ω, 6Ω are connected in series. The equivalent resistance of combination of resistors is ……………………
7. Three resistors of values 2Ω, 4Ω, 6Ω are connected in parallel. The equivalent resistance of combination of resistors is ……………………
8. The power delivered by a battery of emf, 10 V is 10 W. Then the current delivered by the battery is ……………………
Answer:

electrical energy
less
6 Ω
volt
Ampere
12 Ω
$\frac{11}{12} \Omega$
1 ampere

Multiple Choice Questions

1. A uniform wire of resistance 50 Ω. is cut into five equal parts. These parts are now connected in parallel. Then the equivalent resistance of the combination is
A) 2 Ω
B) 12 Ω
C) 250 Ω
D) 6250 Ω
Answer:
A) 2 Ω

2. A charge is moved from a point A to a point B. The work done to move unit charge during this process is called
A) potential at A
B) potential at B
C) potential difference between A and B
D) current from A to B
Answer:
C) potential difference between A and B

3. Joule/ coulomb is the same as
A) 1 – watt
B) 1 – volt
C) 1- ampere
D) 1 – ohm
Answer:
B) 1 – volt

4. The current in the wire depends
A) only on the potential difference applied
B) only on the resistance of the wire
C) on potential difference and resistance
D) none of them
Answer:
C) on potential difference and resistance

5. Consider the following statements.
a) In series connection, the same current flows through each element.
b) In parallel connection, the same potential difference gets applied across each element
A) both a and b are correct
B) a is correct but b is wrong
C) a is wrong but b is correct
D) both a and b are wrong
Answer:
A) both a and b are correct

10th Class Physics 11th Lesson Electric Current InText Questions and Answers

10th Class Physics Textbook Page No. 179

Question 1.
Does motion of charge always lead to electric current?
Answer:
Yes, it does.

Question 2.
Take a bulb, a battery, a switch and few insulated copper wires to the terminals of the battery through the bulb and switch. Now switch on the circuit and observe the bulb. What do you notice?
Answer:
The bulb glows.

10th Class Physics Textbook Page No. 180

Question 3.
Can you predict the reason for the bulb not glowing in situations 2 and 3?
Answer:
Yes, in situation 2 – there is no charge to travel in the circuit as the battery is disconnected. So, the bulb isn’t glowing.

In situation 3, we replaced the copper wires with nylon wires. Nylon is not a conductor. So, the bulb isn’t glowing.

Question 4.
Why do all materials not act as conductors?
Answer:
In conductors the gap between the atoms is very less. So, the transfer of energy is easy. But in other materials the gap is more. So, the transfer of energy is not possible.

Question 5.
How does a conductor transfer energy from source to bulb
Answer:

A source has chemical energy which transfers electrons to the conductor.
The conductor carries the electrons to the bulb when connected.
Thus, the conductor transfers energy from source to bulb.

Question 6.
What happens to the motion of electrons when the ends of the conductor are connected to the battery?
Answer:
The energy transfer takes place from battery to the bulb through conductor.

10th Class Physics Textbook Page No. 181

Question 7.
Why do electrons move in specified direction?
Answer:
The electrons move in specified direction when the ends of the conductpr are connected to the terminals of a battery.
A uniform electric field is set up throughout the conductor. This field makes the electrons move in a specified direction.

Question 8.
In which direction do the electrons move?
Answer:
In a direction opposite to the direction of the field.

Question 9.
Do the electrons accelerate continuously?
Answer:
No, they lose energy and are again accelerated by the electric field.

Question 10.
Do they move with a constant speed?
Answer:
Yes, they move with a constant average Speed.

Question 11.
Why does a bulb glow immediately when we switch on?
Answer:
When we switch on any electric circuit, irrespective of length of the conductor, an electric field is set up throughout the conductor instantaneously due to the voltage of the source connected to the circuit.

Question 12.
How can we decide the direction of electric current?
Answer:
By the signs of the charge and drift speed.

10th Class Physics Textbook Page No. 183

Question 13.
How can we measure electric current?
Answer:
An ammeter is used to measure electric current.

Question 14.
Where do the electrons get energy for their motion from?
Answer:
From an electric field set up throughout the conductor.

Question 15.
Can you find the work done by the electric force?
Answer:
Yes. With the help of the formula W = F_el, we can find the work done by the electric force.

Question 16.
What is the work done by the electric force on unit charge?
Answer:
Work done by the electric force on unit charge $\mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}=\frac{\mathrm{F}_{\mathrm{e}} l}{\mathrm{q}}$. It is called Potential difference.

10th Class Physics Textbook Page No. 184

Question 17.
What is the direction of electric current in terms of potential difference?
Answer:
Electrons move from low potential to high potential.

Question 18.
Do positive charges move in a conductor? Can you give an example of this?
Answer:
No, they don’t move. They are fixed in the lattice.
Eg : battery.

Question 19.
How does a battery maintain a constant potential difference between its terminals?
Answer:
We know that a battery consists electric force (Fe) and chemical force (Fc). These two forces are balanced in a battery. Due to this reason a battery maintains a constant , potential difference between its terminals.

Question 20.
Why does the battery discharge when its positive and negative terminals are connected through a conductor?
Answer:
A conductor permits the charges to pass through it. Due to this the exhaustion of charges is created after completion of all charges. So, when a battery is connected with a conductor it discharges.

10th Class Physics Textbook Page No. 185

Question 21.
What happens when the battery is connected in a circuit?
Answer:
A potential difference is created between the ends of the conductor, when the battery is connected in a circuit.

10th Class Physics Textbook Page No. 186

Question 22.
How can we measure potential difference or emf?
Answer:
With the help of a voltmeter, we measure potential difference or emf.

10th Class Physics Textbook Page No. 187

Question 23.
Is there any relation between emf of battery and drift speed of electrons in the conductor connected to a battery?
Answer:
Yes, when emf of a battery is more the drift speed of electrons will be more.

10th Class Physics Textbook Page No. 189

Question 24.
Can you guess the reason why the ratio of V and I in case of LED is not constant?
Answer:
This is due to forward voltage and maximum continuous current rating characters of LEDs.

Question 25.
Do all materials obey Ohm’s law?
Answer:
No, some materials don’t obey Ohm’s law.

Question 26.
Can we classify the materials based on Ohm’s law?
Answer:
Yes, the materials which obey Ohm’s law are conductors and others are same conductors or non-conductors.

Question 27.
What is resistance?
Answer:
The obstruction offered to the flow of electrons in a conductor is called the resistance.

Question 28.
Is the value of resistance the same for all materials?
Answer:
Yes, it varies.

Question 29.
Is there any application of Ohm’s law in daily life?
Answer:
Yes, this law is used in wiring.

Question 30.
What causes electric shock in the human body – current or voltage?
Answer:
Current with sufficient voltage.

10th Class Physics Textbook Page No. 190

Question 31.
Do you know the voltage of mains that we use in our household circuits?
Answer:
Yes, I know the voltage of mains that we use in our household circuits is 120 V.

Question 32.
What happens to our body if we touch live wire of 240 V?
Answer:
240 V current disturbs the functioning of organs inside the body. It is called electric shock. If the current flow continues further, it damages the tissues of the body which leads to decrease in resistance of the body. When this current flows for a longer time, damage to the tissues increases and thereby the resistance of human body decreases further. Hence, the current through the human body will increase. If this current reaches 0.07 A, it effects the functioning of the heart and if this much current passes through the heart for more than one second it could be fatal.

If this current flows for a longer time, the person in electric shock will be killed.

10th Class Physics Textbook Page No. 191

Question 33.
Why doesn’t a bird get a shock when it stands on a high voltage wire?
Answer:
There are two parallel lines carrying 240 V current. The voltage current will pass through the body if both the wires are touched at the same time. But, when the bird stands on only one wire, there is no potential difference between the legs. So, no current passes through the bird. Hence, it doesn’t feel any electric shock.

10th Class Physics Textbook Page No. 192

Question 34.
What could be the reason for increase in the resistance of the bulb when current flows through it?
Answer:
The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.

Question 35.
What happens to the resistance of a conductor if we increase its length?
Answer:
The resistance of a conductor increases with the increase of its length.

10th Class Physics Textbook Page No. 193

Question 36.
Does the thickness of a conductor influence its resistance?
Answer:
Yes, as the thickness of the conductor increases the resistance decreases.

10th Class Physics Textbook Page No. 195

Question 37.
How are electric devices connected in circuits?
Answer:
Electric devices are connected either in series or parallel in circuits.

Question 38.
When bulbs are connected (resistors) in series, what do you notice
Answer:
We notice that, the sum of the voltages of the bulbs (resistors) is equal to voltage across the combination of the resistors.

10th Class Physics Textbook Page No. 196

Question 39.
What do you notice when bulbs (resistors) are connected in series to the current?
Answer:
The current is not changing

Question 40.
What do you mean by equivalent resistance?
Answer:
If the current drawn by a resistor is equal to the current drawn by the combination of resistors, then the resistor is called equivalent resistor.

Question 41.
What happens when one of the resistors in series breaks down?
Answer:
The circuit becomes open and flow of current will be broken down.

Question 42.
Can you guess in what way household wiring has been done?
Answer:
Parallel connection.

10th Class Physics Textbook Page No. 197

Question 43.
How much current is drawn from the battery if the resistors are connected in parallel?
Is it equal to individual currents drawn by the resistors?
Answer:
It is the sum of currents flowing through each resistor. No, it is the sum of individual currents drawn by the resistors.

10th Class Physics Textbook Page No. 199

Question 44.
How could the sign convention be taken in a circuit?
Answer:
The potential difference across the resistor is taken as negative when we move along the direction of electric current through the resistor, and it is taken as positive when we move against the direction of electric current through the resistor.

10th Class Physics Textbook Page No. 201

Question 45.
You might have heard the sentences like “this month we have consumed 100 units of current”. What does ‘unit’ mean?
Answer:
Unit (or) kilo watt hour is the consumption of electric power in one hour by our electric appliances.

Question 46.
A bulb is marked 60 W and 120 V. What do these values indicate?
Answer:
It means, the resistance of the bulb is

Question 47.
What is the energy lost by the charge in 1 sec.?
Answer:
It is equal to $\frac{\mathrm{W}}{\mathrm{t}}$.

10th Class Physics Textbook Page No. 202

Question 48.
What do you mean by overload?
Answer:
When a high current flows through the wire which is beyond the rating of wire then heating of wire takes place. This phenomenon is called overloading.

Question 49.
Why does it (overloading) cause damage to electric appliances?
Answer:
Due to overload the heat increases in the circuit and this melts the parts of the appliances. Thus overload causdt damage to the electric appliances.

10th Class Physics Textbook Page No. 203

Question 50.
What happens when this current (overloading) increases greatly to the household circuit?
Answer:
It causes fire.

Question 51.
How can we prevent damage due to overloading?
Answer:
To prevent damages due to overloading we connect an electric fuse to the household circuit.

10th Class Physics Textbook Page No. 203

Question 52.
What do you mean by short circuit?
Answer:

The line wires that are entering the meter have a voltage of 240 V.
The minimum and maximum limit of current that can be drawn from the mains is 5 to 20 A.
Thus, the maximum current that we can draw from the mains is 20 A.
When the current drawn from the mains is more than 20 A, overheating occurs and may cause a fire. This is called overloading.
A short circuit is an electrical circuit that allows a current to travel along an unintended path often where essentially no electrical impedance is encountered.

Question 52.
Why does a short circuit damage electric wiring and devices connected to it?
Answer:
In a short circuit the current drawn from the main exceeds the maximum limit 20 A. This will lead to overloading which can damage the electrical appliances.

10th Class Physics 11th Lesson Electric Current Activities

Activity – 1

Question 1.
Write an activity to check when a bulb glows in a circuit.
(OR)
How do you prove a source of energy is required to glow a bulb in a circuit?
Answer:
Aim :
To check when a bulb glows in a circuit.

Materials required:

A bulb
a battery
a switch
few insulated copper wire

Procedure (1) :

Take a bulb, a battery, a switch and few insulated copper wires.
Connect the ends of the copper wires to the terminals of the battery through the bulb and switch.
Now switch on the circuit.
Observation (1) : The bulb glows.

Procedure (2) :

Remove the battery from the circuit and connect the remaining components to make a complete circuit.
Again switch on the circuit and observe the bulb.

Observation (2): The bulb does not glow.

Procedure (3) :
Replace the copper wires with nylon wires and connect the nylon wires to the terminals of the battery through a bulb and switch. Now switch on the circuit. We observe that the bulb does not glow. Because the wires are not conductors.

Observation (3) : The bulb does not glow.

Result:
The battery contains charges which glow the bulb.

Activity – 3

Question 2.
Write an activity to show that the values of current are different for different wires for a constant voltage.
(OR)
The resistance of a conductor depends on the material of the conductor. Prove this through an activity.
(OR)
List out the material required in the experiment to show that the electric resistance depends upon the nature of the material and write experimental procedure.
Answer:
Aim:
To show that the values of current are different for different wires for a constant voltage. Materials required : (wires of the same length and some cross-sectional area).

Copper rod
Nichrome rod
Battery
Ammeter
Key
Manganin Wire

Procedure :

Make a circuit as shown in figure.
Connect one of the wires between the ends P and Q.
Switch on the circuit. Measure the electric current for a fixed voltage, using the ammeter connected to the circuit. Note it in your notebook.
Repeat this experiment with other wires and note the current in your notebook.

Observation :
The values of current are different for different wires for a constant voltage.

Conclusion:
The resistance of a conductor depends on the material of the conductor.

Activity – 5

Question 3.
Write an activity to show that resistance is inversely proportional to the c section area of the conductor.
(OR)
What happens to resistance if the area of a cross-section of conductor is increased? Explain with an activity.
Aim :
To prove that resistance is inversely proportional to the cross-section area of the conductor.

Materials required :

A Battery
Mangnin Wires
Ammeter
Key
Manganin wires with different cross-section areas (lengths are same).

Procedure:

Make the circuit as given figure.
Connect one of the wires between points P and Q.
Switch on me circuit. Note the ammeter reading in your notebook.
Continue the experiment with different wires of same length but different cross-section areas. Note the ammeter readings in your notebook.

Observation :
As the cross-section area of the rods increases, the current increases.

Result (Conclusion) :
Resistance is inversely proportional to cross-section area of the conductor.

Activity – 6

Question 4.
Write an activity to prove that the sum of the potential differences of the bulb is equal to voltage across the combination of the resistors. (OR)
Prove that during series connection potential difference is distributed among the resistors experimentally.
Answer:
Aim:
To prove that the sum of the potential differences of the bulbs is equal to potential difference across the combination of the resistors.

Materials required :

Bulbs
Voltmeters
Insulated wires
Ammeter
Key

Procedure :

Take different bulbs. Using a multimeter measure their resistances. Note them as R,, R2 and Rv
Connect them as shown in figure.
Measure the voltage between terminals of the battery connected to the circuit.
Measure the voltages between the ends of each bulb and note them as Vj, V2 and V3 from voltmeters in your notebook.
Compare them.

Observation :
We notice that the’sum of the voltages of the bulbs is equal to voltages across the combination of the resistors.

Activity – 7

Question 5.
Write an activity to prove that the current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.
(OR)
Prove that during parallel connection the current is distributed among the resistances by using an experimental activity.
Answer:
Aim:
To prove that the current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.
Materials required :

Bulbs
Ammeters
Buttery
Key
Wires

Procedure :

Connect the bulbs in parallel connection as shown in the given circuit.
Measure the voltage across each bulb using a voltmeter or multimeter.
Note these values in your notebook.

Observation :

The voltage at the ends of each bulb is the same.
Measure electric currents flowing through each bulb using ammeters. Note these values.
Measure the current (I) drawn from the battery using the ammeter 1.

Result (Conclusion) :
The current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.

AP SSC 10th Class Chemistry Solutions Chapter 14 Carbon and its Compounds

March 29, 2023February 2, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 14 Carbon and its Compounds Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 14th Lesson Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds Textbook Questions and Answers

Improve Your Learning

Question 1.
Name the simplest hydrocarbon. (AS1)
Answer:
The simplest hydrocarbon is alkane called Methane (CH₄). It’s an aliphatic, saturated compound of Hydrogen and Carbon.

Question 2.
What are the general molecular formulae of alkanes, alkenes and alkynes? (AS1)
Answer:
General molecular formula of alkane is C_nH_2n+2.
General molecular formula of alkene is C_nH_2n.
General molecular formula of alkyne is C_nH_2n-2.

Question 3.
Name the carboxylic acid used as a preservative. (AS1)
Answer:
Vinegar with chemical formula CH₃COOH is used as preservative. 5 – 8% of solution of acetic acid or ethanoic acid in water is called vinegar and it is used widely as preservative in pickles.

Question 4.
Name the product other than water formed on burning of ethanol in air. (AS1)
Answer:
C₂H₃OH + 3O₂ → 2CO₂ + 3H₂O + Energy
So, the product other than water formed on burning of ethanol in air is carbon dioxide (CO₂).

Question 5.
Give the IUPAC name of the following compounds. If more than one compound is possible, name all of them. (AS1)
i) An aldehyde derived from ethane.
ii) A ketone derived from butane.
iii) A chloride derived from propane.
iv) An alcohol derived from pentane.
Answer:
i) An aldehyde derived from ethane is ethanal. Its formula is CH₃CHO.
ii) A ketone derived from butane. Its IUPAC name is Butanone.
Its chemical formula is CH₃COCH₂CH₃
It is also known as methyl ethyl ketone. (Its general name)

iii) A chloride derived from propane.
A) 1-Chloro propane. Its formula is CH₃CH₂CH₂Cl.

iv) An alcohol derived from pentane :
A) 1-Pentanol. Its formula is CH₃CH₂CH₂CH₂CH₂OH.
B) 2-Pentanol. Its formula is CH₃CHOH CH₂CH₂CH₃
C) 3-Pentanol. Its formula is CH₃CH₂ CHOH CH₂CH₃

Question 6.
A mixture of oxygen and ethyne is burnt for welding ; can you tell why a mixture of ethyne and air is not used? (AS1)
Answer:

Ethyne when burnt in the presence of oxygen gives enough heat that can be used for welding.
Whereas if it is burnt in air which contains nitrogen, CO₂ and other inactive gaseous contents, sufficient oxygen is not available for burning ethyne to give the required heat.

Question 7.
Explain with the help of a chemical equation, how an addition reaction is used in vegetable ghee industry. (AS1)
Answer:

The addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon is called hydrogenation. The process of hydrogenation takes place in the presence of nickel or palladium metals as catalyst.
The process of hydrogenation has an important industrial application. It is used to prepare vegetable ghee (or vanaspati ghee) from vegetable oils.
Vegetable oils are unsaturated fats having double bonds between some of their carbon atoms.
When a vegetable oil (like groundnut oil) is heated with hydrogen in the presence of finely divided nickel as catalyst, a saturated oil called vegetable ghee (or vanaspati ghee) is formed. This a reaction is called hydrogenation of oils and it can be represented as follows.

Here vegetable oil is a liquid whereas vegetable ghee is a solid (or a semi solid).

Question 8.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆O? (AS1)
b) Give the IUPAC names of the above possible compounds and represent them in structures. (AS1)
c) What is the similarity in these compounds? (AS1)
Answer:
a) They are CH₃COCH₃and CH₃ CH₂ CHO

b) i) The IUPAC name of CH₃COCH₃ is propanone.
ii) The IUPAC name of CH₃ CH₂ CHO is propanal.

Question 9.
Name the simplest ketone apfl write its molecular formula. (AS1)
Answer:
Acetone is the simplest ketone. Its molecular formula is CH₃COCH₃ Its IUPAC name is propanone.

Question 10.
What do we call the Self linking property of carbon? (AS1)
Answer:
The property of self combination (or linking) of carbon atoms to form long chains is useful to us because it gives rise to an extremely large number of carbon compounds (or organic compounds). This is known as catenation.

Question 11.
Name the compound formed by heating ethanol at 443 K with excess of cone. H₂SO₄. (AS1)
(OR)
What is the compound formed when ethyhalcohol (Ethanol) is dehydrated ? Write the chemical equation of the reaction.
Answer:
1. When ethanol is heated with excess of cone. H₂SO₄ at 443 K (170° C), it gets dehydrated to form ethene (which is an unsaturated hydrocarbon).

2. During dehydration of ethanol molecules (CH₃ – CH₂OH), H from the CH₃ group and OH from CH₂OH group are removed in the form of a water molecule (H₂O) regulating in the formation of this molecule (CH₂ = CH₂).
3. In this reaction concentrated sulphuric acid acts as a dehydrating agent.

Question 12.
Give an example for esterification reaction. (AS1)
Answer:
The reaction between carboxylic acid and an alcohol in the presence of cone. H₂SO₄ to form a sweet odoured substance, ester with the functional group

is called esterification.

Ex: Ethanoic acid (carboxylic acid) reacts with Ethanol (alcohol) and forms ethyl acetate.

Question 13.
Name the product obtained when ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate. (AS1)
(OR)
If the ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate, what is the product obtained from them?
Answer:
Ethanol (Ethyl alcohol) undergoes oxidation to form the product of Acetaldehyde and finally Acetic acid.

Question 14.
Write the chemical equation representing the reaction of preparation of ethanol from ethane. (AS1)
Answer:
1. Ethane in the absence of air on heating forms ethene

2. Then Ethanol is prepared on large scale from ethene by the addition of water vapour to it in the presence of catalyst like P₂O₅, Tungsten oxide at high pressure and temperature.

Question 15.
Write the IUPAC name of the next homologous of CH₃OHCH₂CH₃. (AS1)
Answer:
The IUPAC name of the next homologous of CH₃OHCH₂CH₃ is HO-CH₃CH₂CH₂CH₃ 1 – butanol.

Question 16.
Define homologous series of carbon compounds. Mention any two characteristics of homologous series. (AS1)
Answer:
1. The series of carbon compounds in which two successive compounds differ by – CH₂ unit is called homologous series.
Ex : 1) CH₄, C₂H₆, C₃H₈, ………………..
2) CH₃OH, C₂H₅OH, C₃H₇OH, ………………..

2. If we observe above series of compounds, we will notice that each compound in the series differs by – CH₂ unit by its successive compound.

3. Characteristics of homologous series :
i) They have one general formula.
Ex : alkanes (C_nH_2n+2), alkynes (C_nH_2n-2), alcohols (C_nH_2n+1) OH, etc.
ii) Successive compounds in the series possess a difference of (-CH₂) unit.
iii) They have similar chemical properties.

Question 17.
Give the names of functional groups
(i) – CHO
(ii) – C = O. (AS1)
(OR)
Write the names of the given functional groups
(i) – CHO
(ii) – C = O
Answer:
i) – CHO → aldehyde
ii) – C = O → ketone

Question 18.
Why does carbon form compounds mainly by covalent bonding? (AS1)
Answer:
Since carbon atoms can achieve the inert gas electron arrangements only by the sharings of electrons, therefore, carbon always forms covalent bonds.

Question 19.
Allotropy is a property shown by which class substance: elements, compounds or mixtures? Explain allotropy with suitable examples. (AS1)
Answer:

1. Allotropy is a property shown by the elements.

2. The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy.

3. The different forms of the element are called allotropes and are formed due to the difference in the arrangement of atoms.

4. Example for allotropes : Allotropes of carbon.

Allotropes of carbon are classified into two types. They are
1) Amorphous forms,
2) Crystalline forms.

5) Amorphous forms of carbon:
Coal, coke, wood, charcoal, animal charcoal, lampblack, gas carbon, petroleum coke, sugar charcoal.

6) Crystalline forms of carbon :
Diamond, graphite and buckminsterfullerene.

Question 20.
Explain how sodium ethoxide is obtained from ethanol. Give chemical equations. (AS1)
Answer:
As ethanol is similar to water molecule (H₂O) with C₂H₅ group in place of hydrogen, it reacts with metallic sodium to liberate hydrogen and form sodium ethoxide.

Question 21.
Describe with chemical equation how ethanoic acid may be obtained from ethanol. (AS1)
Answer:
Ethyl alcohol (Ethanol) undergoes oxidation to form the product Acetaldehyde and finally acetic acid (Ethanoic acid).

Question 22.
Explain the cleansing action of soap. (AS1)
Answer:
When a dirty cloth is put in water containing dissolved soap, the hydrocarbon ends of the soap molecules in the micelle attach to the oil or grease particles present on the surface of dirty clothes.

Question 24.
Explain the structure of graphite in terms of bonding and give one property based on this structure. (AS1)
(OR)
Why does graphite act as lubricant?
Answer:

Graphite forms a two dimensional layer structure with C – C bonds within the layers.
There are relatively weak interactions between the layers.
In the layer structure, the carbon atoms are in a trigonal planar environment.
This is consistent with each carbon atom in sp² hybridisation.
Interactions between the sp² orbitals (overlaps) lead to the formation of C – C bonds.
Each carbon atom is with one unhybridised ‘p’ orbital.
The unhybridised ‘p’ orbitals interact to form a π system that is delocalised over the whole layer.
The interactions known as London dispersion forces between the layers which are separated by a distance of 3.35 A° are weakened by the presence of water molecules so that it is easy to cleave graphite.
For this reason graphite is used as lubricant and as the lead in pencils.

Question 25.
Name the acid present in vinegar. (AS1)
Answer:
1) The acid present in vinegar is Ethenoic acid or acetic acid (CH₃COOH).
2) 5 – 8% solution of acetic acid in water is called vinegar.

Question 26.
What happens when a small piece of sodium is dropped into ethanol? (AS2)
Answer:
Ethanol reacts with sodium to liberate hydrogen and form sodium ethoxide.

Question 27.
Two carbon compounds A and B have molecular formula C₃H₈ and C₃H₆ respectively. Which one of the two is most likely to show addition? Justify your answer. (AS2)
Answer:

• It is a saturated hydrocarbon. It shows substitution reaction.

• This is an unsaturated hydrocarbon. Hence it shows addition to become saturated. During the reactions, addition of reagent takes place at the double bonded carbon atoms.

Justification :
In the following, C₃H₆ undergoes addition reaction.

Question 28.
Suggest a test to find the hardness of water and explain the procedure. (AS3)
(OR)
How do you test whether a given water sample is soft or hard?
Answer:

Take about 10 ml hard water (well water or hand pump water) in a test tube.
Add five drops of soap solution to it.
Shake the test tube vigorously.
We see that no lather is formed at first.
Only a dirty white curd like scum is formed on the surface of water.
From this, we conclude that soap does not form lather easily with hard water.
We have to add much more soap to obtain lather with hard water.

Question 29.
Suggest a chemical test to distinguish between ethanol and ethanoic acid and explain the procedure. (AS3)
Answer:

Take ethanol and ethanoic acid in two different test tubes.
Add nearly 18 g of sodium bicarbonate (NaHCO₃) to each test tube.
Lots and lots of bubbles and foam will be observed from the test tube containing ethanoic acid. This is due to release of CO₂.
NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
Ethanol will not react with sodium bicarbonate and thus we won’t observe any change in the test tube containing ethanol.
Thus we can separate ethanol from ethanoic acid.

Question 30.
An organic compound ‘X’ with a molecular formula C₂H₆O undergoes oxidation with alkaline KMnO₄ and forms the compound ‘Y’, that has molecular formula C₂H₄O₂. (AS3)
i) Identify ‘X’ and ‘Y’.
Answer:
X is Ethanol is CH₃CH₂OH and T is Ethanoic acid, i.e., CH₃COOH.

ii) Write your observation regarding the product when the compound X is made to react with compound IT which is used as a preservative for pickles.
Answer:
Ethyl alcohol undergoes oxidation to form the product Acetaldehyde and finally Acetic acid.

Here CH₃COOH is used as preservative for pickles.

When X reacts with Y it forms ethyl acetate and water which is called esterification reaction.
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 31.
Prepare models of methane, ethane, ethene and ethyne molecules using clay balls and matchsticks. (AS4)
Answer:
Stick and ball model :
1) Methane (CH₄) :

2) Ethane (C2H₆):

3) Ethene (C₂H₄):

4) Ethyne (C₂H₂)

Question 32.
Collect information about artificial ripening of fruits by ethylene. (AS4)
Answer:

Seasonal fruits like mango, banana, papaya, sapota and custard apple are often harvested in nature. But due to unripe condition they are subsequently allowed to ripen by natural release of ripening harmone (ethylene) from the fruit.
However, natural ripening in some fruits is a slow process, which leads to high weight loss, desiccation of fruits and under ripening. With the rapid development of fruit trade, artificial ripening has become essential and the methods practised earlier by small traders are smoking and calcium carbide treatment.
Fruits ripened with calcium carbide though seem attractive and colourful are inferior in taste, flavour and spoil faster.
Government of India has banned the use of calcium carbide for artificial ripening of fruits under PFA Act 8-44AA, 1954.
Artificial ripening of fruits by using the above steps spoils the health of consumers, so we should not use such type of fruits.
Government has to take serious action on the fruit sellers who are practising the above said methods.

Question 33.
Draw the electronic dot structure of ethane molecule (C₂H₆). (AS6)
Answer:
C₂H₆:

Question 34.
How do you appreciate the role of esters in everyday life? (AS6)
Answer:

Esters are usually volatile liquids having sweet or pleasant smell.
They are also said to have fruity smell.
Esters are used in making artificial perfumes.
This is because of the fact that most of the esters have a pleasant smell.
Esters are also used as flavouring agents.
This means that esters are used in making artificial flavours and essences used in ice-cream, sweets and cool drinks.
The alkaline hydrolysis of esters is known as saponification (Soap making).
That’s why we can appreciate the role of esters in everyday life.

Question 35.
How do you condemn the use of alcohol as a social practice? (AS7)
Answer:

Consumption of alcohol in the form of beverages is harmful to health.
It causes severe damage to blood circulation system.
Addiction to alcohol drinking leads to heart diseases and damages the liver.
It also causes ulcers in small intestines due to increased acidity and damages the digestive system.
Alcohol which is consumed in raw form under the names liquor, gudumba which is more harmful to health due to adulteration.
Alcohol mixed with pyridine is called denatured spirit. Consumption of denatured spirit causes blindness and death.
Hence use of alcohol is a social evil which harms the society.

Question 36.
An organic compound with molecular formula C₂H₄O₂ produces brisk effervescence on addition of sodium carbonate/bicarbonate.
Answer the following :
a) Identify the organic compound. (AS1)
Answer:
The organic compound is Ethanoic acid (CH₃COOH).

b) Write the chemical equation for the above reaction. (AS1)
Answer:

c) Name the gas evolved. (AS2)
Answer:
CO₂

d) How will you test the gas evolved? (AS3)
Answer:
1) Pass the evolved gas through lime water in a test tube.
2) We will find that lime water turns milky.
3) Only CO₂ gas can turn lime water milky.

e) List two important uses of the above compound. (AS1)
Answer:
1) Dilute ethanoic acid (CH₃COOH) is used as a food preservative in the preparation of pickles and sauces.
2) Ethanoic acid is used for making cellulose acetate which is an important artificial fibre.

Question 37.
1 ml glacial acetic acid and 1 m/of ethanol are mixed together in, a test tube. Few drops of concentrate sulphuric acid is added in the mixture are warmed in a water bath for 5 min.
Answer the following:
a) Name the resultant compound formed.
b) Represent the above change by a chemical equation.
c) What term is given to such a reaction?
d) What are the special characteristics of the compound formed?
Answer:
a) Ethyl acetate.

c) Esterification
d) It has fruity smell or pleasant smell.

Fill In The Blanks

1. Carbon compounds containing double and triple bonds are called ………………….
2. A compound which is basic constituent of many cough syrups ………………………
3. Very dilute solution of ethanoic acid is ………………..
4. A sweet odour substance formed by the reaction of an alcohol and a carboxylic acid is ………………
5. When sodium metal is dropped in ethanol …………………. gas will be released.
6. The functional group present in methanol is …………………….
7. IUPAC name of alkene containing 3 carbon atoms is ………………….
8. The first member of homologous series among alkynes is ……………………
9. The product that is formed by dehydration of ethanol in cone, sulphuric acid is ………………….
10. Number of single covalent bonds in ammonia are ………………..
11. Type of reactions shown by alkanes is ……………….
Answer:

unsaturated compounds
ethanol
vinegar
ester
H₂
– OH (Alcohol)
propene
ethyne (C₂H₂)
ethene (C₂H₄)
3
substitutional

Multiple Choice Questions

1. Which of the four test tubes containing the following chemicals shows the brisk effervescence when dilute acetic acid was added to them?
i) KOH
ii) NaHCO₃
iii) K₂CO₃
iv) NaCl
A) i & ii
B) ii & iii
C) i & iv
D) ii & iv
Answer:
B) ii & iii

2. Which of the following solution of acetic acid in water can be used as preservative?
A) 5-10%
B) 10-15%
C) 15-20%
D) 100%
Answer:
A) 5-10%

3. The suffix used for naming an aldehyde is
A) – ol
B) – al
C) – one
D) – ene
Answer:
B) – al

4. Acetic acid, when dissolved in water, it dissociates into ions reversibly because it is a
A) weak acid
B) strong acid
C) weak base
D) strong base
Answer:
A) weak acid

5. Which one of the following hydrocarbons can show isomerism?
A) C₂H₄
B) C₂H₆
C) C₃H₈
D) C₄H₁₀
Answer:
D) C₄H₁₀

6. Combustion of hydrocarbon is generally accompanied by the evolution of
A) Heat
B) Light
C) Both heat and light
D) Electric current
Answer:
C) Both heat and light

7. 2 ml of ethanoic acid was taken in each of the three test tubes A, B and C and 2 ml, 4 ml and 8 ml water was added to them respectively. A clear solution is obtained in:
A) Test tube A only
B) Test tubes A & B only
C) Test tubes B and C only
D) All the test tubes
Answer:
D) All the test tubes

8. If 2 ml of acetic acid was added slowly in drops to 5 ml of water then we will notice
A) The acid forms a separate layer on the top of water
B) Water forms a separate layer on the top of the acid
C) Formation of a clear and homogenous solution
D) Formation of a pink and clear solution
Answer:
C) Formation of a clear and homogenous solution

9. A few drops of ethanoic acid were added to solid sodium carbonate. The possible results of the reactions are
A) A hissing sound was evolved
B) Brown fumes evolved
C) Brisk effervescence occurred
D) A pungent smelling gas evolved
Answer:
C) Brisk effervescence occurred

10. When acetic acid reacts with ethyl alcohol, we add cone. H₂SO₄, it acts as and the process is called
A) Oxidizing agent, saponification
B) Dehydrating agent, esterification
C) Reducing agent, esterification
D) Acid and esterification
Answer:
B) Dehydrating agent, esterification

10th Class Chemistry 14th Lesson Carbon and its Compounds InText Questions and Answers

10th Class Chemistry Textbook Page No. 254

Question 1.
Can carbon get helium configuration by losing four electrons from the outer shell?
Answer:

If carbon loses four electrons from the outer shell, it has to form C⁴⁺ ions.
This requires huge amount of energy which is not available normally.
Therefore C⁴⁺ formation is also a remote possibility.
Carbon has to satisfy its tetravalency by sharing electrons with other atoms.
It has to form four covalent bonds either with its own atoms or atoms of other elements.

10th Class Chemistry Textbook Page No. 255

Question 2.
How do carbon atoms form bonds in so many different ways?
Answer:
As per valence bond theory, the four unpaired electrons in a carbon atom is main cause to form many bonds.

Question 3.
Explain the four unpaired electrons in carbon atom through excited state.
Answer:
Electronic configuration of carbon (ground state):

Electronic configuration of carbon (excited state):

10th Class Chemistry Textbook Page No. 256

Question 4.
Where does this energy to excite electron come from?
Answer:

We have to understand that free carbon atom would not be in excited state under normal conditions.
When the carbon atom is ready to form bonds with other atoms, the energy required for excitation is taken up from bond energies, which are the liberated energies when bonds are formed between carbon atom and other atoms.

Question 5.
In methane (CH₄) molecule all four carbon – hydrogen bonds are identical and bond angle HCH is 109°28′. How can we explain this?
Answer:
In excited state, carbon atom has three unpaired electrons in p-orbitals and one electron in s-orbital. These four valence electrons are with different energies. These orbitals combine to form four identical orbitals. Four hydrogen atoms form four identical C -H bonds with bond angle 109° 28′. This is called hybridisation.

Question 6.
How do these energetically unequal valence electrons form four equivalent covalent bonds in methane molecule?
Answer:
1) When bonds are formed, energy is released and the system becomes more stable. If carbon forms four bonds rather than two, still more energy is released and so the resulting molecule becomes even more stable.

2) The energy difference between the 2s and 2p orbitals is very small. When carbon atom is ready to form bonds it gets a small amount of energy from bond energies and gets excited to promote an electron from the 2s to the empty 2p to give four unpaired electrons.

3) We have got four unpaired electrons ready for bonding, but these electrons are in two different kinds of orbitals and their energies are different.

4) We are not going to get four identical bonds unless these unpaired electrons are in four identical orbitals.

10th Class Chemistry Textbook Page No. 257

Question 7.
How to explain the four orbitals of carbon containing unpaired electrons as energetically equal?
Answer:
With hybridisation we explai n the four orbitals of carbon containing unpaired electrons are energetically equal.
Ex : Methane (CH₄).

10th Class Chemistry Textbook Page No. 258

Question 8.
How do you explain the ability of C – atom to form two single covalent bonds and one double bond?
Answer:
Ethylene (CH₂ = CH₂) explains the ability of carbon atom to form two single covalent bonds and one double bond.
Ex:

10th Class Chemistry Textbook Page No. 259

Question 9.
How do you explain the ability of carbon atom to form one single bond and one triple bond?
Answer:
Ethyne (HC $\equiv$ CH) explains the ability of carbon atom to form one single bond between one hydrogen and carbon, and one triple bond between carbon and carbon.
Ex : H – C $\equiv$ C – H.

10th Class Chemistry Textbook Page No. 260

Question 10.
What are bond angles H$\widehat{\mathbf{C}}$H in CH₄, C₂H₄ and C₂H₂ molecules?
Answer:

10th Class Chemistry Textbook Page No. 262

Question 11.
How do you understand the markings (writings) of a pencil on a paper?
Answer:

When we write with a pencil, the inter layer attractions breakdown and leave graphite layers on the paper.
It is easy to remove pencil marks from paper with an eraser because, the layers do not bind strongly to the paper.

10th Class Chemistry Textbook Page No. 265

Question 12.
Allotting completely one special branch in chemistry to compounds of only one element. Is it justified when there are so many elements and their compounds but not with any special branches?
Answer:

We understand that all molecules that make life possible carbohydrates, proteins, nucleic acids, lipids, hormones, and vitamins contain carbon.
The chemical reactions that take place in living systems are of carbon compounds.
Food that we get From nature, various medicines, cotton, silk and fuels like natural gas and petroleum almost all of them are carbon compounds.
Synthetic fabrics, plastics, synthetic rubber are also compounds of carbon.
Hence, carbon is a special element with the largest number of compounds:

10th Class Chemistry Textbook Page No. 266

Question 13.
What are hydrocarbons?
Answer:
The compounds containing only carbon and hydrogen in their molecules are called hydrocarbons.

Question 14.
Do all the compounds have equal number of C and H atoms?
Answer:
No. All the compounds do not have equal number of C and H atoms.

10th Class Chemistry Textbook Page No. 269

Question 15.
Observe the following two structures.
a) CH₃ – CH₂ – CH₂ – CH₃
b)

i) How about their structures? Are they same?
Answer:
No, they are not same compounds.

ii) How many carbon and hydrogen atoms are there in (a) and (b) structures?
Answer:
Carbon – 4 ; Hydrogen – 10.

iii) Write the condensed molecular formulae for (a) and (b), do they have same molecular formulae?
Answer:
C₄H₁₀; Yes.

Question 16.
Can carbon form bonds with the atoms of other elements?
Answer:
Carbon forms compounds not only with atoms of hydrogen but also with atoms of other elements like oxygen, nitrogen, sulphur, phosphorus, halogens, etc.

10th Class Chemistry Textbook Page No. 272

Question 17.
What do you mean by nomenclature of Organic componds?
Answer:
Nomenclature of organic chemistry is systematic method of naming organic compound.

Question 18.
What is the basis for nomenclature?
Answer:
The basic of the nomenclature is number of carbons in the parent chain in a compound.

10th Class Chemistry Textbook Page No. 273

Question 19.
What are the word – root and suffix?
Answer:
Word root:
Word root indicates the number of carbon atoms in the longest possible continuous carbon chain also known as parent chain.

Suffix :
Suffix is added immediately after the word root. It is two types

1) Primary Suffix :
It is used to indicate the degree of saturation or unsaturation of the main chain.

2) Secondary Suffix :
It is used to indicate the main functional group in the organic compound.

10th Class Chemistry Textbook Page No. 274

Question 20.
What do you mean by the term ‘alkyl’?
Answer:
Alkyl:
Alkyl is a substituent, that is attached to the molecular fragment.
General formula of alkyl is C_nH_{2n + 1}

10th Class Chemistry Textbook Page No. 278

Question 21.
Can we write the structure of a compound if the name of the compound is given?
Answer:
Yes, we can write the structure of a compound if the name of the compound is given.

10th Class Chemistry Textbook Page No. 279

Question 22.
Why do sometimes cooking vessels get blackened on a gas or kerosene stove?
Answer:
Because of the inlets of air getting closed, the fuel gases do not completely undergo combustion. Hence, it forms a sooty carbon form which gets coated over the vessels.

10th Class Chemistry Textbook Page No. 280

Question 23.
Do you know what is a catalyst?
Answer:
A catalyst is a substance which regulates the rate of a given reaction without itself finally undergoing any chemical change.

10th Class Chemistry Textbook Page No. 281

Question 24.
Do you know how the police detect whether suspected drivers have consumed alcohol or not?
Answer:

The police officer asks the suspect to blow air into a plastic bag through a mouth piece of the detecting instrument which contains crystals of potassium-di-chromate (K₂Cr₂O₇).
As K₂Cr₂O₇ is a good oxidizing agent, it oxidizes any ethanol in the driver’s breath to ethanal and ethanoic acid.
Orange Cr₂O₇^2- changes to bluish green Cr³⁺ during the process of the oxidation of alcohol.
The length of the tube that turned into green is the measure of the quantity of alcohol that had been drunk.
The police even use the IR Spectra to detect the bonds C – OH and C – H of CH₃ – CH₂OH.

10th Class Chemistry Textbook Page No. 283

Question 25.
What are esters?
Answer:

The compounds which contain the functional group and the general formula R – COO – R’, where R and R’ are alkyl groups or phenyl groups, are known as “Esters”.

10th Class Chemistry Textbook Page No. 284

Question 26.
What is a true solution?
Answer:
A true solution is that in which the solute particles dispersed in the solvent are less than 1 nm in diameter.

10th Class Chemistry Textbook Page No. 286

Question 27.
What is the action of soap particles on the greasy cloth?
Answer:

Soaps and detergents make oil and dirt present on the cloth come out into water, thereby making the cloth clean.
Soap has one polar end and one non-polar end.
The polar end is hydrophilic in nature and this end is attracted towards water.
The non-polar end is hydrophobic in nature and it is attracted towards grease or * . ; oil on the cloth, but not attracted towards water.
When soap is dissolved in water, its hydrophobic ends attach themselves to dirt and remove it from the cloth.
The hydrophobic end of the soap molecules move towards the dirt or grease particles. ’
The hydrophobic ends attach to the dirt particle and try to pull out.
The molecules of soap surround the dirt particle at the centre of the cluster and form a spherical structure called micelle.
These micelles remain suspended in water like particles in a colloidal solution.
The various micelles present in water do not come together to form a precipitate as each micelle repels the other because of the ion-ion repulsion.
Thus, the dust particles remain trapped in micelles and are easily rinsed away with water.
Hence, soap micelles remove dirt by dissolving it in water.

10th Class Chemistry Textbook Page No. 280

Question 28.
Why we are advised not to use animal fats for cooking?
Answer:

Animal fats have recently been implicated as the cause of heart disease and obesity. So, we are advised not to use animal fats for cooking.
Excess animal fat is stored in lipocytes, which expand in size until the fat is used for fuel.

Question 29.
Which oil is recommended for cooking? Why?
Answer:
Canola oil :

A recent entrant into the Indian market Canola is flying off the shelves.
Canola oil which is made from the crushed seeds of the Canola plant, is said to be amongst the healthiest of cooking oils.
It has the lowest saturated fat content of any oil.
It’s seen as a healthy alternative as it’s rich in monosaturated fats and is high in omega-3 and omega a fats.
It has a medium smoking point and is an oil that works well for fruits, baking, sauteing, etc.

10th Class Chemistry 14th Lesson Carbon and its Compounds Activities

Activity – 1

Question 1.
Observe the structural formula of the following hydro carbons and write their names in your notebook.
Answer:
1) CH₃ – CH₂ – CH = CH₂
Sol. But-l-ene

Sol. 2-Methyl butane

3) CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – CH₃
Sol. Hexane

Sol. 3-Methyl, but-l-ene

5)

Sol. Prop-l-yne

Activity-2

Question 2.
Read the names of the following hydro carbons and draw their structures in your notebook.
1. 2,2-Dimethyl hexane
Sol.

2. But-l-yne
Sol. CH₃ – CH₂ – C = CH

3. 3-Methyl Pent-2-ene
Sol.

4. But-1.2-diene
Sol. CH₃ – CH₃ = c = CH₂

5. Hept-2 en, 4-yne
Sol.

Activity – 3

Question 3.
Write an activity to show esterification reactions.
Answer:
The compound formed is ester. The process is called esterification.

Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
Warm it in a water bath or a beaker containing water for at least five minutes.
Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
We will notice that the resulting mixture is sweet odoured subatance.
This substance is nothing but ethyl acetate, an ester.
This reaction is called esterification reaction.

Activity – 4

Question 4.
Write an activity to show soap solution separates oil from water.
Answer:

Take about 10 ml of water each in two test tubes.
Add a drop of oil to both the test tubes.
Label them as A and B.
Add a few drops of soap solution to test tube B.
Now shake both the test tubes vigorously for the same period of time.
We can see the oil and water layers separately in both the test tubes immediately after we stop shaking them.
Leave the test tubes undisturbed for sometime and observe.
The oil layer separates out first in which test tube we added drops of soap solution.

AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding

March 29, 2023February 2, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 10th Lesson Chemical Bonding

10th Class Chemistry 10th Lesson Chemical Bonding Textbook Questions and Answers

Improve Your Learning

Question 1.
List the factors that determine the type of bond that will be formed between two atoms. (AS1)
(OR)
How can you identify the type of bond formation between two atoms?
Answer:

The strength of attraction or repulsion between atoms.
Electrons in valence shell (valence electrons).

Question 2.
Explain the difference between the valence electrons and the covalency of an element. (AS1)
(OR)
How are valence electrons different from the covalency of element? Explain with examples.
Answer:
Valence electrons :

Number of electrons in the outermost orbit or an atom is called its valence electrons.
Ex: Na (Z = 11). It has 2e^– in I orbit, 8e^– in II orbit and 1e^– in III orbit.
So number of valence electrons in Na atom are ‘l’.

Covalency of an element:

Number of valance electrons which are taking part in covalent bond is called covalency.
The electron configuration of Boran is 1s² 2s² 2p¹.
It has three valance electrons.
So its covalency is 3.

Question 3.
A chemical compound has the following Lewis notation : (AS1)
a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the elements X and Y.

Answer:
a) 6
b) 2
c) 1
d) two
e) X – is hydrogen and Y – is oxygen. The formed molecule is H₂O.

Question 4.
Why do only valence electrons involve in bond formation? Why not electron of inner shells? Explain. (AS1)
(OR)
Which shell electrons involve in bond formation? Explain. What is the reason behind it?
Answer:

The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
But the electrons in the outermost shell (valence shell) of atoms get affected.
The inner shell electrons are strongly attracted by the nucleus when compared to the valence electrons.
So electrons in valence shell (valence electrons) are responsible for the formation of bond between atoms.

Question 5.
Explain the formation of sodium chloride and calcium oxide on the basis of the concept of electron transfer from one atom to another atom. (AS1)
(OR)
Explain the formation of any two compounds according to Kossel’s theory.
Answer:
I. Formation of sodium chloride (NaCl) :
1) Sodium chloride is formed from the elements sodium (Na) and chlorine (Cl).

2) Cation formation:
i) When sodium (Na) atom loses one electron to get octet electron configuration, it forms a cation (Na⁺).
ii) Now Na⁺ gets electron configuration that of Neon (Ne) atom.

3) Anion Formation :
i) Chlorine has shortage of one electron to get octet in its valence shell.
ii) So it gains the electron that was lost by Na to form anion and gets electron configuration of Argon (Ar).

4) Formation of NaCl :
i) Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–’ ions.
ii) These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–(g) → Na⁺Cl^–_(s) or NaCl

II. Formation of calcium oxide (CaO) :
1. Calcium (Ca) reacts with oxygen (0) to form an ionic compound calcium oxide (CaO).

2. Atomic number of Calcium is 20. Its electronic configuration is 2, 8, 8, 2.
3.

4. Atomic number of Oxygen is 8. Its electronic configuration is 2, 6.

6. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound calcium oxide (CaO).
Ca²⁺ + O^2- → Ca²⁺O^2- (or) CaO

Question 6.
A, B and C are three elements with atomic number 6, 11 and 17 respectively.
i) Which of these cannot form ionic bond? Why? (AS1)
ii) Which of these cannot form covalent bond? Why? (AS1)
iii) Which of these can form ionic as well as covalent bonds? (AS1)
Answer:
i) ‘A’ cannot form ionic bond. Its valence electrons are 4. It is difficult to lose or gain 4e^– to get octet configuration. So it forms covalent bond [Z of A is 6 so it is carbon (C)].

ii) ‘B’ cannot form covalent bond. Its valence electrons are 1 only. So it is easy to donate for other atom and become an ion. So it can form ionic bond [Z of B is 11, so it is sodium (Na)].

iii) Element C can form ionic as well as covalent bonds. Atomic number of Cl is 17. It is able to participate with Na in ionic bond and with hydrogen in HCl molecule as covalent bond.

Question 7.
How do bond energies and bond lengths of molecule help us in predicting their chemical properties? Explain with examples. (AS1)
(OR)
How can you explain with examples that bond energies and bond lengths are used to recognise chemical properties?
Answer:
1. Bond length :
Bond length or bond distance is the equilibrium distance between the nuclei of two atoms which form a covalent bond.

2. Bond energy :
Bond energy or bond dissociation energy is the energy needed to break a covalent bond between two atoms of a diatomic covalent compound in its gaseous state.

3. If the nature of the bond between the same two atoms changes the bond length also changes. For example, the bond lengths between two carbon atoms are C – C > C = C > C = C.

4. Thus the various bond lengths between the two carbon atoms are in ethane 1.54 Å, ethylene 1.34 Å, acetylene 1.20 Å.

5. The bond lengths between two oxygen atoms are in H₂O₂ (O – O) is 1.48 Å and in O₂ (O = O) is 1.21 Å.
6. Observe the table.

7. When bond length decreases, then bond dissociation energy increases.

8. When bond length increases, then bond dissociation energy decreases.

9. Bond length of H – H in H₂ molecule is 0.74 Å and its bond dissociation energy is 436 KJ/mol, whereas bond length of F – F in F₂ molecule is 1.44 Å and its bond dissociation energy is 159 KJ/mol.

10. Melting and boiling points of substances also can be determined by this bond energies and bond lengths.

Question 8.
Predict the reasons for low melting point for covalent compounds when compared with ionic compounds. (AS2)
(OR)
“Covalent compounds have low melting point.” What Is the reason for this statement? Explain.
Answer:
They are covalent compounds.

The melting point is low due to the weak Vander Waal’s forces of attractions between the covalent molecules.
The force of attraction between the molecules of a covalent compound is very weak.
Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compounds have low melting points and low boiling points.
Please note that some of the covalent solids like diamond and graphite have, however very high melting points and boiling points.

Question 9.
Collect the information about properties and uses of covalent compounds and prepare a report. (AS4)
(OR)
Generally these compounds are non-polar in nature. What are those compounds? Explain their properties and uses.
(OR)
Write any two uses and two properties of covalent compounds.
Answer:
The compounds are covalent.
Properties of covalent compounds :

Covalent compounds are usually liquids or gases, only some of them are solids.
The covalent compounds are usually liquids or gases due to the weak force of attraction between their molecules.
Covalent compounds have usually low melting and low boiling points.
Covalent compounds are usually in soluble in water but they are soluble in organic solvents.
Covalent compounds do not conduct electricity.

Uses of covalent compounds :

Covalent compounds form 99% of our body.
Water is a covalent compound. We know its many uses.
Sugars, food substances, tea and coffee are all covalent compounds.
Air we breathe in contains covalent molecules of oxygen and nitrogen.
Almost everything on earth other than most simple in organic salts are covalent.

Question 10.
Draw simple diagrams to show how electrons are arranged in the following covalent molecules : (AS5)
a) Calcium oxide (CaO)
b) Water (H₂O)
c) Chlorine (Cl₂)
Answer:
a) Calcium oxide (CaO) :

b) Water (H₂O):

The formation of water molecule can be shown like this also

c) Chlorine (Cl₂):

We can explain the formation of Cl₂ molecule in this way also.

Question 11.
Represent the molecule H₂O using Lewis notation. (AS5)
(OR)
How can you explain the formation of H₂O molecule using dot structure?
Answer:
One atom of oxygen shares its two electrons with two hydrogen atoms to form a water molecule.

Question 12.
Represent each of the following atoms using Lewis notation : (AS5)
a) Beryllium
b) Calcium
c) Lithium
Answer:

Question 13.
Represent each of the following molecules using Lewis notation : (AS5)
a) Bromine gas (Br2)
b) Calcium chloride (CaCl₂)
c) Carbon dioxide (CO₂)
d) Which of the three molecules listed above contains a double bond?
Answer:
a) Bromine gas (Br₂) :

b) Calcium chloride (CaCl₂)

c) Carbon dioxide (CO₂) :

d) CO₂, contains double bond in above list. Its structure is like this : O = C = O.

Question 14.
Two chemical reactions are described below. (AS5)
♦ Nitrogen and hydrogen react to form ammonia (NH₃).
♦ Carbon and hydrogen bond to form a molecule of methane (CH₄).
For each reaction give :
a) The valency of each of the atoms involved in the reaction.
b) The Lewis structure of the product that is formed.
Answer:
a) ♦ Nitrogen and hydrogen react to form ammonia (NH₃):
i) The valency of nitrogen is 3 and hydrogen is 1.
ii) The chemical formula of the product is NH₃

♦ Carbon and hydrogen bond to form a molecule of methane (CH₄):
i) The valency of carbon is 4 and hydrogen is 1.
ii) The chemical formula of the product is CH₄.

b) ♦ The Lewis structure of the product that is formed (: NH₃)

♦ The Lewis structure of the product that is formed (CH₄)

Question 15.
How does Lewis dot structure help in understanding bond formation between atoms? (AS6)
(OR)
What is the use of Lewis dot structure in bond formation? Explain.
Answer:

Only the outermost electrons of an atom take part in chemical bonding.
They are known as valence electrons.
The valence electrons in an atom are represented by putting dots (•) on the symbol of the element, one dot for each valence electron.
For example, sodium atom has 1 valence electron in its outermost shell, so we put 1 dot with the symbol of sodium and write Na• for it.
Sodium atom loses this 1 electron to form sodium ion.
By knowing the valence electrons of two different atoms by Lewis dot structure, we can understand which type of bond is going to establish between them and forms corresponding molecule.

Question 16.
What is octet rule? How do you appreciate role of the ‘octet rule’ in explaining the chemical properties of elements? (AS6)
(OR)
Which rule decides whether given element is chemically stable or not? Appreciate that rule.
Answer:
Octet rule decides whether given element is stable or not.
Octet rule :

‘The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer shell electrons.”
It was found that the elements which participate in chemical reaction get octet (or) ns2 np6 configuration similar to that of noble gas elements.

Role of octet in chemical properties of elements :

Except He remaining inert gas elements have 8 electrons in their outermost orbit. Since these elements are having stable octet configuration in their outermost orbit, they are very stable.
They do not allow the outermost electrons to take part in chemical reactions.
So by having octet configuration for these elements we can conclude these are chemically inertial.
If any group of elements (take halogens) which contain 7 electrons in their outermost orbit, they require only 1 e^– to get octet configuration.
So they try to participate in chemical reaction to get that 1 difference electron for octet configuration.
Similarly, Na contains 2, 8, 1 as its electronic configuration.
So it loses le^– from its outermost shell; it should have 8e^– in its outer shell and get the octet configuration.
Thus the octet rule helps in explaining the chemical properties of elements.

Question 17.
Explain the formation of the following molecules using valence bond theory.
a) N₂ molecule
b) O₂ molecule
(OR)
Write the formation of double bond and triple bond according to valence bond theory.
(OR)
Who proposed Valence Bond Theory? Explain the formation of N₂ molecule by using this theory.
Answer:
Linus Pauling was proposed valence bond theory.
Formation of N₂ molecule :

Electronic configuration of Nitrogen is 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹.
Suppose that p_x orbital of one Nitrogen atom overlaps the p_x orbital of other ‘N’ atom giving σ p_x – p_x bond along the inter nuclear axis.
The p_y and p_z orbitals of one ‘N’ atom overlaps with the p_y and p_z orbital of other ‘N’ atom laterally giving π p_y – p_y and π p_z – p_z bonds.
Therefore, N₂ molecule has a triple bond between two Nitrogen atoms.

Formation of O₂ molecule :

Electronic configuration of ‘O’ is 1s² 2s² 2p_x² 2p_y¹ 2p_z¹.
If the P_y orbital of one ‘O’ atom overlaps the p_y orbital of other ‘O’ atom along inter- nuclear axis, a σ p_y – p_y bond is formed.
p_z orbital of oxygen atom overlaps laterally, perpendicular to inter nuclear axis giving a π p_y – p_z bond.
So O₂ molecule has a double bond between the two oxygen atoms.

Question 18.
What is hybridisation? Explain the formation of the following molecules using hybridisation.
a) BeCl₂
b) BF₃
Explain the formation of sp and sp² hybridisation using examples.
(OR)
What is the name given to inter mixing of atomic orbitals to form new orbitals. Explain the formation of following molecules by using that process,
a) BeCl₂
b) BF₃
Answer:
This process is called hybridisation.
Hybridisation :
It is a phenomenon of inter mixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

a) Formation of BeCl₂ (Beryllium chloride) molecule :

₄Be has electronic configuration 1s² 2s².
It has no unpaired electrons.
It is expected not to form covalent bonds, but informs two covalent bonds one each with two chlorine atoms. „
To explain this, an excited state is suggested for Beryllium in which an electron from ‘2s’ shifts to 2p_x level.
Electronic configuration of ₄Be is 1s² 2s¹ 2p_x¹].
Electronic configuration of ₁₇Cl is 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹.
If Be forms two covalent bonds with two chlorine atoms, one bond should be σ 2s-3p due to the overlap of ‘2s’ orbital of Be, the ‘3p_z‘ orbital of one chlorine atom.
The other bond should be σ 2p-3p due to the overlap of ‘2p_x’ orbital of Be atom the 3p orbital of the other chlorine atom.
As the orbitals overlapping are different, the bond strengths of two Be-Cl must be different.
But, both bonds are of same strength and Cl$\hat{\mathrm{Be}}$ Cl is 180°.

The Hybridisation of BeCl₂ can be explained in this way also :
a) Be atom in its excited state allows its 2s orbital and 2p_x orbital which contain unpaired electrons to intermix and redistribute to two identical orbitals.
b) As per Hund’s rule each orbital gets one electron.
c) The new orbitals based on the types of orbitals that have undergone hybridisation are called sp orbitals.
d) The two sp orbitals of Be get separated by 180°.
e) Now each chlorine atom comes with its 3p_z¹ orbital and overlaps it the sp orbitals of Be forming two identical Be-Cl bonds (σ sp-p bonds).
Cl$\hat{\mathrm{Be}}$ Cl = 180°.

f) Both the bonds are of same strength.

b) Formation of BF₃ molecule :

₅B has electronic configuration 1s² 2s² 2pxh
The excited electronic configuration of ₅B is 1s² 2s¹ 2p_x¹ 2p_y¹
As it forms three identical B-F bonds in BF₃.
It is suggested that excited ‘B’ atom undergoes hybridisation.
There is an intermixing of 2s, 2p_x, 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one election.
Now three fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (F₉ 1s² 2s² 2p_x² 2p_y² 2p_z¹) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three σsp²-p bonds.

Fill in the Blanks

1. Electrons in the outermost orbit are called …………………… .
2. Except …………………… gas all other noble gases have octet in their valence shell.
3. Covalency of elements explains about member of …………………… formed by the atom.
4. Valence bond theory was proposed by …………………… .
5. In …………………… bonding the valence electrons are shared among all the atoms of the metallic elements.
Answer:

valence electrons
Helium
covalent bonds
Linus Pauling
covalent

Multiple Choice Questions

1. Which of the following elements is electronegative?
A) Sodium
B) Oxygen
C) Magnesium
D) Calcium
Answer:
B) Oxygen

2. An element ₁₁X²³ forms an ionic compound with another element ‘Y’. Then the charge on the ion formed by X is
A) +1
B) +2
C) -l
D) – 2
Answer:
A) +1

3. An element ‘A’ forms a chloride ACl₄. The number of electrons in the valence shell of ‘A’
A) 1
B) 2
C) 3
D) 4
Answer:
D) 4

10th Class Chemistry 10th Lesson Chemical Bonding InText Questions and Answers

10th Class Chemistry Textbook Page No. 153

Question 1.
How do elements usually exist?
Answer:
They may exist as a single atom or as a group of atoms.

Question 2.
Do atoms exist as a single atom or as a group of atoms?
Answer:
Atoms exist as a single atom, sometimes as a group of atoms also.

Question 3.
Are there elements which exist as atoms?
Answer:
Yes. There are elements which exist as atoms.

Question 4.
Why do some elements exist as molecules and some as atoms?
Answer:
By following different laws of chemical combination the chemical compounds take place as a result of combination of atoms of various elements in different ways.

Question 5.
Why do some elements and compounds react vigorously while others are inert?
Answer:
1) Number of electrons in their outermost shell.
2) Bond strength between the atoms in compound.

Question 6.
Why is the chemical formula for water H₂O and for sodium chloride NaCl, why not HO₂ and NaCl₂?
Answee:
Valencies of the atoms participating in the molecules.

Question 7.
Why do some atoms combine dille tl do not?
Answer:
1) Atoms which have 8e“ in their outer shell will not combine.
2) Atoms which have more than or less than 8e“ in their outer shell will combine.

Question 8.
Are elements and compounds simply made up of separate atoms Individually arranged?
Answer:
No. Elements and compounds are not simply made up of separate atoms individually arranged.

Question 9.
Is there any attraction between atoms?
Answer:
Yes. There is some attraction betwen atoms.

Question 10.
What is that holding them together?
Answer:
Force of attraction between them.

10th Class Chemistry Textbook Page No. 155

Question 11.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
Answer:’
Because of bond energy between the atoms in a molecule.

Question 12.
Where does the absorbed energy go?
Answer:
For breaking chemical bonds between atoms in a molecule.

Question 13.
Is there any relation to energy and bond formation between atoms?
Answer:
Yes. There is some relation to energy and bond formation between atoms.

Question 14.
What could be the reason for the change in reactivity of elements?
Answer:
Number of electrons in their outermost orbit.

Question 15.
What could be the reason for this?
Answer:
They have 8 (e^–) electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 157

Question 16.
What did you notice in Lewis dot structure of noble gases and electronic configurations of the atoms of these elements shown in table – 1?
Answer:
Except He remaining Ne, Ar, Kr have 8 electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 158

Question 17.
What have you observed from the above conclusions about the main groups?
Answer:

Number of gained electrons of non-metals in their valency.
Number of lost electrons of metals in their valency.

Question 18.
Why do atoms of elements try to combine and form molecules?
Answer:
To attain stable electronic configuration.

10th Class Chemistry Textbook Page No. 159

Question 19.
Is it accidental that IA to VIIA main group elements during chemical reactions get eight electrons in the outermost shells of their ions, similar to noble gas atoms?
Answer:
No, it cannot be simply accidental.

Question 20.
Explain the formation of ionic compounds NaCl, MgCl₂, Na₂O and AlCl₃ through Lewis electron dot symbols (formulae).
Answer:
1) Lewis electron dot symbol for NaCl:

Formation of sodium chloride (NaCl) :
Sodium chloride is formed from the elements sodium and chlorine. It can be explained as follows.
Na_(s) + ½Cl_2(g) → NaCl₂

Cation formation :
When sodium (Na) atom loses one electron to get octet electron configuration it forms a cation (Na+) and gets electron configuration that of Neon (Ne) atom.

Anion formation :
Chlorine has shortage of one electron to get octet in its valence shell. So it gains the electron from Na atom to form anion and gets electron configuration as that of argon (Ar).

Formation of the compound NaCl from its ions :
Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–‘ ions. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–_(g) → Na⁺Cl^–_(s) or NaCl

2) Lewis electron dot symbol for MgCl₂:
MgCl₂

Formation of magnesium chloride (MgCl₂):
Magnesium chloride is formed from the elements magnesium and chlorine. The bond formation MgCl₂ in brief using chemical equation is as follows :
Mg_(s) + Cl_2(g) → MgCl_2(g)
Cation formation:

Anion formation :

The compound MgCl₂ formation from its ions :
Mg²⁺ gets ‘Ne’ configuration and
Each Cl^– gets ‘Ar’ configuration
Mg²⁺_(g) + 2 Cl^–_(g) → MgCl_2(s)
One ‘Mg’ atom transfers two electrons one each to two ‘Cl’ atoms and so formed Mg²⁺ and 2Cl^– attract to form MgCl₂.

3) Lewis electron dot symbol for (Na₂O) :

Formation of di sodium monoxide (Na₂O):
Di sodium monoxide formation can be explained as follows:
Cation formation (Na⁺ formation):

Two ‘Na’ atoms transfer one electron each to one oxygen atom to form 2 Na⁺ and O^2-
Each Na⁺ gets ‘Ne’ configuration and O^2- gets ‘Ne’ configuration.
These ions (2Na⁺ and O^2-) attract to form Na₂O.

4) Lewis electron dot symbol for (AlCl₃):

Formation of aluminium chloride (AlCl₃):
Aluminium chloride formation can be explained as follows:
Formation of aluminium ion (Al³⁺), the cation:

Each aluminium atom loses three electrons and three chlorine atoms gain them, one electron each.
The compound AlCl₃ is formed from its component ions by the electrostatic forces of attractions.
Al³⁺_(g) + 3 Cl^–_(g) → AlCl_3(s)

10th Class Chemistry Textbook Page No. 163

Question 21.
How do cations and anions of an ionic compound exist in its solid state?
Answer:
Ionic compounds exist in crystalline state.

Question 22.
Do you think that pairs of Na⁺ Cl^– as units would be present in the solid crystal?
Answer:
Yes. I think that pairs of Na⁺ Cl^– as units would be present in the solid crystal.

10th Class Chemistry Textbook Page No. 164

Question 23.
Can you explain the reasons for all these?
Answer:
Ionic bond is formed between atoms of elements with electronegativity, difference equal to or greater than 1.9.

10th Class Chemistry Textbook Page No. 165

Question 24.
Can you say what type of bond exists between atoms of nitrogen molecule?
Answer:
Triple bond exists between atoms of nitrogen molecule.

10th Class Chemistry Textbook Page No. 168

Question 25.
What do you understand from bond lengths and bond energies?
Answer:
Bond lengths and bond energies are not same when the atoms that form the bond are different.

Question 26.
Are the values not different for the bonds between different types of atoms?
Answer:
Yes. The values are not different for the bonds between different types of atoms.

10th Class Chemistry Textbook Page No. 170

Question 27.
What is the bond angle in a molecule?
Answer:
It is the angle subtended by two imaginary lines that pass from the nuclei of two atoms which form the covalent bonds with the central atom through the nucleus of the central atom at the central atom.

10th Class Chemistry Textbook Page No. 172

Question 28.
How is MCI molecule formed?
Answer:
The ‘1s’ orbital of ‘H’ atom containing unpaired electron overlaps the ‘3p’ orbital of chlorine atom containing unpaired electron of opposite spin.

10th Class Chemistry 10th Lesson Chemical Bonding Activities

Activity – 1

1. Write the Lewis structures of the given elements in the table. Also, consult the periodic table and fill in the group number of the element.
Answer:

AP SSC 10th Class Chemistry Solutions Chapter 9 Classification of Elements- The Periodic Table

March 29, 2023February 2, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 9 Classification of Elements- The Periodic Table Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 9th Lesson Structure of Atom

10th Class Chemistry 9th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
Newlands proposed the law of octaves. Mendeleeff suggested eight groups for elements in his table. How do you explain these observations in terms of modem periodic classification? (AS1)
(OR)
Correlate various tables proposed on classification of elements.
Answer:

According to Newlands, every eighth element starting from a given element jsembles in its properties to that of the starting element, when elements are ranged in ascending order of their atomic weights.
According to Newlands, the properties of fluorine and chlorine are similar and sodium and potassium are similar. Same aspect is given by modern periodic table.
Mendeleeff divided it into horizontal rows and vertical columns. He called them peribds and groups respectively. Modem periodic table also gives the same.
According to Mendeleeff, the elements of same group have similar properties. Modern periodic table also proposed the same thing.
Mendeleeff gave the general formula for first group elements as R,0, and general formula for second group elements as RO. We can find the same thing in modern periodic table.
The elements of particular group possess same common valency. Same was proposed by modern periodic table.

Question 2.
What are the limitations of Mendeleeff’s periodic table? How could the modern periodic table overcome the limitations of Mendeleeff’s table? (AS1)
(OR)
How can the limitations of Mendeleeffs table be overcome with the help of modern periodic table?
Answer:
Limitations of Mendeleeffs periodic table :
1) Anomalous pair of elements :
Certain elements of highest atomic weights precede those with lower atomic weights.
Eg : Tellurium (atomic weight 127.6) precedes iodine (atomic weight 126.9).

2) Dissimilar elements placed together :
a) Elements with dissimilar properties were placed in same group as sub-group A and sub-group Bt
Eg : Alkali metals like Li, Na, K, etc. of IA group have little resemblance with coinage metals like Cu, Ag, Au of IB group.

b) Cl of VII A group is a non-metal and Mn of VII B group is a metal.

Method of overcoming the limitations of Mendeleeffs periodic table by modern periodic table :
1. In modern periodic table, elements are arranged in the ascending order of their atomic numbers. So this arrangement eliminated the problem of anomalous series.
Eg : Though Tellurium (Te) has more atomic weight than Iodine (I), its atomic number is one unit less compared to Iodine.

2. The elements with similar outer shell (valence shell) electronic configurations in their atoms are in the same column called group in modern periodic table. So the elements have similar properties overcoming the Mendeleeffs second limitation.

Question 3.
Define the modern periodic law. Discuss the construction of the long form of the periodic table. (AS1)
(OR)
What are the salient features of modern periodic table?
Answer:
Modern periodic law :
‘The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms”.

Construction of the long form periodic table :

Based on the modern periodic law, the modern periodic table is proposed.
This periodic table is known as long form of the periodic table.
Long form periodic table is the graphical representation of Aufbau principle.
The modern periodic table has 18 vertical columns called groups and 7 horizontal rows known as periods.
There are 18 groups, represented by using Roman numerals I to VIII, with letters A and B in traditional notation, (or) 1 to 18 by Arabic numerals.
There are 7 periods. These periods are represented by Arabic numerals 1 to 7.
The number of main shells present in the atom of particular atom decides to which period it belongs.
First period consists 2 elements, 2^nd and 3rd periods contains 8 elements each, 4^th and 5^th periods contains 18 elements each, 6 period contains 32 elements and 7^th period is incomplete.
The elements are classified into s, p, d and f block elements.
Inert gases are placed in 18^th group.

Question 4.
Explain how the elements are classified into s, p, d and f-block elements in the periodic table and give the advantage of this kind of classification. (AS1)
(OR)
How is the periodic table classified based upon the entering of differenciating electron? Explain that classification. What is the advantage of such classification?
Answer:
1) Depending upon which sub-shell the differentiating electron enters, the elements are classified into s, p, d and f-block elements. They are

s – block elements,
p – block elements,
d – block elements,
f – block elements.

2) s – block elements :
i) If the differentiating electron enters in s-sub-shell, then the elements are called s-block elements.
ii) IA (1), IIA (2) group elements belong to this block.

3) p – block elements :
i) If the differentiating electron enters in p-sub-shell, then the elements are called p-block elements.
ii) IIIA(13), IV A (14), V A (15), VIA (16), VIIA (17) belong to p-block.

4) d – block elements :
i) If the differentiating electron enters in d-sub-shell, then the elements are called d – block elements.
ii) I B, II B, III B, IV B, V B, VI B, VII B, VIII B belong to d-block elements.
iii) They are also called transition elements.

5) f – block elements :
i) If the differentiating electrons enter in f-sub-shell, then the elements are called f-block elements.
ii) These are divided into two types
a) Lanthanides (41 elements),
b) Actinides (5f elements).
iii) These are also called as inner transition elements.

Advantage of this classification :
1) The systematic grouping of elements into groups made the study simple.
2) Each period begins with the electron entering a new shell and ends with the complete filling of s and p-sub-shells of that shell.

Question 5.
Given below is the electronic configuration of elements A, B, C, D. (AS1)

A) 1s² 2s²	1. Which are the elements coming within the same period?
B) 1s² 2s² 2p₆ 3s²	2. Which are the elements coming within the same group?
C) 1s² 2s² 2p6 3s² 3p³	3. Which are the noble gas elements?
D) 1s² 2s² 2p6	4. To which group and period does the element ‘C’ belong?

Answer:
According to electronic configuration
A = Be B = Mg C = P D = Ne
1. Which are the elements coming within the same period?
Answer:
A and D i.e. Be and Ne coming within the same period. [They have same valence shell (n = 2)]

2. Which are the ones coming within the same group?
Answer:
A and B i.e., Be and Mg coming within the same group. [They have same valence subshell with same valency (2s² and 3s²)]

3. Which are the noble gas elements?
Answer:
D, i.e. Ne is the noble gas element. [It has valency as ‘O’ and it has ‘8’ electrons in valence shell].

4. To which group and period does the element ‘C’ belong?
Answer:
Element ‘C’ i.e. ‘P’ belongs to 3^rd period and VA group.

Question 6.
Write down the characteristics of the elements having atomic number 17. (AS1)
1) Electronic configuration ___________
2) Period number _____________
3) Group number _____________
4) Element family ____________
5) No. of valence electrons ___________
6) Valency _____________
7) Metal or non-metal ____________
Answer:

1s² 2s² 2p⁶ 3s² 3p⁵
3
VII A or 17
Halogen family
7
1
Non-metal

Question 7.
a) State the number of valence electrons, the group number and the period number of each element given in the following table : (AS1)

Answer:

b) State whether the following elements belong to a Group (G), Period (P) or neither Group nor Period (N). (AS1)

Answer:

Question 8.
Elements in a group generally possess similar properties, but elements along a period have different properties. How do you explain this statement? (AS1)
(OR)
Elements in a group possess similar properties, but elements along a period have different properties. Explain the reason.
Answer:

Physical and chemical properties of elements are related to their electronic configurations, particularly the outer shell configurations.
Therefore, all the elements in a group should have similar chemical properties.
Similarly, across the table from left to right in any period, elements get an increase in the atomic number by one unit between any two successive elements.
Therefore, the electronic configuration of valence shell of any two elements in a period is not same. Due to this reason, elements along a period possess different chemical properties.

Question 9.
s – block and p – block elements except 18^th group elements are sometimes called as ‘Representative elements’ based on their abundant availability in the nature. Is it justified? Why? (AS1)
(OR)
Which elements are called representative elements? Why?
Answer:

s, p – block elements are called representative elements because these are the elements which take part in chemical reactions because of incompletely filled outermost shell.
These elements undergo chemical reactions to acquire the nearest noble gas configuration by losing or gaining or sharing of electrons.
So they are called representative elements.

Question 10.
Complete the following table using periodic table. (AS1)

Answer:

Question 11.
Complete the following table using the periodic table. (AS1)

Answer:

Question 12.
The electronic configuration of the elements X, Y, and Z are given below.
a) X = 2
b) Y = 2, 6
c) Z = 2, 8, 2
i) Which element belongs to second period?
Answer:
Y belongs to second period.

ii) Which element belongs to second group?
Answer:
Z belongs to second group,

iii) Which element belongs to 18^th group?
Answer:
X belongs to 18^th group.

Question 13.
Identify the element that has the larger atomic radius in each pair of the following and mark it with a symbol (✓). (AS1)
(i) Mg or Ca
(ii) Li or Cs
(iii) N or P
(iv) B or Al
Answer:

Question 14.
Identify the element that has the lower ionization energy in each pair of the, following and mark it with a symbol (✓). (AS1)
(i) Mg or Na (ii) Li or O (iii) Br or F (iv) K or Br
Answer:

Question 15.
In period 2, element X is to the right of element Y. Then, find which ofitheydements have : (AS1)
i) Low nuclear charge
Answer:
Y has low nuclear charge.

ii) Low atomic size
Answer:
X has lower atomic size,

iii) High ionization energy
Answer:
X has higher ionization energy.

iv) High electronegativity
Answer:
Xhas high electronega^vity.

v) More metallic,character
Answer:
Y has more metallic character.

Question 16.
How does metallic character change when we move
i) Down a group?
ii) Across a period?
Answer:
i) Down a group :
When we move from top to bottom in a group, the metallic character increases.

ii) Across a period:
When we move left to right in a period, the metallic character decreases.

Question 17.
Why was the basis of classification of elements changed from the atomic mass to the atomic number? (AS1)
(OR)
Which atomic property is more suitable for classification of elements? Why?
Answer:

The first attempt to classify elements was made by Dobereiner.
Dobereiner’s attempt gave a clue that atomic masses could be correlated with properties of elements:
Newlands’ law of octaves also followed the same basis for classification but this law is not valid for the elements that had atomic masses higher than calcium.
Mendeleeff’s classification also based on the atomic masses of elements, but it lead to some limitations like Anomalous pair of elements and Dissimilar elements placed together.
Moseley by analyzing the X-ray patterns of different elements was able to calculate the number of positive charges in the atoms of respective elements.
With this analysis, Moseley realized that the atomic number is more fundamental
characteristic of an element than its atomic weight. ,
So, he arranged the elements in the periodic table according to the increasing order of their atomic number.
This arrangement eliminated the problem of anomalous series and dissimilar elements placed together in Mendeleeff’s classification.

Question 18.
What is a periodic property? How do the following properties change in a group and period? Explain. (AS1)
I. a) Atomic radius
b) Ionization energy
c) Electron affinity
d) Electronegativity
II. Explain the ionization energy order in the following sets of elements: (AS1)
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
Periodic property:
The property in which there shall be a regular gradation is called periodic property.

I. a) Atomic radius :
Period :
Atomic radius of elements decreases across a period from left to right because the nuclear charge increases due to increase in atomic number.

Group :
Atomic radius increases from top to bottom in a group due to addition of new shell.

b) Ionization energy:
Period :
When we move from left to right it does not follow a regular trend but generally increases due to increase in atomic number.

Group :
In a group from top to bottom, the ionization energy decreases due to increase in atomic size. –

c) Electron affinity:
Period :
Electron affinity values increase from left to right in a period.

Group :
Electron affinity values decrease from top to bottom in a group.

d) Electronegativity :
Period :
Electronegativity increases from left to right in a period.

Group :
Electronegativity decreases from top to bottom in a group.

II. Ionization energy order :
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
a) In a period ionisation energy increases so the order is Na < kl < Cl.
b) Beryllium has stable configuration 1s² 2s². So it has more ionisation energy. So the order is Li < B < Be.
c) Nitrogen has half-filled p-orbitals. So it has greater ionisation energy. So the order is C < O < N.
d) Ne is inert gas right to F. Whereas Na is a metal ion in third period. So, the order is Na < F < Ne. e) In a group ionisation energy decreases. So the order is Be > Mg > Ca.

Question 19.
Name two elements that you would expect to have chemical properties similar to Mg. What is the basis for your choice? (AS2)
Answer:

The two elements which have chemical properties similar to Magnesium are Beryllium and Calcium.
The basis for my expectation is that they belong to same group as we know elements belonging to same group have similar properties.

Question 20.
On the basis of atomic numbers predict to which block the elements with atomic number 9, 37, 46 and 64 belong to? (AS2)
Answer:

The element with atomic number 9 belongs to p-block.
The element with atomic number 37 belongs to s-block.
The element with atomic number 46 belongs to d-block.
The element with atomic number 64 belongs to f-block.

Question 21.
Using the periodic table, predict the formula of compound formed between and element X of group 13 and another element Y of group 16. (AS2)
Answer:
The valency of 13^th group elements is 3.
The valency of 16^th group elements is 2.
The formula of compound is X₂Y₃.

Question 22.
An element X belongs to 3rd period and group 2 of the periodic table. State (AS2)
a) The no. of valence electrons
b) The valency.
c) Whether it is metal or a non-metal.
Answer:
a) The number of valence electrons are 2.
b) The valency of element is +2.
c) It is a metal.

Question 23.
An element has atomic number 19. Where would you expect this element in the periodic table and why? (AS2)
Answer:
The clement with atomic number 19 is in 4^th period and first group of the periodic table.
Reason :

Electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s or [Ar]4s¹
The differentiating electron enters into 4^th shell. Hence it belongs to 4^th period.
The differentiating electron is in ‘s’ orbital. So it belongs to ‘s’ block.
The outermost orbital has only one electron. Hence it belongs to first group.

Question 24.
Aluminium does not react with water at room temperature but reacts with both dil. HCl and NaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that Al is a metalloid? (AS3)
Answer:

Aluminium reacts with dil. HCl and releases hydrogen gas with formation of Aluminium chloride.
Aluminium reacts with NaOH solution and releases hydrogen gas.
The above two reactions says that Aluminium is amphoteric.
Aluminium does not react with water at room temperature.
This concludes that the properties of Aluminium are in between a metal and non¬metal. So it behaves like a metalloid.

Question 25.
Collect the information about reactivity of VIIIA group elements (noble gases) from internet or from your school library and prepare a report on their special character when compared to other elements of periodic table. (AS4)
Answer:
Reactivity of Noble gases :

The noble gases show extremely low chemical reactivity.
He and Ne do not form chemical compounds.
Xenon, krypton and argon show only minor reactivity.
The reactivity order follows like this : Ne < He < Ar < Kr < Xe < Rn.
Xenon can form compounds like XeF₂, XeF₄ and XeF₆, etc.

Reasons for low reactivity :

The extremely low reactivity of noble gases is due to stable electronic configuration.
But as we move from top to bottom the reactivity increases. So xenon can form some compounds with high electronegative elements.

Question 26.
Collect information regarding metallic character of elements of IA group and prepare report to support the idea of metallic character increases in a group as we move from top ro bottom. (AS4)
Answer:
Metallic character of IA group elements :

Alkali metals exhibit many of the physical properties common to metals but their densities are lower than those of other metals.
Alkali metals have one electron in their outer shell which is loosely bound.
They have largest atomic radii of the elements in their respective periods.
The lower ionization energies result in their metallic properties and high reactivities.
An alkali metal can easily lose its valence electron to form positive ion.
So they have greater metallic character.
The metallic character increases as we move from top to bottom in group due to addition of another shell, it is easy to lose electron.

Question 27.
How do you appreciate the role of electronic configuration of the atoms of elements in periodic classification? (AS6)
(OR)
How does electronic configuration help in the classification of elements in modern periodic table?
Answer:
The quantity is electronic configuration.

Modern periodic table is based on electronic configuration. So elements are arranged in ascending order of their atomic numbers.
The chemical properties of elements depend on valence electrons. The elements in same group have same number of valence electrons. So the elements belonging to same group have similar properties.
So the construction of modern periodic table mainly depends on electronic configuration.
Thus electronic configuration plays a major role in the preparation of modern periodic table. So its role is thoroughly appreciated.

Question 28.
Without knowing the electronic configurations of the atoms of elements Mendeleeff still could arrange the elements nearly close to the arrangements in the Modern periodic table. How can you appreciate this? (AS6)
Answer:

Mendeleeff took consideration about chemical properties while arranging the elements. So the arrangement of elements is close to arrangement of elements in Modern periodic table.
For this, he violated his periodic law.
He left some gaps for elements, later those elements are discovered.
So the efforts of Mendeleeff should be thoroughly appreciated.

Question 29.
Comment on the position of hydrogen in periodic table. (AS7)
Answer:

Hydrogen is the element which has easier atomic structure than any other element.
Electron configuration of hydrogen is Is1. It has one proton in its is nucleus and one electron in its is orbital.
Hydrogen combines with halogens, oxygen and sulphur to form compounds having similar formulae just like alkali metals.
Similarly, just like halogens, hydrogen also exists as diatomic molecule and combine with metals and non-metals to form covalent compounds.
As alkali metals hydrogen can lose one electron and accept one electron as halogens.
So in periodic table, its place may be in IA or VIIA group.
But based on electronic configuration of hydrogen, it is placed in IA group.

Question 30.
How do the positions of elements in the periodic table help you to predict its chemical properties? Explain with an example. (As7)
Answer:
1) The physical and chemical properties of atoms of the elements depend on their electronic configuration, particularly the outer shell configurations.

2) Elements are placed in the periodic table according to the increasing order of their electronic configuration.

3) The elements in a group possess similar electronic configurations. Therefore all the elements in a group should have similar chemical properties.
Ex : Consider K

It is the element in 4^th period 1^st group.
Electron configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
Differentiating electron enters into s-orbital. Hence it belongs to s-block.
It is on the left side of the periodic table. Hence it is a metal.
It is ready to lose one electron to get octet configuration. Hence its reactivity is more.
It is Alkali metal.
All alkali metals react with both acids and bases and releases H₂ gas.

Fill In The Blanks

1. Lithium, ……………… and potassium constitute a Dobereiner’s triad.
2. ……………… was the basis of the classifications proposed by Dobereiner, Newlands, and Mendeleeff.
3. Noble gases belong to ……………… group of periodic table.
4. The incomplete period of the modern periodic table is
5. The element at the bottom of a group would be expected to show …………….. metallic character than the element at the top

Answer:

Sodium
Atomic weight
VIIIA or 18 group
7<sup>th</sup>
higher

Multiple Choice Questions

1. Number of elements present in period – 2 of the long form of periodic table …………
A) 2
B) 8
C) 18
D) 32
Answer:
B) 8

2. Nitrogen (Z = 7) is the element of group V of the periodic table. Which of the following is the atomic number of the next element in the group?
A) 9
B) 14
C) 15
D) 17
Answer:
C) 15

3. Electronic configuration of an atom is 2, 8, 7. To which of the following elements would it be chemically similar?
A) Nitrogen (Z = 7)
B) Fluorine (Z = 9)
C) Phosphorous (Z – 15)
D) Argon (Z = 18)
Answer:
B) Fluorine (Z = 9)

4. Which of the following is the most active metal?
A) lithium
B) sodium
C) potassium
D) rubidium
Answer:
D) rubidium

10th Class Chemistry 9th Lesson Classification of Elements-The Periodic Table InText Questions and Answers

10th Class Chemistry Textbook Page No. 129

Question 1.
Can you establish the same relationship with the set of elements given in the remaining rows?
Answer:
Yes, we can establish the approximately same relationship between other elements given in the table.

Question 2.
Find average atomic weights of the first and third elements in each row and compare it with the atomic weight of the middle element. What do you observe?
Answer:
The atomic weight of middle element is arithmetic mean coverage of first and third elements.

10th Class Chemistry Textbook Page No. 135

Question 3.
What is atomic number?
Answer:
The number of positive charges (protons) in the atom of element is the atomic number of element.

10th Class Chemistry Textbook Page No. 142

Question 4.
How does the valency vary in a period on going from left to right?
Answer:
It does not follow a regular trend when we move from left to right in a period. First, it increases and then decreases and finally ‘O’ for inert gases.

Question 5.
How does the valency vary on going down a group?
Answer:
The valency is constant when we move from top to bottom in a group because the number of valence electrons are same for same group elements.

10th Class Chemistry Textbook Page No. 144

Question 6.
Do the atom of an element and its ion have same size?
Answer:
No, the positive ion has smaller size than neutral atom whereas negative ion has greater size than neutral atom.

Question 7.
Which one between Na and Na⁺ would have more size? Why?
Answer:

The atomic number of Sodium is 1 and it has 11 protons and 11 electrons with outer electron as 3s¹ whereas Na⁺ ion has 11 protons but only 10 electrons.
The 3s shell of Na⁺ has no electron in it.
So the outer shell configuration is 2s²2p⁶.
As proton number is more than electrons, the nucleus of Na⁺ ion attracts outer shell electrons with strong nuclear force.
As a result the Na⁺ ion shrinks in size.
Therefore, the size of Na⁺ ion is less than Na atom.

Question 8.
Which one between Cl and Cl^– would have more size? Why?
Answer:

The electronic configuration of chlorine (Cl) atom is 1s2 2s² 2p⁶ 3s² 3p⁵ and the electronic configuration of chloride (Cl^–) ion is 1s² 2s² 2p⁶ 3s² 3p⁶.
Both chlorine and chloride ions have 17 protons each but there are 17 electrons in chlorine atom, whereas 18 electrons in chloride ion.
Therefore, the nuclear attraction is less in Cl^– ion when compared with chlorine atom.
Therefore the size of the chlorine (Cl) atom is less size than chloride of Cl^– ion.

Question 9.
Which one in each of the following pairs is larger in size? Why?
a) Na, Al
b) Na, Mg⁺²
c) S^2-, Cl^–
d) Fe²⁺, Fe³⁺
e) C^4-, F^–.
Answer:
a) Na has larger size because Sodium and Aluminium are third period elements in which Na is left to Al. As we move from left to right in a period atomic size decreases.

b) Mg²⁺ has smaller size because Mg²⁺ has 10 electrons and 12 protons whereas Na has 11 electrons and 11 protons. So the distance between nucleus and outermost orbital is less in Mg²⁺ due to greater nuclear attraction.

c) S^2- has, larger size because S^2- has 18 electrons and 16 protons and Cl^– has 18 electrons and 17 protons. So nuclear attraction over outermost orbital is more in Cl^– when compared with S^2-. So S^2- has larger size.

d) In Fe²⁺ it has 26 protons and 24 electrons whereas for Fe³⁺ it has 26 protons and 23 electrons. So nuclear attraction over outermost orbital is more in Fe³⁺. So Fe³⁺ has smaller size (or) Fe²⁺ has larger size.

e) C^4- has 6 protons and 10 electrons whereas F^– has 9 protons and 10 electrons. So nuclear attraction is less in C^4-. So size of C^4- is more than F^–.

10th Class Chemistry Textbook Page No. 129

Question 10.
What relation about elements did Dobereiner want to establish?
Answer:
Dobereiner wanted to give a relationship between the properties of elements and their atomic weights.

Question 11.
The densities of calcium (Ca) and barium (Ba) are 1.55 and 3.51 gem-3 respectively. Based on Dobereiner’s law of triads can you give the approximate density of strontium (Sr)?
Answer:
Molecular weight is directly proportional to density.

So density of strontium is mean of calcium and barium according to Dobereiner.
∴ Density of strontium = $\frac{1.55+3.51}{2}$ = 2.53.

10th Class Chemistry Textbook Page No. 130

Question 12.
Do you know why Newlands proposed the law of octaves? Explain your answer in terms of the modern structure of the atom.
Answer:

John Newlands found that when elements were arranged in the ascending order of their atomic weights, they appeared to fall into seven groups.
Each group contained elements with similar properties.
If we start with hydrogen and move down, the next eighth element is fluorine, and then next eighth element is chlorine and the properties of these elements are similar.
Similarly, if we start from Lithium their eighth element is Sodium and next eighth element is potassium. These show similar properties.

Question 13.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, there are some limitations of Newlands’ model:

There are instances of two elements fitted into the same slot. Eg : Cobalt and Nickel.
Certain elements, totally dissimilar in their properties, were fitted into the same group.
Law of octaves holds good only for the elements up to Calcium.
Newlands’ periodic table was restricted to only 56 elements and did not leave any room for new elements.
Newlands had taken consideration about active pattern sometimes without caring the similarities.

10th Class Chemistry Textbook Page No. 134

Question 14.
Why did Mendeleeff have to leave certain blank spaces in his periodic table? What is your explanation for this?
Answer:
1) Mendeleeff predicted that some elements which have similar properties with the elements in a group are missing at that time.
2) So he kept some blanks in the periodic table by writing ’eka’ to the name of the element immediately above the empty space.
3) Later these elements are discovered and they are fitted into those empty spaces.

Question 15.
What is your understanding about Ea₂O₃, EsO₂?
Answer:

Mendeleeff predicted that after aluminium there was another element namely eka- aluminium (Ea) and after silicon, there was another element namely eka-silicon (Es).
He also gave the formulae of those oxides as Ea203 and Es02.
Later those elements are discovered namely gallium and germanium and Ea₂O₃ and EsO₂ as Ga₂O₃ and GeO₂.

10th Class Chemistry Textbook Page No. 135

Question 16.
All alkali metals are solids but hydrogen is a gas with diatomic molecules. Do you justify the inclusion of hydrogen in first group with alkali metals?
Answer:
No, hydrogen shows the properties of both alkali metals and halogens. Still the position of hydrogen has some questions. So it was kept just above alkali metals in first group.

10th Class Chemistry Textbook Page No. 141

Question 17.
Why are lanthanoids and actinoids placed separately at the bottom of the periodic table?
Answer:
The properties of these elements do not coincide with other elements because the valence electron enters 4f and 5f orbitals respectively. So they are placed separately at the bottom of the periodic table.

Question 18.
If lanthanoids and actinoids are inserted within the table, imagine how the table would be?
Answer:
It looks very big in size, and it is difficult to identify, as these elements have similar properties.

10th Class Chemistry Textbook Page No. 145

Question 19.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:

The energy required to remove an electron from unipositive ion is called second ionisation energy.
It is difficult to remove an electron from unipositive ion when compared with neutral atom due to an increase in nuclear attraction.
So always second ionisation energy is higher than first ionisation energy.

10th Class Chemistry Textbook Page No. 146

Question 20.
The calculated electron gain enthalpy values for alkaline earth metals and noble gases are positive. How can you explain this?
Answer:

Generally alkaline earth metals having one, two or three valence electrons prefer to lose electrons in order to get inert gas configuration. So it is difficult to add electron to alkaline earth metals. So they have positive electron gain enthalpy values.
Noble gases are stable. So they do not prefer to take electrons. So they have positive electron gain enthalpy.

Question 21.
The second period element, for example, ‘F’ has less electron gain enthalpy than the third period element of the same group for example ‘Cl’. Why?
Answer:

Electron gain enthalpy values decrease in a group as we go down and increase from left to right along a period.
But the size of Fluorine is small compared chlorine.
So it is difficult to add electron to fluorine.
So fluorine has less electron gain enthalpy.

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table Activities

Activity – 1

Question 1.
Observe the following table. Establish the relationship of other elements given in the table.

Answer:

Activity – 2

Question 2.
Some main group elements of s-block and p-block have family names as given in the following table.
Observe the long form of a periodic table and complete the table with proper information.

Answer:

Activity – 3

Question 3.
Collect valencies of first 20 elements.
Answer:

AP SSC 10th Class Chemistry Solutions Chapter 8 Structure of Atom

March 29, 2023February 1, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 8 Structure of Atom Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 8th Lesson Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
What information does the electronic configuration of an atom provide? (AS1)
Answer:

The distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.
It provides the information of position of an electron in the space of atom.
The distribution of electrons in various atomic orbitals provides an understanding of the electronic behaviour of the atom and in turn its reactivity.
The short hand notation is as shown below.

Question 2.
a) How many maximum number of electrons that can be accommodated in a principal energy shell?
Answer:
The maximum number of electrons that can be accommodated in a principal energy shell is 2n². Here n is principal quantum number.

b) How many maximum number of electrons that can be accommodated in a sub-shell?
Answer:
The maximum number of electrons that can be accommodated in a sub-shell is 2(2l +1) (where l is orbital quantum number).

c) How many maximum number of electrons can that be accommodated in an orbital?
Answer:
The maximum number of electrons that can be accommodated in an orbital is 2.

d) How many sub-shells are present in a principal energy shell?
Answer:
The number of sub-shells in a principal energy shell is n (n is principal quantum number).

e) How many spin orientations are possible for an electron in an orbital?
Answeer:
The spin orientations possible for an electron in an orbital are 2.

Question 3.
In an atom the number of electrons in M-shell is equal to the number of electrons in the K and L-shell. Answer the following questions. (AS1)
a) Which is the outermost shell?
Answer:
The outermost shell is N shell.

b) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

c) What is the atomic number of element?
Answer:
The atomic number of element is 22.

d) Write the electronic configuration of the element.
Answer:
The element is Ti (Titanium). Its electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d².

Reason :

Electrons enter M shell after completion of K and L.
So the number of electrons in M shell is 10.
But after completion of 3p orbital electron enters 4s before entering to 3d.
So outermost orbit or shell is N shell.
So the atomic number of element is 22.
Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d².

Question 4.
Rainbow is an example for continuous spectrum – explain. (AS1)
(OR)
Which is naturally occurring continuous spectrum ? Explain.
Answer:

Rainbow is a spectrum of different colours (VIBGYOR) with different wavelengths.
These colours are continuously distributed.
There is no fixed boundary for each colour.
Hence, rainbow is a continuous spectrum.

Question 5.
How many elliptical orbits are added by Sommerfeld in third Bohr’s orbit ? What was the purpose of adding these elliptical orbits? (AS1)
Answer:
Sommerfeld added two elliptical orbits to Bohr’s third orbit.

Purpose of adding elliptical orbits :

Bohr’s model failed to account for splitting of line spectra and line spectrum.
In an attempt to account for the structure of line spectrum, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.

Question 6.
What is absorption spectrum?
Answer:
Absorption spectrum: The spectrum formed by the absorption of energy when electron jumps from lower energy level to higher energy level is called absorption spectrum. It contains dark lines on bright background.

Question 7.
What is an orbital? How it is different from Bohr’s orbit? (AS1)
(OR)
Comparison between orbit and orbital.
Answer:
The region of space around the nucleus where the probability of finding electron is maximum is called orbital. Whereas orbit is the path of the electron around the nucleus.

These two are differentiated like this.

Orbit	Orbital
1. Path of electron around nucleus.	1) Probability of finding electron around nucleus.
2. Orbits are represented by letters K, L, M, N, 0, …….etc.	2. Orbitals are represented by letters s, p, d, f, g, …….etc.
3) Its information is given by principal	3) Its information is given by orbital quantum number.
4) It is two dimensional.	4) It is three dimensional.
5) It does not satisfy Heisenberg’s uncertainty principle.	5) It satisfies the Heisenberg’s principle of uncertainty.

Question 8.
Explain the significance of three quantum numbers in predicting the positions of an electron in an orbit. (AS1)
(OR)
How are quantum numbers helpful to understand the atomic structure?
Answer:
Significance of three quantum numbers in predicting the positions of an electron in an orbit.

1) Principal quantum number (n) :
The principal quantum number explains about the size and energy of shells (or) orbitals. It is denoted by n.

As ‘n’ increases, the orbitals become larger and the electrons in those orbitals are farther from the nucleus.

It takes values 1, 2, 3, 4, ……………. for that the shells are represented by letters K, L, M, N, ……….

The number of electrons in a shell is limited to 2n².

2) The Angular – momentum quantum number (l) :
The angular momentum quantum number defines the shape of the orbital occupied by the electron and the orbital angular momentum of the electron, is in motion.

l takes values from 0 to n – 1 for these values the orbitals are designated by letters s, p, d, f, ………….. etc.

l also governs the degree with which the electron is attached to nucleus. The larger the value of l, the smaller is the bond with which it is maintained with the nucleus.

3) Magnetic orbital quantum number (m_l) :
The orientation of orbital with external magnetic field determines magnetic orbital quantum number.

m_l has integer values between – l and l including zero.

The number of values for m, are 2l + l, which give the number of orbitals per sub-shell. The maximum number of electrons in orbitals in the sub-shell is 2 (2l + l).

Question 9.
What is nl^x method? How is it useful? (AS1)
(OR)
What is nl^x method? How is it useful in electronic configuration?
Answer:
The shorthand notation consists of the principal energy level (n value) the letter representing sub – level (l value), and the number of electrons (x) in the sub-shell is written as superscript nl^x.

It is useful in writing electron configuration of elements. For example, in Hydrogen (H), the set of quantum numbers is n = 1, l = 0, m_l = 0, m_s = ½ or – ½. The electronic configuration is

Question 10.
Following orbital diagram shows the electronic configuration of nitrogen atom. Which rule does not support this? (AS1)

(OR)
Write the correct electronic configuration of the given nitrogen atom with the help of Hund’s rule.
Answer:

This electron configuration does not support Hund’s rule.
According to Hund’s rule, the orbitals of equal energy are occupied with one elec-tron each before pairing of electrons starts.
Here, pairing of electrons in 2p_x orbital was taken place without filling of an elec-tron in 2p_z orbital.
Hence the correct electron configuration is as follows.

Question 11.
Which rule is violated in the electronic configuration 1s⁰ 2s² 2p⁴?
Answer:

Aufbau principle is violated in this electronic configuration because according to Aufbau principle, electron enters orbital of lowest energy.
Among 1s, 2s and 2p, Is has least energy.
So Is orbital must be filled before the electron should enter 2s.

Question 12.
Write the four quantum numbers for the differentiating electron of sodium (Na) atom. (AS1)
Answer:
The electronic configuration of sodium (Na) is 1s² 2s² 2p⁶ 3s¹. So the differentiating electron enters 3s. Therefore the four quantum numbers are

Question 13.
What is emission spectrum?
(OR)
When radiation is emitted what is the name given to such spectrum? Explain such spectrum.
Answer:

The spectrum produced by the emitted radiation is known as emission spectrum.
This spectrum corresponds to liberation of energy when an excited electron returns back to ground state.

Emission spectrum is of two types :

1) Continuous spectrum :
When white light passes through a prism it dissociates into seven colours. This spectrum is called continuous spectrum.

2) Discontinuous spectrum :
Discontinuous spectrum is of two types.

a) Line spectrum :
The spectrum with sharp and distinct lines. It is given by gaseous atoms.

b) Band spectrum :
The spectrum very closely spaced lines is known as band spectrum. It is given by molecule.

Question 14.
i) An electron in an atom has the following set of four quantum numbers to which orbital it belong to : (AS2)
Answer:

This electron belongs to 2s orbital.
Spin is in clockwise direction. ⇒ 2s¹

ii) Write the fojur quantum numbers for Is1 electron. (AS1)
Answer:
The four quantum numbers for Is1 electron are

Question 15.
Which electronic shell is at a higher energy level K or L? (AS2)
Answer:
L – shell is at higher energy level, because it is far from nucleus than K shell.

Question 16.
Collect the information regarding wavelengths and corresponding frequencies of three primary colours red, blue and green. (AS4)
Answer:
The wavelengths and corresponding frequencies of three primary colours red, blue and green are given below.

Primary colours	Wavelength in nm (1 nm = 10^-9m)	Frequency in Hz (Hertz)
Red	700	4.29 × 10¹⁴
Green	530	5.66 × 10¹⁴
Blue	470	6.38 × 10¹⁴

Question 17.
The wavelength of a radio wave is 1.0 m. Find its frequency. (AS7)
Answer:
c = 3 × 10⁸ m/s ; λ = 1m ; c = vλ ⇒ v = $\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^{8}}{1}$ = 3 × 10⁸ Hz.

Question 18.
Why are there exemptions in writing the electronic configurations of Chromium and Copper?
Answer:
1. Elements which have half-filled or completely filled orbitals have greater stability.

2. So in chromium and copper the electrons in 4s and 3d redistributes their energies to attain stability by acquiring half-filled and completely filled d-orbitals.

3. Hence the actual electronic configuration of chromium and copper are as follows.

Fill In The Blanks

1. If n = 1, then angular momention quantum number (l) = …………………
2. If a sub-shell is denoted as 2p, then its magnetic quantum number values are …………………, …………………, …………………
3. Maximum number of electrons that an M-shell contain is / are …………………
4. For ‘n’, the minimum value is ………………… and the maximum value is …………………
5. For?, the minimum value is ………………… and the maximum value is …………………
6. For’m/ the minimum value is ………………… and the maximum value is …………………
7. The value of ‘ms’ for an electron spinning in clockwise direction is ………………… and for anti-clockwise direction is …………………
Answer:

0
– 1, 0, + 1
18
1, – ∞
0, (n – 1)
– l, + l
+ ½, – ½

Multiple Choice Questions

1. An emission spectrum consists of bright spectral lines on a dark back ground. Which one of the following does not correspond to the bright spectral lines?
A) Frequency of emitted radiation
B) Wavelength of emitted radiation
C) Energy of emitted radiations
D) Velocity of light
Answer:
D) Velocity of light

2. The maximum number of electrons that can be accommodated in the L-shell of an atom is
A) 2
B) 4
C) 8
D) 16
Answer:
C) 8

3. If l = 1 for an atom, then the number of orbitals in its sub-shell is
A) 1
B) 2
C) 3
D) 0
Answer:
C) 3

4. The quantum number which explains about size and energy of the orbit or shell is
A) n
B) l
C) m_l
D) m_s
Answer:
A) n

10th Class Chemistry 8th Lesson Structure of Atom InText Questions and Answers

10th Class Chemistry Textbook Page No. 112

Question 1.
How many colours are there in a rainbow? What are they?
Answer:
There are seven colours in a rainbow. They are Violet, Indigo, Blue, Green, Yellow,Orange and Red.

Question 2.
What are the characteristics of electromagnetic waves?
A.nswer:
Electromagnetic energy is characterised by wavelength (l) and frequency (u).

10th Class Chemistry Textbook Page No. 113

Question 3.
Can we apply this equation c = uA, to a sound wave?
Answer:
Yes. It is a universal relationship and applies to all waves.

10th Class Chemistry Textbook Page No. 114

Question 4.
What happens when you heat an iron rod on a flame? Do you find any change in colour while heating an iron rod?
Answer:

When we heat an iron rod, some amount of heat energy that was absorbed by iron rod is emitted as light.
First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).
If we go on heating the rod, it turns into white light which includes all visible wavelengths.
So we find some changes in colour while heating an iron rod.

Question 5.
Do you observe any other colour at the same time when one colour is emitted?
Answer:
While heating the rod if the temperature is high enough, other colours will also be emitted, but due to higher intensity of one particular emitted colour (eg.: red), others cannot be observed.

Question 6.
How do various colours come from fire works?
(OR)
Do you enjoy Deepavali fire works? Variety of colours is seen from fire works. How do these colours come from fire works?
Answer:
Yes. The electrons present in atoms of elements absorb energy and move to excited states and they return to ground state with emission of energy in visible spectrum. So the colours observed during fire works are the emitted energy by various elements in different fire works.

10th Class Chemistry Textbook Page No. 115

Question 7.
Do you observe yellow light in street lamps? Which will produce yellow light?
Answer:
Yes, sodium vapours produce yellow light in street lamps.

Question 8.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:

All the materials are made up of atoms and molecules. These atoms and molecules possess certain fixed energy.
An atom or molecule having lowest possible energy is said to be in ground state.
When we heat the materials the electrons of these atoms gain energy and move to excited states (higher energy state).
An atom of molecule in excited state can emit light to lower its energy in order to get stability and come back to ground state.
Light emitted in such process has certain fixed wavelength for one kind of atoms.
The light emitted by different kinds of atoms is different because the excited states electrons will go are different. So different elements produce different flame colours.

Question 9.
What happens when an electron gains energy?
Answer:
The electron moves to higher energy level called the excited state.

10th Class Chemistry Textbook Page No. 116

Question 10.
Does the electron retain the energy forever?
Answer:
The electron loses the energy and comes back to its ground state. The energy emitted by the electron is seen in the form of electromagnetic energy.

Question 11.
Did Bohr’s model account for the splitting of line spectra of a hydrogen atom into finer lines?
Answer:
No, Bohr’s model failed to account for splitting of line spectra.

Question 12.
Why is the electron in an atom restricted to revolve around the nucleus at certain fixed distances?
Answer:
In order to explain the atomic spectra, Bohr-Sommerfeld model proposed that the electrons are restricted to revolve around the nucleus at certain fixed distances.

10th Class Chemistry Textbook Page No. 117

Question 13.
Do the electrons follow defined paths around the nucleus?
Answer:
No, they revolve around the nucleus in a region called orbital.

Question 14.
What is the velocity of the electron?
Answer:
It is very close to light.

Question 15.
Is it possible to find exact position of electron? How do you find the position and velocity of an electron?
Answer:
No, as the electrons are very small, light of very short wavelength is required for this task.

This short wavelength light interacts with the electron and disturbs the motion of electron. So it is not possible to find exact position and velocity of electron simultaneously. Whereas we can find the region where the probability of finding electron is more.

Question 16.
Do atoms have a definite boundary, as suggested by Bohr’s model?
Answer:
Yes, atoms have definite boundary.

Question 17.
What do we call the region of space where the electron might be, at a given time?
Answer:
The region of space around the nucleus where the probability of finding an electron is maximum, called an orbital.

10th Class Chemistry Textbook Page No. 118

Question 18.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nucleus where the electrons are found and also their energies.

Question 19.
What does each quantum number signify?
Answer:
The quantum numbers signify the probability of finding electron in the space around nucleus.

10th Class Chemistry Textbook Page No. 119

Question 20.
What is the maximum value of/for n = 4?
Answer:
The maximum value of / for n = 4 is 3.

Question 21.
How many values can l have for n = 4?
Answer:
l takes values from 0 to n – 1. So l has 4 values for n = 4. Those values are 0, 1,2, 3.

Question 22.
Do all the p-orbitals have the same energy? A. Orbitals in the sub-shell belonging to same shell possess same energy but they differ in their orientations.

10th Class Chemistry Textbook Page No. 121

Question 23.
How are two electrons in the Helium atom arranged?
Answer:
They are arranged in pair in Is orbital and the electronic configuration is 1s².

10th Class Chemistry Textbook Page No. 122

Question 24.
What are the spins of two electrons in an orbital?
Answer:
The two electrons in an orbital have opposite spins. If one is clockwise spin, then other electron has anti-clockwise spin.

Question 25.
How many electrons can occupy an orbital?
Answer:
An orbital can hold only two electrons.

10th Class Chemistry 8th Lesson Structure of Atom Activities

Activity – 1

Question 1.
Explain the wave nature of light.
(OR)
How does light behave? Explain.
Answer:

Light is an electromagnetic wave.
Electromagnetic waves are produced when an electric charge vibrates.
This vibrating electric charge creates a change in the electric field. The changing electric field creates a changing magnetic field.
This process continues with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.
This electromagnetic wave is produced.

Activity – 2

Question 2.
Write an activity which shows metal produces colour in flame.
(OR)
‘Metal produces colour in a flame.’ Prove the statement by giving examples.
Answer:
A)

Take a pinch of cupric chloride in a watch glass and make a paste with concentrated hydrochloric acid.
Take this paste on a platinum loop and introduce it into a non-luminous flame.
Cupric chloride produces a green colour flame.

Take a pinch of strontium chloride in a watch glass and make a paste with concentrated hydrochloric acid.
Take this paste on a platinum loop and introduce it into a non-luminous flame.
Strontium chloride produces a crimson red flame.

Activity – 3

Question 3.
Complete the electronic configuration of the following elements.

Answer:

AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World

March 29, 2023February 1, 2023 by AP Board

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What is the function of lens in human eye?
Answer:
The eye lens forms a real image on retina then we can see the objects.

Question 2.
How does it help to see objects at long distances and short distances?
Answer:
Eye lens can adjust itself in shape, so that it helps to see the objects at long and short distances.

Question 3.
How is it possible to get the image at the same distance on the retina?
Answer:
When the eye is focussed at distant object, the ciliary muscles are relaxed. So the focal length of eye lens adjusted itself which is equal to the distance of object from the retina. Then we can see the object clearly.
When the eye is focussed on a closer object the ciliary muscles adjust the focal length in such a way that the image is formed on retina and we see the object clearly.

Question 4.
Are we able to see all objects in front of our eye clearly?
Answer:
Yes. By maintaining the 25 cm distance of the object from our eye, we can see the objects in front of our eye.

Question 5.
How do the lenses used in spectacles correct defects of vision?
Answer:
To form image on retina.
To correct
myopia- Concave lens
hypermetropia i- Convex lens

Improve Your Learning

Question 1.
How do you correct the eye defect Myopia? (AS1)
Answer:

Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect in vision is called ‘Myopia’ or ‘Near sightedness’.
Myopia is corrected by using a con-cave lens of focal length equal to the distance of the far point F from the eye.
This lens diverges the parallel rays from distant object as if they are coming from the far point.
Finally the eye lens forms a clear im-age at the retina.

Question 2.
Explain the correction of the eye defect Hypermetropia. (AS1)
Answer:

A person with hypermetropia can see distant objects clearly but cannot see objects at near distances. This is also known as ‘far sightedness’.
Eye lens can form a clear image on the retina when any object is placed beyond near point.
To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point at H, when the object is between H and L. This is possible only when a double con¬vex lens is used.
The image acts like an object for the eye lens. Hence final image due to eye is formed at retina.

Question 3.
How do you find experimentally the refractive index of material of a prism? (AS1)
(OR)
Write the experimental procedure in finding the refractive index of a prism.
(OR)
Which quantity will decide whether a given medium is denser or rarer? How do you find that quantity of prism experimentally?
Answer:
Refractive index decides whether a medium is denser or rarer.
Aim :
To find the refractive index of a prism.

Materials required :
Glass prism, white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Procedure :

Keep a prism on white chart.
Draw the triangular base of the prism with pencil.
Remove the prism.
The shape of the outline drawn prism is triangle and name its vertices as P, Q and R.
PQ and PR be the refracting surfaces.
Find the angle between PQ and PR. This is the angle of the prism (A) (or) Refracting angle.
Mark M on PQ and draw a perpendicular line to PQ at M.
Place the centre of the protractor at M, along the normal and mark an angle of 30° and then draw a line up to M. This line denotes incident ray. This angle is called angle of incidence.
Place the prism in its position (triangle) again.
Now fix two pins vertically on the line at points A and B .
See the images of pins through the 2nd refracting side (PR).
Fix another two pins at points C and D such that all the four pins appear to lie along the same line.
Remove the pins and prism, join the pin-holes. Draw the incident and emergent rays.
The angle between the normal and the emergent ray at N is the angle of emergence.
The line passing through the points A, B, M, N, C and D represents the path of light when it suffers refraction through prism.

The angle of deviation :

Extend incident and emergent rays are intercept at a point ‘O’.
The angle between these two rays is the angle of deviation (d).
Note the emergent deviation angles for different values of i, in the given table.

Draw the graph between angle of incidence on X – axis and the angle of deviation on Y – axis.
We notice that the angle of deviation decreases first and then increases with increase of the angle of incidence.
Mark points on a graph paper and join the points to obtain a graph (smooth curve).
Draw a tangent line to the curve, which parallel to X-axis, at the lowest point of the graph.

The point where this line cuts the Y- axis gives the angle of minimum deviation. It is denoted by D.
From the graph, we notice that, at angle of minimum deviation, the angle of incidence is equal to the angle of emergence.
By finding A and D we can find refractive index of prism by using formula .

Question 4.
Explain the formation of rainbow. (AS1)
(OR)
What is the natural spectrum occuring in sky? Explain the formation of that spectrum.
Answer:

A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on rain drops.
It is combined result of reflection, refraction and dispersion of sunlight from water droplets, in atmosphere.
It always forms in the direction opposite to the sun.
To see a rainbow, the sun be must behind us and the water droplets falls in front of us.
When a sunlight enters a spherical raindrop, it is refracted and dispersed. The different colours of light are bent in different angles.
When different colours of light falls on the back inner surface of drop, it (Water drop) reflects (different colours of light) interwnally (total internal reflection).
The water drops again refract the different colours, when it comes out from the raindrop.
After leaving this different colours from the raindrop as rainbow, reach our eye. Thus, we see a rainbow.

Question 5.
Explain briefly the reason for the blue of the sky. (AS1)
Answer:

The reason for blue sky is due to the molecules N₂ and O₂.
The size of these molecules are comparable to the wavelength of blue light.
These molecules act as scattering centres for scattering of blue light.
So scattering of blue light by molecules of N₂ and O₂ is responsible for blue of the sky.

Question 6.
Explain two activities for the formation of artificial rainbow. (AS1)
(OR)
Give two activities for the formation of the artificial rainbow? And explain it.
(OR)
Suggest an experiment to create a rainbow in your classroom and explain the procedure.
Answer:
Activity -1 :

Select a white coated wall on which the sun rays fdll.
Stand in front of a wall in such a way that the sun rays fall on your back.
Hold a tube through which water is flowing.
Place your finger in the tube to obstruct the flow of water.
Water comes out from small gaps between the tube and finger like a fountain.
Observe the changes on wall while showering the water.

Activity – II : (Activity – 4)

Take a metal tray and fill with water.
Place a mirror in water such that it makes an angle to the water surface.
Now focus white light on the mirror through the water.
Keep a white card board sheet above the water surface.
We may observe the colours VIBGYOR on the board.
The splitting of white light into different colours (VIBGYOR) is called dispersion.
So consider a white light is a collection of waves with different wavelengths.
Violet has shortest wavelength, and red has longest wavelength.

Question 7.
Derive an expression for the refractive index of the material of a prism. (AS1)
Answer:
Derivation of formula for refractive index of a prism :
PQ and PR are refracting surfaces of prism, i, is an angle of incidence, i2 is an angle of emergence, is an angle of refraction of first surface and r2 is an angle of incidence on second surface.

1. From triangle OMN, we get
d = i₁ – r₁ + i₂ – r₂
∴ d = (i₁ + i₂)-(r₁+ r₂) ……… (1)
2. From triangle PMN, we have
∠A + (90° – r₁) + (90° – r₂) = 180°
⇒ r₁ + r₂ = A ……… (2)
3. From (1) and (2),
⇒ d = (i₁ + i₂) – A
⇒ A + d = i₁ + i₂
4. This is the relation between angle of incidence, angle of emergence, angle of deviation and angle of prism.

5. From Snell’s law, we know that
n₁ sin i = n₂ sin r

6. Let ‘n’ be the refractive index of the prism.
7. Using Snell’s law at M, refractive index of air n₁ = 1; i = i₁; n₂ = n; r = r₁.
⇒ sin i₁ = n sin r₁ …………. (4)
8. Similarly, at N, n₁ = n; i = r₂; n₂ = 1; i₁ = i₂
⇒ n sin r₂ = n sin i₂ ………… (5)
9. We know that at the angle of minimum deviation position (D), i.e. i₁ = i₂
10. We will notice that MN is parallel to the base of prism QR.

14. This is the formula for the refractive index of the prism.

Question 8.
Light of wavelength λ₁ enters a medium with refractive index n₂ from a medium with refractive index n₁ What is the wavelength of light in second medium? (AS1)
Answer:
Wavelength of first medium =λ₁
Refractive index of first medium = n₁
Let the wavelength of second medium = λ₂,
Refractive index of second medium = n₂
Light enters from first medium to second medium = $\Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}$ (∵ υ = const)
$\Rightarrow \lambda_{2}=\frac{\lambda_{1} \mathrm{n}_{1}}{\mathrm{n}_{2}}$

Note:
For the below questions the following options are given. Choose the correct option by making hypothesis based on given assertion and reason. Give an explanation.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.

Question 9.
Assertion (A) : The refractive index of a prism depends only on the kind of glass of which it is made of and the colour of light. (AS2)
Reason (R) : The refractive index of a prism depends on the refracting angle of the prism and the angle of minimum deviation.
Answer:
(b) Both A and R are true and R is not the correct explanation of A.

Question 10.
Assertion (A) : Blue colour of sky appears due to scattering of light.
Reason (R) : Blue colour has shortest wavelength among all colours of white light. (AS2)
Answer:
(C) A is true but R is false.

Question 11.
Suggest an experiment to produce a rainbow in your classroom and explain the procedure. (AS3)
Answer:
Activity to produce a rainbow in classroom :

Take a prism and place it on the table near a vertical white wall.
Take a thin wooden plank, make a small hole in it and fix it vertically on the table.
Place the prism between the wooden plank and wall.
Place a white light source behind the hole of the wooden plank. Switch on the light.
Adjust the height of the prism such that the light falls on one of the lateral surfaces.
Observe the changes in the emerged ray of the prism.
Adjust the prism by slightly rotating it till you get an image on the wall.
We observe a band of different colours on the wall.
These colours are nearly equal to the colours of the rainbow, i.e., VIBGYOR.

Question 12.
Prisms are used in binoculars. Collect information why prisms are used in binoculars. (AS4)
Answer:

Binoculars consists of a pair of identical or mirror symmetrical telescope mounted side by side and aligned to point accurately in the same direction, allowing the viewer to use both eyes, when viewing distant objects.
The size of binoculars is reduced by using prisms.
We get good image with more brightness.
Objective size and optical quality should be increased by using prisms in binoculars.

Question 13.
Incident ray on one of the face (AB) of prism and emergent ray from the face AC are given in figure. Complete the ray diagram. (AS5)

Answer:

Question 14.
How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky? (AS6)
(OR)
How do you appreciate the nature of molecules, responsible for the blue of the sky?
Answer:

The sky appear blue due to atmospheric refraction and scattering of light through molecules.
Molecules are scattering centres.
The reason to blue sky is due to the molecules N2 and 02.
The sizes of these molecules are comparable to the wavelength of blue light.
In the absence of molecules there will be no scattering of sunlight and the sky will appear dark.
We should appreciate the molecules which are scattering centres.

Question 15.
Eye is the only organ to visualise the colourful world around us. This is possible due to accommodation of eye lens. Prepare a six line stanza expressing your wonderful feelings. (AS6)
Answer:
“Many people simply see
The world in black and white
But through my eyes the world’s alive
So colourful and bright
Don’t close your mind to the sights
That light up the night and day
There’s so much to see here on this earth
And not a rupee do you have to pay”
The most obvious things aren’t the ones
There is beauty in the unknown
with willing eyes and an open mind
The true wonders you will be shown

Question 16.
How do you appreciate the working of ciliary muscles in the eye? (AS6)
(OR)
Which muscles are helpful in changing the focal length of eye lens? How do you appreciate those muscles?
Answer:

Ciliary muscles are helpful in changing focal length of eye lens.
The ciliary muscles which attached with eye lens help to change the focal length of eye lens.
When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
The parallel rays coming into the eye are focussed on the retina then we can see the object clearly.
When the eye is focussed on a nearer object the muscles are strained so the focal length of eye-lens decreases. The ciliary muscles adjust the focal length and the image is formed on retina then we can see the object.
Their process of adjusting focal length of eve lens is called accomodation.
So ciliary muscles must be appreciated for its accomodation of eye lens.

Question 17.
Why does the sky sometimes appear white? (AS7)
(OR)
The sky sometimes appears white. What is the reason behind it?
Answer:

Our atmosphere contains atoms and molecules of different sizes.
According to their sizes, they are able to scatter different wavelengths of light.
For example, the size of the water molecule is greater than the size of the N₂ or O₂.
It acts as a scattering centre for light other frequencies which are lower than the frequency of blue light.
On a hot day due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies (other than blue).
All such colours of other frequencies reach our eye and white colour is appeared to us.

Question 18.
Glass is known to be transparent material. But ground glass is opaque and white in colour. Why? (AS7)
Answer:

Ground glass is glass whose surface has been ground that produces a flat but rough finish.
Ground glass has the effect of rendering the glass translucent by scattering of light during transmission thus blurring visibility while still transmitting light.
To get more permanent frost, the glass may be ground by rubbing with some gritty substance.

Question 19.
If a white sheet of paper is stained with oil, the paper turns transparent. Why? (AS7)
(OR)
What is the reason behind the paper turns transparent when it is immersed in oil?
Answer:

The paper fibres have higher index of refraction probably much greater than 1.5.
The oil or fat also has a high index of refraction so that it nearly matches the index of refraction of the paper fibres and it reduces the scattering significantly.
The fat adhering to the cellulose fibers lowers the index of refraction of the cellulose and also fills in air voids, so that visible light passes through the bag with significantly less scattering.
The oil connects the fibres in the paper with a liquid which can transmit by refraction (rather than scatter) light that falls upon it. As a result, the paper stained with oil is turned transparent.

Question 20.
A light ray falls on one of the faces of a prism at an angle 40° so that it suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface. (AS7)
Answer:
Given that, incident ray on one of the prisms (i₁) = 40°
Angle of minimum deviation (D) = 30°
Angle of prism (A) = ?
A+D = 2i₁
⇒ A = 2i₁ – D
⇒ A = 2(40°) -30°
= 80° – 30°
= 50°
∴ Angle of prims (A) = 50°
Angle of refraction (r₁) = ?

Question 21.
The focal length of a lens suggested to a person with Hypermetropia is 100 cm. Find the distance of near point and power of the lens. (AS7)
Answer:
i) The distance of near point :
If the distance of near point is ‘d’ and focal length is ‘f’ then the relation between

Question 22.
A person is viewing an extended object. If a converging lens is placed in front of his eye, will he feel that the size of object has increased? Why? (AS7)
Answer:

A simple magnifier allows us to put the object closer to the eye than we could normally focus and forms an enlarged virtual images.
The principle behind this is angular magnification.
The magnification is Ma = $\frac{25}{f}$ for the close focus point, but since that causes eye strain, it is usually desirable to put the images at infinity giving Ma = $\frac{25}{f}$
So he fees the size of object is increased. The reason is mentioned.
Angular magnification: ‘The ratio of the angle substended at the eye by the image formed by an optical instrument to the angle substended at the eye by the object being viewed.”

Fill In The Blanks

1. The value of least distance of distinct vision is about ……………………… .
2. The distance between the eye lens and retina is about ……………………… .
3. The maximum focal length of the eye lens is about ……………………… .
4. The eye lens can change its focal length due to working of ……………………… muscles.
5. The power of lens is ID then focal length is ……………………… .
6. Myopia can be corrected by using ……………………… lens.
7. Hypermetropia can be corrected by using ……………………… lens.
8. In minimum deviation position of prism, the angle of incidence is equal to angle of ……………………… .
9. The splitting of white light into different colours (VIBGYOR) is called ……………………… .
10. During refraction of light, the character of light which does not change is ……………………… .
Answer:

25 cm
2.5 cm
2.5 cm
ciliary
100 cm
bi-concave
bi-convex
emergence
dispersion of light
frequency

Multiple Choice Questions

1. The size of an object as perceived by an eye depends primarily on
A) actual size of the object
B) distance of the object from the eye
C) aperture of the pupil
D) size if the image formed on the retina
Answer:
B) distance of the object from the eye

2. When objects at different distances are seen by the eye which of the following remains constant?
A) focal length of eye-lens
B) object distance from eye-lens
C) the radii of curvature of eye-lens
D) image distance from eye-lens
Answer:
D) image distance from eye-lens

3. During refraction, …………….. will not change.
A) wavelength
B) frequency
C) speed of light
D) all the above
Answer:
B) frequency

4. A ray of light falls on one of the lateral surfaces of an equilateral glass prism placed on the horizontal surface of a table as shown in figure. For minimum deviation of ray, which of the following is true?

A) PQ is horizontal
B) ‘QR’ is horizontal
C) ‘RS’ is horizontal
D) Either ‘PQ’ or ‘RS’ is horizontal
Answer:
B) ‘QR’ is horizontal

5. Far point of a person is 5 m. In order that he has normal vision what ki nd of spectacles should he use?
A) concave lens with focal length 5 m
B) concave lens with focal length 10 m
C) convex lens with focal length 5 m
D) convex lens with focal length 2.5 mm
Answer:
A) concave lens with focal length 5 m

6. The process of re-emission of absorbed light in all directions with different intensi¬ties by the atom or molecule is called ……………
A) scattering of light
B) dispersion of light
C) reflection of light
D) refraction of light
Answer:
A) scattering of light

10th Class Physics 7th Lesson Human Eye and Colourful World InText Questions and Answers

10th Class Physics Textbook Page No. 104

Question 1.
Can you imagine the shape of rainbow when observed during travel in an airplane?
Answer:

The shape of rainbow, when observed during travel in an airplane is circular as shown in the following figure.

10th Class Physics Textbook Page No. 88

Question 2.
Why do the values of least distance of distinct vision and angle of vision change with person and age?
Answer:

The ciliary muscle helps the eye lens to change its focal length by changing radii of curvature of eye lens.
When the eye is focussed on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
The working of ciliary muscle in eye changes from person to person.
So, the values of least distance of distinct vision and angle of vision changes with person and age.

10th Class Physics Textbook Page No. 89

Question 3.
How can we get same image distance for various positions of objects?
Answer:
For different positions of object, the image distance remains constant only when focal length of lens changes.

Question 4.
Can you answer above question using concepts of refraction through lenses?
Answer:
The focal length of a lens depends on the material by which it has been made and radii of curvatures of lens. We need to change focal length of eye lens to get same image distance for various positions of object.

10th Class Physics Textbook Page No. 90

Question 5.
How does eye lens change its focal length?
(OR)
What is the role of ciliary muscles in the eye ? Write the answer in one or two sentences only.
Answer:
The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing the radii of curvature of eye lens.

Question 6.
How does this change (eye lens changes its focal lengths) take place in eye ball?
Answer:
When the eye is focussed on a distant object the ciliary muscles are relaxed. So the focal length of eye lens has its maximum value which is equal to its distance from the retina. The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.

Question 7.
Does eye lens form a real image or virtual image?
Answer:
The eye lens forms a real and inverted image of an object on the retina.

Question 8.
How does the image formed on retina help us to perceive the object without change in its shape, size and colour?
Answer:

The eye lens forms a real and inverted image of an object on the retina.
This retina is a delicate membrane, which contains large number of receptors (125 million) called ‘rods’ and ‘cones’.
They receive the light signals and identify the colour, and the intensity of light.
These signals are transmitted to the brain through optic-nerve fibres.
The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 9.
Is there any limit to change of focal length of eye lens?
Answer:
Yes. When the object is at infinity, the parallel rays from the object falling on the eye lens are refracted. They form a point-sized image on retina. In this situation, eye lens has a maximum focal length.

Question 10.
What are the maximum and minimum focal lengths of the eye lens?
Answer:
Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.

Question 11.
How can we find the maximum and minimum focal lengths of the eye lens?
(OR)
Calculate the maximum and minimum focal lengths of the eye lens.
Answer:’
a) When the object is at infinity,
u = – ∞ ; v = 2.5 cm (image distance which is equal to distance between eye lens and retina)

b) Object at a distance of 25 cm from eye. In this case, eye has minimum focal length.
Here u = -25 cm ; v 2.5 cm

10th Class Physics Textbook Page No. 91

Question 12.
What happens if eye lens is not able to adjust its focal length?
Answer:
In this case the person cannot see the object clearly and comfortably.

Question 13.
What happens if the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?
Answer:
The vision (image) becomes blurred due to defects of eye lens.

10th Class Physics Textbook Page No. 92

Question 14.
What can we do to correct myopia?
Answer:
To correct myopia, we use concave lens.

10th Class Physics Textbook Page No. 93

Question 15.
How can you decide the focal length of the lens to be used to correct myopia?
Answer:
Assume that the object distance (u) is infinity and image distance (v) is equal to distance of far point.
u = – ∞ ; v = distance of far point = – D
Let ‘f be the focal length of bi-concave lens.

Here ‘f is negative showing that it is a concave lens.

Question 16.
What happens when the eye has a minimum focal length greater than 2.27 cm?
Answer:
In this case, the rays coming from the nearby object after refraction at eye lens, forms image beyond the retina.

10th Class Physics Textbook Page No. 94

Question 17.
How can you decide the focal length of convex lens to be used?
Answer:
Here object distance (u) = -25 cm
Image distance (v) = – d (distance of near point)
Let ‘f be the focal length of bi-convex lens.

If d > 25 cm, then ‘f’ becomes positive then we need to use bi-convex lens to correct hypermetropia.

10th Class Physics Textbook Page No. 95

Question 18.
Have you ever observed details in the prescription?
Answer:
A prescription that contains some information regarding type of lens to be used to correct vision.

Question 19.
What does it (“my sight is increased or decreased”) mean?
Answer:
Usually doctors, after testing the defects of vision prescribe corrective lenses indicating
their power which determines the type of lens to be used and its focal length.

Question 20.
What do you mean by power of lens ? flftorH)
Answer:
The reciprocal of focal length is called power of lens.

Question 21.
Doctor advised to use 2D lens. What is its focal length?
Answer:
Given power of lens P = 2D

∴ Focal length of lens (f) = 50 cm.

10th Class Physics Textbook Page No. 96

Question 22.
How could the white light of the sun give us various colours of the rainbow?
Answer:
Due to reflection, refraction and dispersion of sunlight.

Question 23.
What happens to a light ray when it passes through a transparent medium bounded by plane surfaces which are inclined to each other?
Answer:
When light incident on one of the plane surfaces and emerges from the other.

Question 24.
What is a prism?
Answer:
A prism is a transparent medium separated from the surrounding medium by consisting two refracting plane surfaces which are inclined.

10th Class Physics Textbook Page No. 97

Question 25.
What is the shape of the outline drawn for a prism?
Answer:
The outline drawn for a prism is in a triangle shape.

10th Class Physics Textbook Page No. 98

Question 26.
How do you find the angle of deviation?
Answer:
The angle between the extended incident and emergent rays is called angle of deviation.
(OR)
Extend both incident and emergent rays till they meet at a point ‘O’. Measure the angle between these two rays. This is the angle of deviation.

Question 27.
What do you notice from the angle of deviation?
Answer:
The angle of deviation decreases first and then increases with increase in the angle of incidence.

Question 28.
Can you draw a graph between angle of incidence and angle of deviation?
Answer:

Yes, we can draw the graph between angle of incidence and angle of deviation.

Question 29.
From the graph, can you find the minimum of the angle of deviation?
Answer:
Yes, we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts Y – axis gives the angle of minimum deviation.

Question 30.
Is there any relation between the angle of incidence (i) angle of emergence (r) and piangle of deviation (d)?
Answer:
(i₁ + i₂)= A + D
i + r = A + D

10th Class Physics Textbook Page No. 101

Question 31.
In activity-3,we noticed that light has chosen different paths. Does this mean that the refractive index of the prism varies from colour to colour?
Answer:
Yes, refractive index of the prism varies from colour to colour.

Question 32.
Is the speed of light of each colour different?
Answer:
In vacuum – No, the speed of each colour is constant.
In medium – Speed is different for different colours.

Question 33.
Can you guess now, why light splits into different colours when it passes through a prism?
Answer:
Due to dispersion of light and different wavelength of colours in medium.

10th Class Physics Textbook Page No. 102

Question 34.
Does it (light passing through a prism) split into more colours? Why?
Answer:
We know the frequency of light is the properly of the source and it is equal to number of waves leaving the source per second. This cannot be changed by any medium. Hence frequency doesn’t change due to refraction. The coloured light passing through any transparent medium retains its colour.

Question 35.
Can you give an example in nature, where you observe colours as seen in activity 3?
Answer:
Yes, in rainbow. It is a good example of dispersion of light.

Question 36.
When do you see a rainbow in the sky?
Answer:
Dut to the refraction, reflection and dispersion of sunlight, when the sunlight passes through the rain drops, we can see the rainbow in the sky.

Question 37.
Can we create a rainbow artificially?
Answer:
Yes, we can create a rainbow artificially.

10th Class Physics Textbook Page No. 104

Question 38.
Why does the light dispersed by the raindrops appear as a bow?
Answer:

A rainbow is not the flat two dimensional arc as it appears to us.
The rainbow we see is actually a three dimensional cone with the tip of our eye.
All the drops that disperse the light towards us lie in the shape of the cone – a cone of different layers.
The drops that disperse red colour to our eye are on the outermost layer of the cone, similarly, the drops that disperse orange colour to our eye are on the layer of the cone beneath the red colour cone.
In this way the cone responsible for yellow lies beneath orange and so on it till the violet colour cone becomes the innermost cone.

Question 39.
Why is the sky blue?
Answer:
A clear cloudless day – time sky is blue because molecules in the air scatter blue light from the sun more than thev scatter red liuht.

Question 40.
What is scattering?
A. Atoms or molecules which are exposed to absorb light energy and emit some part of the light energy in different directions is called scattering of light.

10th Class Physics Textbook Page No. 106

Question 41.
Why is that the sky appears white sometimes when you view it in certain direction on hot days?
Answer:

On a hot day, due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere.
These water molecules scatter the colours of other frequencies (other than blue).
All such colours of other frequencies reach our eye and the sky appears white.

Question 42.
Do you know the reasons for appearance the red colour of sun during sunrise and at sunset?
(OR)
Sun appears red in colour during sunrise and sunset. Give reason.
(OR)
Why does sky appear red at Sunshine and Sunset?
Answer:

The light rays from the sun travel more distance in atmosphere to reach our eye in morning and evening times.
During sunrise and sunset except red light all colours scatter more and vanish before they reach us.
Since scattering of red light is very less, it reaches us.
As a result sun appears red in colour during sunrise and sunset.

Question 43.
Can you guess the reason why sun does not appear red during noon hours?
Answer:
During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence the sunlight appears white in noon hours.

10th Class Physics 7th Lesson Human Eye and Colourful World Activities

Activity – 1

Question 1.
How do you find least distance of distinct vision?
(OR)
What is the least distance a person can see an object comfortably and distinctly known as ? Write an activity to find that (least, distance of distinct vision) distance.
Answer:

The least distance a person can see an object comfortably and distinctly is known as least distance of distinct vision.
Hold the textbook at certain distance with your hands.
Try to read the contents on the page.
Gradually move the book towards eye, till it reaches very close to your eyes.
You may see that printed letters on the page appear blurred or you feel strain to read.
Now move the book backwards to a position where you can see clear printed letters without strain.
Ask your friend to measure distance between your eye and textbook at this position.
Note down its value.
Repeat this activity with other friends and note down the distances for distinct vision in each case.
Find the average of all these distances of clear vision.
You will notice that to see an object comfortably and distinctly, you must hold it at a distance about 25 cm from your eyes.
This 25 cm distance is called least distance of distinct vision.
This value varies from person to person and with age.

Activity – 2

Question 2.
How can you find angle of vision?
(OR)
What is maximun angle a person is able to see whole object? Write an activity to find that angle.
Answer:

The maximum angle, at which we are able to see the whole object is called angle of vision.
Arrange a retort stand.
Collect a few wooden sticks (or) PVC pipes that are used for electric wiring.
Prepare sticks or pipes of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
Keep the retort stand on a table and stand near the table such that vour head is beside the vertical stand.
Adjust the clamp on horizontal rod and fix at a distance of 25 cm from the eyes.
Ask one of your friends to fix a wooden stick of 30 cm height to the clamp in vertical position.
Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
If you are not able to see both ends of stick at this distance (25 cm), adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick at nearest possible distance from your eye.
Fix the clamp to the vertical stick at this position.
Without changing the position of the clamp on horizontal rod, replace this stick of 30 cm length.
Try to see the top and bottom of the stick simultaneously without any change in the position of eye.
Try the same activity with various lengths of the sticks.
You can see the whole object AB which is at a distance of 25 cm (least distance of clear vision) because the rays coming from the ends A and B of the object will enter the eye.
Similarly we can also see complete object CD with eye as explained above.
Let us assume that AB is moving closer to eye to a position A B .
You notice that you will be able to see only the part (EF) of the object A B because the rays coming from E and F enter your eye.

The rays coming from A and B cannot enter your eye.
The rays coming from the extreme ends of an object forms an angle at the eye.
If this angle is below 60°, we can see whole object.
If this angle is above 60°, then we see only the part of the object.
This maximum angle, at which we are able to see the whole object is called angle of vision.
The angle of vision for a healthy human being is about 60°.
It varies from person to person and with age.

Activity – 3

Question 3.
Describe an activity for dispersion of light.
(OR)
What is the name given to process when white light passes through a prism it splits into different colours ? Explain the process with an activity.
Answer:

The splitting of white light into different colours is called despersion of light.
Do this experiment in the dark room.
Take a prism and place it on the table near a vertical white wall.
Take a thin wooden plank.
Make a small hole in it and fix it vertically on the table.
Place the prism between the wooden plank and wall.
Place a white light source behind the hole of wooden plank.
Switch on the light.
The rays coming out of the hole of plank become a narrow beam of light.
Adjust the height of the prism such that the light falls on one of the lateral surfaces.
Observe the changes in emerged rays of the prism.
Adjust the prism by slightly rotating it till you get an image on the wall.
You will observe that white light is splitting into certain different colours.

Activity – 6

Question 4.
Describe an experiment for scattering of light.
(OR)
What is the principle involved in blue of the sky ? Explain the principle with an experiment?
Answer:

Take the solution of sodium-thio-Sulphate (hypo) with sulphuric acid in a glass beaker.
Place the beaker in which reaction is taking place in an open place where abundant sunlight is available.
Watch the formation of grains of sulphur and observe changes in beaker.
You will notice that sulphur precipitates as the reaction is in progress.
At the beginning, the grains of sulphur are smaller in size as the reaction progresses, their size increases due to precipitation.
Sulphur grains appear blue in colour at beginning and slowly their colour becomes white as their size increases.
The reason for this is scattering of light.
At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
Hence they appear blue in the beginning.
As the size of grains increases, their size becomes comparable to the wavelengths of other colours.
As a result of this, they act as scattering centres for other colours.
The combination of all these colours appears as white.

AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces

March 29, 2023February 1, 2023 by AP Board

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 6th Lesson Refraction of Light at Curved Surfaces

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Have you ever touched a magnifying glass with your hand?
Answer:
Yes.

Question 2.
Have you touched the glass in the spectacles used for reading with your hand?
A. Yes.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 3.
Is it a plane or curved surface?
Answer:
Curved surface.

Question 4.
Is it thicker in the middle or at the edge?
Answeer:
Magnifying glass and some spectacle are thicker in middle whereas some spectacles are thicker at edge.

Improve Your Learning

Question 1.
A man wants to get a picture of a zebra. He photographed a white donkey after fitting a glass, with black stripes, on to the lens of his camera. What photo will he get? Explain. (AS1)
(OR)
A person wants to get a picture of zebra and he photographed a white donkey fitting a glass with black stripes. Does he get photo of zebra? Explain.
Answer:
The person was unable to gel the picture <>l zebra because only two rays are enough to form complete image after convergence. So he will get the image of white donkey but the intensity may be less.
(OR)
He will get a picture of while donkey because e\ery part of lens forms an image so if you cover lens with stripes still it forms a complete image. However, the intensity of the image will be reduced.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 2.
Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram. (AS1)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 17$

Two lenses are placed in the path of parallel rays as shown in figure.
The first lens is placed in the direction of parallel lines, which converges at focus.
The second lens is arranged so that it is the focus of 2nd then emerging rays will be parallel.

Question 3.
The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. Where will the image be formed and what kind of image is it? (AS1)
Answer:
f = 20 cm (by sign conversion f = + 20 cm)
u = 60 cm (by sign conversion u = – 60 cm)
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 1$
Image will be formed at 30cm in between F₁, and 2F1. Image is real, inverted and diminished.

Question 4.
A double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5. Find the focal length ‘f’. (AS1)
(OR)
What is the focal length ‘f, when its double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5?
Answer:
R₁ = R₂ = R (suppose)
Focal length (f) = ?; Refractive index (n) = 1.5
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 2$
∴ Focal length of lens = Radius of curvature of surface.

Question 5.
Write the lens maker’s formula and explain the terms in it. (AS1)
(OR)
Ravi wants to make a lens. Which formula he has to follow ? Write the formula and explain the terms in it.
(OR)
Write lens formula.
Answer:
Lens maker’s formula:
$\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$
n = Refractive index of the medium
R₁ = Radius of curvature of 1 st surface
R₂ = Radius of curvature of 2nd surface
f = Focal length

Question 6.
How do you verify experimentally that the focal length of a convex lens is increased when it is kept in water? (AS1)
(OR)
Write an activity to show that the focal length of a lens depends on its surrounding medium.
Answer:
Aim :
To prove focal length of convex lens is increased when it is kept in water.

Apparatus :
Convex lens, water, cylindrical vessel, circular lens holder, stone.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 18$

Procedure :

Take a cylindrical vessel like glass tumbler.
Its height must be greater than the focal length of lens, (the around four times focal length of lens).
Keep a black stone inside the vessel at its bottom.
Pour the water into the vessel such that the height of the water level from the top of the stone is greater than the focal length of lens.
Now dip the lens horizontally using a circular lens holder.
Set the distance between stone and lens that is equal to or less than focal length of lens.
Now see the stone through the lens.
We can see the image of the stone.
If we dip the lens to a certain height which is greater than the focal length of lens in air, still we can see the image.
This shows that the focal length of convex lens has increased in water.
Thus we conclude that the focal length of lens depends upon the surrounding medium.

Note : For convenience, use 5 or 10 cm focal length convex lens.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 7.
How do you find the focal length of a lens experimentally? (AS1)
Answer:

Take the lens (Ex : Convex), which focused towards the distant object.
A white coated screen (Ex : White paper) is placed on the other side of the lens.
Adjust the screen till you get a clear image of the object.
At this position measure the distance between the lens and screen which is equal to the focal length of the lens.

Question 8.
Harsha tells Siddhu that the double convex lens always behaves like a convergent lens. But Siddhu knows that Harsha’s assertion is wrong and corrected Harsha by asking some questions. What are the questions asked by Siddhu? (As2)
Answer:
The questions asked by Siddhu :

Is the object placed beyond 2f point?
Is the object located at 2f point?
Is the object located in between the 2f and the focal point?
Is the object located at the focal point?
Is the object located in front of the focal point?
Is the lens kept in a medium with refractive index less than lens or more than lens?

Question 9.
Assertion (A): A person standing on the land appears taller than his actual height to a fish inside a pond. (AS2)
Reason (R) : Light bends away from the normal as it enters air from water.
Which of the following is correct? Explain.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.
Answer:
Answer a is correct.
Explanation :
Because the light travelling from water to air it bends away from the normal so the fish observes the apparent image of the person, appears taller than his original.

Question 10.
A convex lens is made up of three different materials as shown in the figure Q-10. How many of images does it form? (AS2)
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 3$
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 4$

A lens made of three different materials of refractive indices say n₁, n₂ and n₃.
These three materials will have three different refractive indices. Thus for a given object it forms three images.

Question 11.
Can a virtual image be photographed by a camera? (AS2)
Answer:
Yes, we can.
Ex : – A plane mirror forms a virtual image, we can able to take photograph of that image in plane mirror.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 12.
You have a lens. Suggest an experiment to find out the focal length of the lens. (AS3)
(OR)
Through an experiment, find out the focal length of the lens.
Answer:
Aim :
To find focal length of given lens.

Apparatus :
Object (candle), convex lens, v – stand, screen.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 19$

Procedure :

Take a v-stand and place it on a long table at the middle.
Place a convex lens on the v-stand. Imagine the principal axis of the lens.
Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.
Measure the distance of the image from the v-stand of lens (image distance V) and also measure the distance between the candle and stand of lens (object distance ‘u’). Record the values in the table.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 20$

Now place the candle at a distance of 60 cm from the lens, try to get an image of the candle flame on the other side on a screen. Adjust the screen till you get a clear image.
Measure the image distance V and object distance ‘u’ and record the values in table.
Repeat the experiment lor various object distances like 50 cm, 40 cm, 30 cm, etc. Measure the image distances in all cases and note them in table.
Using the formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$, find f in all the cases. We will observe the value ‘f is equal in all cases. This value off is the focal length of the given lens.

Question 13.
Let us assume a system that consists of two lenses with focal length f₁, and f₂ respectively. How do you find the focal length of the system experimentally, when
i) two lenses are touching each other
ii) they are separated by a distance ‘d’ with common principal axis? (AS3)
Answer:
Experimental Proof:
i) Two lenses are touching each other :
Aim :
To find focal length of combination of two convex lenses, touching each other. Material required : Convex lenses – 2 (with known focal lengths say f, and f2); V-stands – 2, candle, screen scale.

Procedure:

Place two V-stands with two convex lenses as they touch each other on a table.
Place a candle (object) far away from the lenses.
Adjust a screen, which is placed other side of the lenses until we get a clear image on it.
At that position, measure the image distance (v) and object distance (u).
Do this experiment for several object distances and record in the given table.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 21$

ii) They are separated by a distance of ‘d’ :
Procedure :

Now place v-stands along with lenses with distance’d’.
Do the same procedure again.
Record the observations in the given table.
Find the average of the ‘f’_comb.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 22$

Question 14.
Collect the information about the lenses available in an optical shop. Find out how the focal length of a lens may be determined by the given power’ of the lens. (AS4)
Answer:
I had collected the information regarding different lenses available at optical shops.
The relationship between power and focal length is power (D) = $\frac{1}{f}$. f is in meters.

Power of lens in diopters	Type of lens	Focal length
0.25	Convex	400 cm
0.5	Convex	200 cm
1	Convex	100 cm
-2	Concave	50 cm
– 1	Concave	– 100 cm
-0.5	Concave	– 200 cm
-0.25	Concave	– 400 cm

Question 15.
Collect the information about lenses used by Galileo in his telescope. (AS4)
(OR)
What lenses are used by Galileo in his telescope?
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 5$
A Galilean telescope is defined as having one convex lens and one concave lens. The concave lens serves as the ocular lens or the eye piece, while the convex lens serves as the objective. The lens are situated on either side of a tube such that the focal point of the ocular lens is the same as the focal point for the objective lens.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 16.
Use the data obtained by activity – 2 in table-1 of this lesson and draw the graphs of u vs v and $\frac{1}{u}$ vs $\frac{1}{v}$ (AS5)
(OR)
By obtaining data from activity – 2 in table – 1 of this lesson, draw the graphs of u vs v and $\frac{1}{u}$ vs $\frac{1}{v}$
Answer:
Graph of u – v using data obtained by activity – 2. Take lens with focal length 30 cm.

Object distance (u)	Image distance (v)	Focal length (f)
60 cm	60 cm	30 cm
50 cm	75 cm	30 cm
40 cm	120 cm	30 cm

The graph looks like this
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 23$

The shape of the graph is rectangular hyperbola.

Graph of $\frac{1}{u}$ – $\frac{1}{v}$
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 24$
For these values the graph is straight line which touches the axis as shown in figure.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 25$

Question 17.
Figure shows ray AB that has passed through a divergent lens. Construct the path of the ray up to the lens if the F position of its foci is known. (AS5)
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 6$
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 7$
The path of the ray up to the lens if the position of foci is known for ray AB is diverging lens or concave lens path.

Question 18.
Figure shows a point light source and its image produced by a lens with an optical axis N₁, N₂. Find the position of the lens and its foci using a ray diagram. (AS5)
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 8$
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 9$

The object is in between focus and optic centre.
The image is virtual, erect and magnified. Nv
l is the lens, ‘O’ is the object and T is the image.

Question 19.
Find the focus by drawing a ray diagram using the position of source S and the image S’ given in the figure. (AS5)
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 10$
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 11$

Image is real.
l’ is lens, ‘O’ is object and T is image.
Lens is convex.

(Or)
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 12$

Image is real.
l’ is lens, ‘O’ is object and ‘I’ is image.
Lens is convex.

Question 20.
A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm. Where should a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses? Draw a ray diagram. (AS5)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 26$
1. A parallel beam of rays when incident on a convergent lens, after refraction they meet at the focus of the lens.

2. A beam of rays which is incident on a divergent lens, after refraction, pass parallel to the principal axis. If we extend these incident rays, they seems to meet at focus of the lens.

3. Hence the divergent lens should be kept at 25 cm distance from convergent lens (40 – 15 = 25 cm) as shown in the figure.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 27$
PF = 40 cm (Focal length of convergent lens)
P’F = 15 cm (Focal length of divergent lens)
PP’ = 40 – 15 = 25 cm (Position of divergent lens)

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 21.
Draw ray diagrams for the following positions and explain the nature and position of image.
i) Object is placed at 2F₂
ii) Object is placed between F₂ and optic centre P. (AS5)
Answer:
i) Object is placed at 2F₂:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 13$
Nature : Real, inverted and diminished.
Position : Image is formed on the principal axis between the points F₁, and 2F₁.

ii) Object is placed between F₂ and optic centre P :
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 14$
Nature : Virtual, erect and magnified.
Position : Same side of the lens where object is placed.

Question 22.
How do you appreciate the coincidence of the experimental facts with the results obtained by a ray diagram in terms of behaviour of images formed by lenses? (AS6)
Answer:

Ray diagrams are very useful in optics.
By the ray diagrams, we can easily find the values of image distance, object distance, focal length, radius of curvature, magnification, etc.
These results are exactly equal to the result gotten by an experiment.
For example : In the experiment, with a convex lens, we get clear image of an object, on a screen by adjusting the screen.

Then, we measure the image distane (v) practically. This takes more time and requires equipped lab also.

But, by simply draw a ray diagram on a paper, we can get exact image distance (v) very easily, without lab.

So, ray diagrams are very useful in the construction of microscopes, telescopes, etc.
Hence, one can trust and depend on the result of ray diagrams instead of several lab experiments.
So, I appreciate the ray diagrams.

Question 23.
Find the refractive index of the glass which is a symmetrical convergent lens if its focal length is equal to the radius of curvature of its surface. (AS7)
Answer:
Given that lens is convergent symmetrical
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 15$

Question 24.
Find the radii of curvature of a convexo – concave convergent lens made of glass with refractive index n = 1.5 having focal length of 24 cm. One of the radii of curvature is double the other. (AS7)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 28$

Question 25.
The distance between two point sources of light is 24 cm. Where should a convergent lens with a focal length of f = 9 cm be placed between them to obtain the images of both sources at the same point? (AS7)
Answer:
For Source S₁ :
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 16$
∴ The convex lens may be placed between the two sources, such that a distance of 18 cm from one source, and 6 cm from other source.

Question 26.
Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height? Why? (AS7)
(OR)
If your friend is standing near an edge of the swimming pool and you are in the water, do you find he is taller or shorter than his usual height?
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 29$

My friend appears to be taller because the light is travelling from rarer to denser.
The rays bend in such away that they seems to be coming from long distance.
So it is actually apparent image of my friend which appears to be taller due to refraction.

Fill in the Blanks

1. The rays from the distant object, falling on the convex lens pass through ……………….. .
2. The ray passing through the ……………….. of the lens is not deviated.
3. Lens formula is given by ……………….. .
4. The focal length of the plano-convex lens is 2R where R is the radius of curvature of the surface. Then the refractive index of the material of the lens is ……………….. .
5. The lens which can form real and virtual images is ……………….. .
Answer:

Tocus
optical centre
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
1.5
convex lens

Multiple Choice Questions

1. Which one of the following materials cannot be used to make a lens?
A) water
B) glass
C) plastic
D) clay
Answer:
D) clay

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

2. Which of the following is true?
A) The distance of virtual image is always greater than the object distance for convex lens.
B) The distance of virtual image is not greater than the object distance for convex lens.
C) Convex lens always forms a real image.
D) Convex lens always forms a virtual image.
Answer:
B) The distance of virtual image is not greater than the object distance for convex lens.

3. Focal length of the plano-convex lens is when its radius of curvature of the surface is R and n is the refractive index of the lens.

4. The value of the focal length of the lens is equal to the value of the image distance when the rays are
A) passing through the optic centre
B) parallel to the principal axis
C) passing through the focus
D) in all the cases
Answer:
D) in all the cases

5. Which of the following is the lens maker’s formula?
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 44$
Answer:
C

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Additional Questions and Answers

Question 1.
Derive a relation between refractive indices of two media (n₁, n₂), object distance (u), image distance (v) and radius of curvature (R) for a curved surface.
(OR)
Derive $\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}$
(OR)
Derive curved surface formula.
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 30$

Consider a curved surface separating two media of refractive indices n₁, and n₂.
A point object is placed on the principal axis at point ‘O’.
The ray which travels along the principal axis passes through the pole undeviated.
The second ray, which forms an angle with a princi¬pal axis, meets the interface at A. The angle of incidence is Q₁. The ray bends and passes through the second medium along the line AI. The angle of refraction is Q₂.
The two refracted rays meet at I and the image is formed there.
6) Let the angle made by the second refracted ray with principal axis be γ and the angle between the normal and principal axis be β.
From figure,

PO = u (object distance), PI = v (Image distance),
PC = R (radius of curvature) and n₁, n₂ are refractive indices of the media.
From ∆ACO, θ₁ = α + β
∆ACI, β = θ₂ + γ
⇒ θ₂ = β – γ
According to Snell’s law, n₁sin θ₁ = n₂ sin θ₂.
∴ n₁ sin (α + β) = n₂ sin (β – γ) …………….. (1)
As per paraxial approximation,
sin (α + β) = α + β and sin (β – γ) = β – γ.
∴ (1) ⇒ n₁(α + β) = n₂ (β – γ)
⇒ n₁ α + n₁β = n₂ β – n₂ γ — (2)
Since all angles are small, we can write
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 31$
∴ This is the required relation for curved surfaces.

Question 2.
Derive expression for lens maker’s formula.
(OR)
Prove $\frac{1}{\mathbf{f}}=(\mathbf{n}-\mathbf{1})\left(\frac{1}{\mathbf{R}_{1}}-\frac{1}{\mathbf{R}_{2}}\right)$.
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 32$
Procedure :

Imagine a point object ‘O’ placed on the principal axis of the thin lens
Let this lens be placed in a medium of refractive index n_a and let refractive index of lens be n_b.
Consider a ray, from ‘O’ which is incident on the convex surface of the lens with radius of curvature R₁ at A.
The incident ray refracts at A.
It forms image at Q, if there were no concave surface.
From figure Object distance PO = – u;

Image distance PQ = v = x
Radius of curvature R = R₁
n₁ = n_a and n₂ = n_b.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 33$

But the ray that has refracted at A suffers another refraction at B on the concave surface with radius of curvature (R₂).
At B the ray is refracted and reaches I.
The image Q of the object due to the convex surface. So I is the image of Q for concave surface.
Object distance u = PQ = + x
Image distance PI = v
Radius of curvature R = – R₂
The refraction of the concave surface of lens is medium -1 and surrounding is medium – 2.
∴ n₁ = n_b and n₂ = n_a

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 34$

Question 3.
Derive the lens formula.
Answer:
1. Consider an object 00′ placed on the principal axis in front of a convex lens as shown in the figure. Let II’ be the real image formed by the lens, i.e. the other side of it.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 35$
2. From the figure : PO, PI, PFt are the object distance, image distance and focal length respectively.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 36$
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 37$

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 64

Question 1.
What happens to a ray that is incident on a curved interface separating the two media? Are the laws of refraction still valid?
Answer:
It undergoes deviation from its path. Yes, the laws of reflection are still valid.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 2.
How do rays betid when they are incident on a curved surface?
Answer:
A ray will bend towards the normal when it travels from rarer to denser medium and bends away from the normal when it travels from denser to a rarer medium.

10th Class Physics Textbook Page No. 65

Question 3.
What happens to ray that travels along the principal axis? Similarly, a ray that travels through the centre of curvature?
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 40$
Answer:
According to Snell’s law the ray which travels along the normal drawn to the surface does not deviate from its path. Hence both rays in the given condition travel along normal, so they do not deviate.

Question 4.
What difference do you notice in the refracted rays in 4 (a) and 4 (b)? What could be the reason for that difference?
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 41$
Answer:

In figure 4 (a) ray travelling parallel to the principal axis strikes a convex surface and passes from a rarer medium to a denser medium.
In figure 4 (b) a ray travelling parallel to the principal axis strikes a convex surface passes from a denser medium to a rarer medium.
Figure 4 (a) : The refracted ray moves towards the normal.
Figure 4 (b) : The refracted ray moves away from the normal.
Reason : The main reason is that light passes through different media.

10th Class Physics Textbook Page No. 66

Question 5.
What difference do you notice in refracted rays in 4 (c) and 4 (d)? What could be the reasons for that difference?
(OR)
Draw the ray diagrams when the incident ray passes through the curved surfaces.
a) Rarer medium to denser medium.
b) Denser medium to rarer medium.
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 42$
Answer:

In figure 4 (c) a ray travelling parallel to the principal axis strikes a concave surface and passes from a denser medium to a rarer medium.
In figure 4 (d) a ray travelling parallel to the principal axis strikes a concave surface and passes from a rarer medium to a denser medium.

Reasons :

Figure 4 (c) :The refracted ray reaches a particular point on the principal axis.
Figure 4 (d) : The refracted ray moves away from the principal axis.
The main reason is that light passes through different media.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 6.
You might have observed that a lemon in the water of a glass tumbler appears bigger than its actual size, when viewed from the sides of tumbler.
1) How can you explain this (appeared) change in size of lemon?
Answer:
It can be explained by using refraction. When light travels from one medium to another medium it undergoes refraction.

2) Is the lemon that appears bigger in size an image of lemon or is it the real lemon?
Answer:
That is image of lemon.

3) Can you draw a ray diagram to explain this phenomenon?
Answer:
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 43$

10th Class Physics Textbook Page No. 70

Question 7.
What happens to the light ray when a transparent material with two curved surfaces is placed in its path?
Answer:
The light ray undergoes refraction.

Question 8.
Have you heard about lenses?
Answer:
Yes, we have heard about lenses. A transparent material bounded by two spherical v surfaces is called lens.

Question 9.
How does a light ray behave when it is passed through a lens?
Answer:
A light ray will deviate from its path in some cases and does not deviate in some other cases.

10th Class Physics Textbook Page No. 72

Question 10.
How does the lens form an image?
Answer:
Lens forms an image through converging light rays or diverging light rays.

Question 11.
If we allow a light ray to pass through the focus, which path does it take?
Answer:
The ray passing through the focus takes a parallel path to principal axis after refraction.

10th Class Physics Textbook Page No. 73

Question 12.
What happens when parallel rays of light fall on a lens making some angle with the principal axis?
Answer:’
The rays converge at a point (or) appear to diverge from a point lying on the focal plane.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 13.
What do you mean by an object at infinity? What type of rays fall on the lens?
Answer:
The distance between the lens and the object is very much greater than when compared to object size is known as object at infinity. Parallel rays fall on the lens.
The object at infinity means distant object. The rays falling on the lens from an object at infinity are parallel to principal axis.

10th Class Physics Textbook Page No. 77

Question 14.
Could you get an image on the screen for every object distance with a convex lens?
Answer:
No, when the object is placed between pole and focus we will get virtual, erect and enlarged image on the other side of the- object.

Question 15.
Why don’t you get an image for certain object distances?
Answer:
Because at those distances the light rays diverge each other.

$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces$

Question 16.
Can you find the minimum limiting object distance for obtaining a real image? What do you call this minimum limiting object distance?
Answer:
Yes, this minimum limiting object distance is called focal length.

Question 17.
When you do not get an image on the screen, try to see the image with your eye directly from the place of the screen. Could you see the image? What type of image do you see?
Answer:
Yes, we can see the image. This is a virtual image which we cannot capture on screen.

Question 18.
Can you find the image distance of a virtual image? How could you do it?
Answer:
We can find the image distance of virtual image by using lens formula $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$ (if we know the focal length of lens and object distance.)

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Activities

Activity – 1

Question 1.
Write an activity to observe the light refraction at curved surface.
Answer:
Procedure and observation :
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 38$

Draw an arrow of length 4 cm usfng a black sketch pen on a thick sheet of paper.
Take an empty cylindrical-shaped transparent vessel.
Keep it on the table.
Ask your friend to bring the sheet of paper on which arrow was drawn behind the vessel while you look at it from the other side.
We will see a diminished image of the arrow.
Ask your friend to fill vessel with water.
Look at the arrow from the same position as before.
We can observe an inverted image.

Explanation :

In the first case, when the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass, enters in to air then it again undergoes refraction on the opposite curved surface of vessel and comes out into the air.
In this way light travels through two media, comes out of the vessel and forms a diminished image.
In the second case, light enters the curved surface, moves through water, comes out of the glass and forms an inverted image.

Lab Activity

Question 2.
Write an activity to know the characteristics of image due to convex lens at various distances.
Answer:
Aim:
Determination of focal length of bi-convex lens using UV method.

Material Required :
V Stand, convex lens, light source, screen, meter scale. Take a V-stand and place it on a long (nearly 2m) table at the middle. Place a convex lens on the v-stand. Imagine the principal axis of the lens. Light a candle and ask your friend to take the candle far away from the lens along the principal axis. Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.

Procedure :

Take a V-stand and place a convex lens on this stand.
Imagine the principal axis of the lens.
Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
We use a screen because it forms a real image generally which will form on a screen. Real images cannot be seen with an eye.
Adjust the screen, on other side of lens until clear image forms on it.
Measure the distance of the image from the stand and also measure the distance between the candle and stand of lens.
Now place the candle at a distance of 60 cm from the lens such as the flame of the candle lies on the principal axis of the lens.
Try to get an image of candle flame on the other side on a screen.
Adjust the screen till you get a clear image.
Measure the distance of image (v) from lens and record the value of’u’ and V in the table.
Repeat this for various distances of images; in all cases note them in the table.

Observation :
$AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces 39$
Conclusion : From this we conclude that a convex lens forms both real and virtual images when object is placed at various positions.

AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces

March 29, 2023February 1, 2023 by AP Board

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
Why is it difficult to shoot a fish swimming in water? (AS1)
(OR)
If the fish is swimming in water, why it is difficult to shoot?
(OR)
A shooter finds it difficult to shoot a fish swimming in water. Why?
Answer:
Due to refraction of light, it is difficult to shoot a fish swimming in water.

Reason :
The light rays coming from the fish towards shooter, bend at water-air interface. So, shooter sees only image of the fish, but not actual fish.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 2.
The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)
Answer:
Speed of light in diamond = 1,24,000 km/s
Speed of light in vacuum = 3,00,000 km/s
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 1$

Question 3.
Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)
Answer:
Refractive index of glass relative to water = $\frac{n_{g}}{n_{w}}=\frac{9}{8}$
∴ Refractive index of water relative to glass = $\frac{\mathrm{n}_{\mathrm{w}}}{\mathrm{n}_{\mathrm{g}}}=\frac{8}{9} \cdot\left[\because \mathrm{n}_{12}=\frac{1}{\mathrm{n}_{21}}\right]$

Question 4.
The absolute refractive index of water is 4/3. What is the critical angle ? (AS1)
Answer:
Absolute refractive index of water = 4/3
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 2$
Critical angle of water = C = 48°5′ = 48.5°.

Question 5.
Determine the refractive index of benzene if the critical angle is 42°. (AS1)
Answer:
Critical angle of benzene = 42°.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 3$

Question 6.
Explain the formation of mirage. (AS1)
(OR)
How is the mirage formed? Explain.
(OR)
A person walking on a road observed some water being present on the road but when he went there actually he did not find water. What is that actually formed called? Explain that process.
(OR)
Sometimes during the hot summer at noon time on tar roads, it appears that there is water on the road, but there would really be no water. What do you call this phenomenon? Explain why it happens.
(OR)
Why do you see a mirage the road on a hot summer day?
Answer:

During hot summer day, air just above the road surface is very hot and the air at higher altitudes is cool.
We know that refractive index of air increases with density.
So, the cooler air at the top has greater refractive index than hotter air just above the road.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 4$

Light travels faster through the thinner hot air than the denser cool air above it.
On hot days, the temperature decreases with height.
Thus the refractive index of air increases with height.
When the light from a tall object such as tree or from the sky passes, through a medium just above the road whose refractive index decreases towards ground, suffers refraction and takes a curved path because of total internal reflection.
This refracted light reaches the observer in a direction shown as in second figure.
This appears to the observer that the ray is reflected from ground.
Hence we will see water on road, which is the virtual image of sky and an inverted image of tree on the road.
Such virtual images of distant high objects cause the optical illusion called ‘mirage’.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 7.
How do you verify experimentally that $\frac{\sin i}{\sin r}$ is a constant? (AS1)
(OR)
Explain the experiment that shows the relation between angle of incidence and angle of refraction through figure.
(OR)
Write an experiment to obtain the relation between angle of incidence and angle of refraction.
Answer:
Aim:
Identifying relation between angle of incidence and angle of refraction.

Materials required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2 cm pencil and laser light.

Procedure :
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 5$ $AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 6$

Take a wooden plank which is covered with white chart.
Draw two perpendicular lines, passing through the middle of the paper as shown in the figure (a).
Let the intersecting point be O.
Mark one line as NN which is normal to the another line marked as MM.
Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’.
Take a protractor and place it along NN in such a way that its centre coincides with ‘O’ as shown in figure (b).
Then mark the angles from 0° to 90° on both sides of the line NN.
Repeat the same on the other side of the line NN.
The angles should be represented on circular line.
Now place semi circular glass disc so that its diameter coincides with the interface line (MM) and its centre coincides with the point O.
Take the laser light and send it along NN in such a way that the laser propagates from air to glass through the interface at point O and observe the way of laser light coming from other side of disc.
There is no deviation.
Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it must pass through point O.
Measure its corresponding angle of refraction.
Repeat the experiment with angle of incidences of 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

Observation :

Find sin i, sin r for every i and r note down the values in table.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 7$

Evaluate $\frac{\sin i}{\sin r}$ for every incident angle i.
We get $\frac{\sin i}{\sin r}$ as constant.
That is the relationship between angle of incidence and angle of refraction.
The ratio of sin i and sin r is called refractive index.

Question 8.
Explain the phenomenon of total internal reflection with one or two activities. (AS1)
Answer:
Procedure :

Place the semi-circular glass disc in such a way that its diameter coincides with interlace line MM and its center coincides with point O’.
Now send light from the curved side of the semicircular glass disc.
The light travels from denser medium to rarer medium.
Start with angle of incidence (i), equals to 0° and observe for refracted on other side of the disc.
It does not deviate into its path when entering rarer medium.
Send laser light along angles of incidence 5°, 10°, 15°, etc. and measure the angle of refraction.
And tabulate the results in the given table.

Observation :

Make a table shown below and note the values ‘i’ and ‘r’.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 8$

At particular angle of incidence, the refracted ray does not come out but grazes the interface separating the air and glass. This angle is called critical angle.
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e. light never enters rarer medium. This phenomenon is called total internal reflection.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 9.
How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium? (AS1)
(OR)
When the light rays travel from denser to rarer medium, how can you explain, the angle of refraction is more than angle of incidence experimentally?
Answer:
Procedure :
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 9$
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 10$

Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure.
Arrange two straws at the centre of the disc, in such a way that they can be rotated freely about the centre of the disc.
Adjust one of the straws to make an angle 10°.
Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10° must be inside the water.
From the top of the vessel, try to view the straw which is inside the water as shown in the figure.
Then adjust the other straw which is outside of the water until both straws look like they are in a single straight line.
Then take the disc out of the water and observe the two straws on it. You will find that they are not in a single straight line.
Measure the angle between the normal and second straw. Note the values in the following table.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 11$

Do the same for various angles and find the corresponding angles of refraction and note them in the table.

Observation :
We will find the angle of refraction is more than angle of incidence.
i. e., r > i.

Conclusion :
When light travels from denser (water) to rarer (air) it bends away from the normal.

Question 10.
Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment) (AS2)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 12$

The black metallic .ball appears to be raised up in the water because the path of the ray changes its direction at the interface, separating the two media, i.e., water and air.
This path is chosen by light ray so as to minimize time of travel between ball and eye.
This can be possible only when the speed of light changes at interface of two media.
In another way the speed of light is different in different media.

Hypothesis :
Speed of light changes when it travels from one medium to another medium.

Question 11.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
1) What changes do you notice?
2) What could be the reasons for these changes? (AS2)
Answer:

We cannot see the glass rod in glycerine but we can see the rod in water.
We can also observe an apparent image of glass rod in water.
Reasons:
i) Glycerine has essentially same refractive index as glass.
ii) So, any light passing through these is bent equally.
iii) Since both are transparent, it is not possible for our eye to distinguish the boundary by a change in the angle of reflection, and the glass seems to vanish.
iv) But, the refractive index of glass and water are different.
v) So the glass rod is visible to our eye in water. .

Question 12.
Do Activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 13$
Procedure:

Take a cylindrical transparent vessel.
Place a coin at the bottom of the vessel.
Now pour water until you get the image of the coin on the water surface.
This is due to total internal reflection.

Critical angle of water :

Refractive index of water = 1.33
The sine of critical angle of water = $\frac{1}{\text { Refractive index }}$
Sin C = $\frac{1}{\text { 1.33 }}$ ⇒ sin C = 0.7518.
∴ C = 8.7°
∴ The critical angle of water = 48.7°.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 13.
Collect the values of refractive index of the following media. (AS4)

Water, coconut oil, flint glass, crown glass, diamond, benzene and hydrogen gas.

Answer:

Medium	Refractive Index
1. Water	1.33
2. Coconut oil	1.445
3. Flint glass	1.65
4. Crown glass	1.52
5. Diamond	2.42
6. Benzene	1.50
7. Hydrogen gas	1.000132

Question 14.
Collect information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life. (AS4)
(OR)
What do you know about the working of optical fibres and make a report of various uses of optical fibres in our daily life?
(OR)
How are the optical fibres working? What are the various uses of optical fibres in our daily life?
Answer:
1) An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m).
2) A bunch of such thin fibres form a light pipe.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 14$

Working :
1. Optical fibre having three parts-namely core (n = 1, 7), clading (n = 1, 6) and shielding.
2. The ray of light AB gets refracted at point ‘B’ into core and incident at ‘C’ with angle of incidence i (i > c).

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 15$
3. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
4. The light is thus transmitted along the fibre.
5. The optical fibre is also based on ‘Fermat’s principle.

Uses :

Optical fibres are used in ‘endoscopy’ to see the internal organs like throat, stomach, intestines, etc.
Optical fibres are used in transmitting communication signals through light pipes.
Optical fibres are used in international telephone cables laid under the sea, in large computer networks, etc.
In optical fibre about 2000 telephone signals appropriately mixed with light waves may be simultaneously transmitted through a typical optical fibre.

Question 15.
Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc. and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in figure. Now try to view the free end (head) of the needle from surface of the water.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 17$
1) Are you able to see the head of the needle?
Now do the same with other discs of different radii. Try to see the head of the needle, each time.
Note : The position of your eye and the position of the disc on water surface should not be changed while repeating the Activity with’other discs.
2) At what maximum radius of disc, were you not able to see the free end of the needle ?
3) Why were you not able to view the head of the nail for certain radii of the discs ?
4) Does this Activity help you to find the critical angle of the medium (water) ?
5) Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 29$
1. Yes, we can see head of the needle.

2. Height of the pin = 6 cm
Radius of the disc = x cm
Critical angle of water = C = 48.7°
Tan C = 48.7°
$\frac{x}{6}$ = 1.138 ⇒ x = 6.828 cm
So, at radius of 6.8 cm we cannot see the free end of the needle.

3. Because the light rays coming from object undergoing total internal reflection by touching the surface of disc.

4. Yes, we can find critical angle.
Refractive index of air (n₂) = 1.003 ; Refractive index of water (n₁) = 1.33
Sin C = $\frac{n_{2}}{n_{1}}=\frac{1.003}{1.33}$ = 0.7541 ⇒ C = 48.7°

5.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 18$

Question 16.
Explain the refraction of light through the glass slab with a neat ray diagram. (AS5)
(OR)
Draw a glass slab diagram and explain the refraction of light through glass slab.
(OR)
Write the procedure of a lab Activity to understand lateral shift of light rays through a glass slab.
(OR)
How can you find lateral shift using glass slab?
Answer:
Aim :
A) Determination of position and nature of image formed by a glass slab.
B) Understanding lateral and vertical shift.
C) Determination of refractive index of given glass slab.

Materials required :
Plank, chart paper, clamps, scale, pencil, thin glass slab and pins.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 16$

Procedure :

Place a piece of chart on a plank. Clamp it. Place a glass slab in the middle of the paper.
Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle.
Name the vertices of the rectangle as A, B, C and D.
Draw a perpendicular at a point on the longer sides (AB) of the rectangle.
Now draw a line, from the point of intersection where side AB of rectangle and perpendicular meet, in such a way that it makes 30° angle with the normal.
This line represents the incident ray falling on the slab and the angle it makes with normal represents angle of incidence.
Now place the slab on the paper in such a way that it fits in the rectangle drawn. Fix two identical pins on the line making 30° angle with normal, such that they stand vertically with same height.
By looking at the two pins from the other side of the slab, fix two pins in such a way that all pins appear to be along a straight line.
Remove the slab and take out pins. Draw a straight line by joining the dots formed by the pins up to the edge CD of the rectangle.
This line represents emergent ray of the light.
Draw a perpendicular to the line CD where our last line drawn meets the line CD.
Measure the angle between emergent ray and normal.
This is called angle of emergence.
The angle of incidence and angle of emergence are equal.
Incident emergent rays are parallel.
The distance between the parallel rays is called shift.

Question 17.
Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 19$

Question 18.
What is the reason behind the shining of diamond and how do you appreciate it? (AS6)
(OR)
For which reason is the diamond shining and how is it appreciable?
Answer:

The critical angle of diamonds is very low, i.e., 24.4°.
So if a light ray enters diamond, it undergoes total internal reflection.
It makes the diamond shine brilliant.
So total internal reflection is main cause of brilliance of diamonds.
Majority of people are attracted towards diamonds due to this property.
So we have to thoroughly appreciate total internal reflection for brilliance of diamonds.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 19.
How do you appreciate the role of Fermat’s principle in drawing ray diagrams? (AS6)
(OR)
How do you admire the role of Fermat’s principle in drawing ray diagrams? Fermat’s principle: The light ray always travels in a path which needs shortest possible time to cover distance between two points.
This principle has lot of importance on optics. This is used in

Laws of reflection (i.e., angle of incidence = angle of reflection)
Laws of refraction (Snell’s law)
To derive refractive index of a medium.
To derive refractive index of glass slab.
So, I appreciate the Fermat’s principle.

Question 20.
A light ray is incident on air-liquid interface at 45° and is refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90°? (AS7)
Answer:
i) Given that angle of incidence (i) = 4S°
angle of refraction (r) = 30°
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 20$

ii) Given that angle between reflected and refracted ray is 90°.
We know angle of incidence = angle of reflection
∴ Angle of refraction (r) = 90 – angle of incidence
= 90 – i
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 21$
Critical angle = 54.7°. This angle is also known as polarising angle.

Question 21.
Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface from a certain viewing position. (AS7)
Answer:
When a test tube is immersed at a certain angle in a tumbler of water appears to have
a mirror surface from a certain viewing positions due to total internal reflection.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 22$
Explanation :

The critical angle for glass is 42°.
The glass and air in test tube works as denser and rarer mediums.
The rays of light while travelling through water strike glass – air interface of test tube at an angle of more than 42° (i > c) they get totally internal reflected as shown figure.
When these reflected rays reach the eye, they appear to come from the surface of test tube itself.
Now the test tube appears like silvary.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 23.
In what cases does a light ray not deviate at the interface of two media? (AS7)
Answer:

When a light ray incident is perpendicular to the interface of surface, it does not undergo deviation.
When a light ray incident is more than critical angle, it does not undergo deviation (refraction) but it undergoes reflection to come back into the original medium.

Question 25.
When we sit at camp fire, objects beyond the fire seen swaying. Give the reason for it. (AS7)
(OR)
What are the reasons for the objects beyond the fire seen swaying, when we sit at camp fire?
Answer:

The temperature of the surrounding air changes due to convection of heat by the camp fire.
This leads to chang in density and refractive index of air, continuously.
The continuous change in refractive index of air changes the refracted path of the light ray.
This is the cause for swaying of an object.

Question 26.
Why do stars appear twinkling? (AS7)
(OR)
What is the reason for the appearance of stars like twinkling?
Answer:

The twinkling of a star is due to atmospheric refraction of star light.
The atmosphere consists of a number of layers of varying densities.
When light rays coming from a star pass through this layers and undergo refraction for several times.
Thats why stars appear twinkling.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

Question 27.
Why does a diamond shine more than a glass piece cut to the same shape? (AS7)
(OR)
What is the reason for shining of diamond brightly as compared to glass piece cut?
Answer:

The critical angle of a diamond is very low (i.e., 24.4°).
So if a light ray enters a diamond it definitely undergoes total internal reflection.
Whereas it is not possible with glass piece cut to the same shape.
So diamond shines more than a glass piece.

Fill In The Blanks

1. At critical angle of incidence, the angle of refraction is ……………… .
2. n₁ sin i = n₂ sin r, is called ……………… .
3. Speed of light in vacuum is ……………… .
4. Total internal reflection takes place when a light ray propagates from …………. to …………… medium.
5. The refractive index of a transparent material is 3/2. The speed of the light in that medium is …………… .
6. Mirage is an example of ……………… .
Answer:

90°
Snell’s law
3 × 10⁸ m/s
denser, rarer
2 × 10⁸ m/s
optical illusion / total internal reflection

Multiple Choice Questions

1. Which of the following is Snell’s law?
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 30$
Answer:
B)

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass air interface is ………………….
A) 0°
B) 45°
C) 30°
D) 60°
Answer:
C) 30°

3. Total internal reflection takes place when the light ray travels from …………….. .
A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
Answer:
C) denser to rarer medium

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

4. The angle of deviation produced by the glass slab is …………… .
A) 0°
B) 20°
C) 90°
D) depends on the angle formed by the light ray and normal to the slab
Answer:
D) depends on the angle formed by the light ray and normal to the slab

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Additional Questions and Answers

Question 1.
Derive Snell’s law.
(OR)
Prove n₁ sin i = n₂ sin r.
(OR)
Derive the Snell’s formula from Fermat’s principle.
(OR)
Derive the formula in realtion with and of incidence and angle of refraction.
Answer:
Let X be the path and A be the point above X and B be the point below X.
Now, we have to find the way from A to B.

Let us try to calculate how long it would take to go from A to B by the two paths through point D and another through point C.
If we draw a perpendicular DE, between two paths at D, we see that the path on line is shortened by the amount EC.
On the other hand, in the water, by drawing corresponding perpendicular CF we find that we have to go to the extra distance DF in water. These times must be equal since we assumed there was no change in time between two paths.
Let the time taken by the man to travel from E to C and D to F be ∆t and v₁ and v₂ be the speeds of the running and swimming. From figure we get,
EC = v₁ ∆t and DF = v₂ ∆t
⇒ $\frac{\mathrm{EC}}{\mathrm{DF}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}$ ………….. (1)
Let i and r be the angles measured between the path ACB and normal NN, perpendicular to shore line X.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 23$ $AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 24$

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 56

Question 1.
Why should you see a mirage as a flowing water?
Answer:

A mirage is a naturally occuring optical phenomenon, in which light rays are bent to produce a displaced image of distant objects or the sky.
As light passes from colder air (higher place) to warmer air (lower place), the light ray bends away from the direction of the temperature gradient.
Once the rays reach the viewer’s eye, the visual cortex interprets it as if it traces back along a perfectly straight “line of sight”. However this line is at a tangent to the path the ray takes at the point it reaches the eye.
The result is that an “inferior image ” of the sky above appears on the ground.
The viewer may incorrectly interprets this sight as water that is reflecting the sky, which is to the brain, a more reasonable and common occurrence.

Question 2.
Can you take a photo of a mirage?
Answer:

Yes, I can take a photo of a mirage.
Our eye can catches the total internal reflected rays.
So, camera lens also catches the same.

10th Class Physics Textbook Page No. 46

Question 3.
What difference do you notice in fig 2(a) and Fig 2(b) with the respect to refracted rays?
(OR)
Draw the ray diagram of refraction in between denser and rarer medium.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 27$
Answer:
In figure 2(a) the light ray bends towards normal whereas in 2(b) the light ray bends away from the normal.

Question 4.
Is there any relation between behaviour of refracted rays and speed of the light?
Answer:
Yes. The speed of light changes when it travels from one medium to another medium. So the light may bend towards normal or away from normal.

10th Class Physics Textbook Page No. 47

Question 5.
Why db different material media possess different values of refractive indices?
Answer:
Refractive index depends on nature of material. So different media have different values of refractive indices.

10th Class Physics Textbook Page No. 48

Question 6.
On what factors does the refractive index of a medium depend?
Answer:
Refractive index depends on (1) Nature of material and (2) Wavelength of light used.

10th Class Physics Textbook Page No. 49

Question 7.
Can we derive the relation between the angle of incidence and the angles of refraction theoretically?
Answer:
Yes, we can derive the relation between angle of incidence and angles of refraction theoretically. We can get nt sin i = n2 sin r.

10th Class Physics Textbook Page No. 53

Question 8.
Is there any chance that angle of refraction is equal to 90° ? When does this happen?
Answer:
Yes, when angle of incidence is equal to critical angle then angle of refraction is equal to 90°.

10th Class Physics Textbook Page No. 54

Question 9.
What happens to light when the angle of incidence is greater than critical angle?
Answer:
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e., light never enters rarer medium. This phenomenon is called total internal reflection.

10th Class Physics Textbook Page No. 57

Question 10.
How does light behave when a glass slab is introduced in its path?
Answer:
The light ray undergoes refraction two times.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Activities

Activity – 1

Question 1.
Procedure :
Take some water in a glass tumbler. Keep a pencil in it. See the pencil from one side of glass and also from the top of the glass.
Observation:

1. How does it look?
Answer:
From the side it appears to be bent. From the top it appears as it is raising.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

2. Do you find any difference between two views?
Answer:
Yes, the position of pencil is different.

Activity – 2

Question 2.
Procedure :

Go to a long wall (of length of 30 feet) facing the Sun. Go to the one end of a wall and ask someone to bring a bright metal object near the other end of the wall.
When the object is few inches from the wall, it will distort and we will see a reflected image on the wall as though the wall were a mirror.

Observation:

Why is there an image of the object on the wall?
Answer:
The image is due to refraction of light.

Activity-3 Refraction

Question 3.
Procedure: –

Take a shallow vessel with opaque walls such as a mug, a tin or a pan.
Place a coin at the bottom of the vessel.
Move away from the vessel until we cannot see the coin (fig. 2). Ask someone to fill the vessel with water. When the vessel is filled with water the coin comes back into view (fig. 3).

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 25$

1. Why are you able to see the coin when the vessel is filled with water?
Answer:
The ray of light originated from the coin does not reach your eye when the vessel is empty. Hence you are not able to see the coin. But the coin becomes visible after the vessel is filled with water.

2. How is it possible? Do you think that the ray reaches your eye when the vessel is filled with water?
Answer:
Yes, it reaches the second instance.

3. What happens to the light ray at interface between water and air?
Answer:
It bends towards the normal.

$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces$

4. What could be the reason for this bending of the light ray in the second instance?
Answer:
It is based on Fermat’s principle, which states that the light ray always travels in a path which needs shortest possible time to cover the distance between the two points.

Activity – 4

Question 4.
Prove that when light ray travels from denser to rarer medium it bends away from the normal.
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 9$
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 10$
Procedure :

Take a metal disc. Use protractor and mark angles along its edge as shown in the figure.
Arrange two straws from the centre of the disk.
Adjust one of the straws to the angle 10°.
Immerse half of the disc vertically into the water, filled in a transparent vessel.
Inside the water the angle of straw should be at 10°.
From the top of the vessel try to view the straw which is inside the water.
Then adjust the other straw which is outside the water until both straws are in a single straight line.
Then take the disc out of the water and observe the two straws on it.
We will find that they are not in a single straight line.
It could be seen from the side view while half of the disc is inside the water.
Measure the angle between the normal and second straw. Draw table for various angles and corresponding angles of refraction.

Observation :
We observe that ‘r’ is greater than ‘i’ in all cases and when light travels from
denser to rarer medium it bends away from the normal.

Activity – 6

Question 5.
Why can we not see a coin placed in water from the side of glass?
Answer:
Procedure :

Take a transparent glass tumbler and coin.
Place a coin on a table and place glass on the coin.
Observe the coin from the side of the glass. We can see the coin.
Now fill the glass with water and observe the coin from the side of the glass tumber.
Now we cannot see the coin because the coin rises up due to refraction.

Activity – 7

Question 7.
Why can we see the coin in water from top? What is the phenomenon behind that?
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 13$
Procedure :

Take a cylindrical transparent vessel. Place a coin at the bottom of the vessel.
Now pour water until we will get the image of the coin on the water surface.
This is due to total internal reflection.

Activity – 8

Question 8.
Write an Activity to find refractive index of glass slab by calculating vertical shift.
(OR)
Explain the experiment with glass slab in determination of refraction through vertical shift.
Answer:
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 26$
Procedure :

Take a glass slab and measure the thickness of the slab.
Take a white chart and fix it on the table.
Place the slab in the middle of the chart.
Draw line around it.
Remove the slab from its place.
The lines form a rectangle. Name the vertices of it as A. B, C and D. ‘
Draw a perpendicular to the longer line AB of the rectangle at any point on it.
Place slab again in the rectangle ABCD.
Take a needle. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 20 cm from the slab.
Now take another needle and by seeing at the first needle from the other side of the slab, try to keep the needle so that it forms a straight line with the first needle. 1 1)
Remove the slab and observe the positions of the needles.
They are not in same line.
Draw a perpendicular line from the second needle to the line on which the first needle is placed.
Take the intersection point as Q.
The distance between P and Q is vertical shift.
We will get the same vertical shift placing needle at different distances.
$AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 28$