Mahesh

Students can use TS 10th Class English Model Papers and TS 10th Class English Question Paper April 2023 as a tool for exam preparation.

TS 10th Class English Question Paper April 2023

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. Read the question paper carefully.
2. Answer the questions under Part – A in the answer booklet provided.
3. Write the answer to the questions under Part – B on the question paper itself.
4. Avoid overwriting.

Part – A (60 Marks)

Question No. (1 – 4) : Read the following passage.

‘When I was 13, I read a newspaper article about a disabled man who had managed to achieve great things and helped others, ’said Nick.
‘I realised why God had made us like this – to give hope to others. It was so inspirational to me that I decided to use my life to encourage other people and give them the courage that article had given me.’
‘I decided to be thankful for what I do have, not get angry about what I don’t’.
‘I looked at myself in the mirror and said: ‘You know what the world is right that I have no arms or legs, but they’ll never take away the beauty of my eyes. ‘I wanted to concentrate on something good that I had.’
“The challenges in our lives are there to strengthen our convictions. They are not there to run us over,” said Nick. In 1990 Nick won the Australian Young Citizen of the Year award for his bravery and perseverance. (Attitude is Altitude)

Answer each of the following questions in three to four sentences. 4 × 3 = 12 M
Question 1.
What change did the newspaper article bring in Nick?
Answer:
After reading the newspaper article there was a great change in Nick’s attitude. A pessimist changed into an optimist. Nick who hated God for making him a torso realised that God made him so to give hope to others. The article was so inspirational to him that he decided to use his life to encourage other people and give them the courage that the article had given him.

Question 2.
‘I decided to be thankful for what I do have, not get angry about what I don’t’, -is it applicable to Nick ? Explain.
Answer:
Yes, it is applicable to Nick. The newspaper article changed his attitude. Before reading the article he was a pessimist. He even tried to end his life by committing suicide. He hated God for making him a torso. After reading the article he became an optimist. He realised the purpose of his life. He realised that God made him a torso to give hope to others. He decided to use his life to encourage others. Hence he decided to be thankful for what he had, not get angry about what he didn’t have.

Question 3.
According to Nick, how can we overcome the challenges in our life ?
Answer:
Nick feels that the challenges in our lives are there to strengthen our convictions. They are not there to run us over. According to Nick, we can overcome challenges in our life with a positive attitude, optimistic nature, self-confidence, bravery and perseverance.

Question 4.
If you were Nick, how would you face the challenges in your life ? Explain.
Answer:
The life of Nick Vujicic teaches Us to have confidence on ourselves. It is a motivation for physically challenged people. There is none who is less than the others. Despite various difficulties everyone can achieve goals.

If I were Nick, I would face the challenges in my life efficiently. With positive traits such as never-give up attitude, optimism, bravery and perseverance. I would overcome the challenges in my life.

Question No. (5 – 8) : Read the following passage.

There is little else one could expect when all time greats like Marcus Bartley (cinematography), Ghantasala (music), M.L. Vasantha Kumari, Leela, Suseela and Madfravapeddi (playback), Gokhale (art), Pasumarthy (choreography) and Pitambaram (make-up) got together to weave magic around an episode from Mahabharatha, Sasirekha parinayam.

However, the greatness of Maya Bazaar, about which much is said and written, is hot just because of these facets alone.

It is a tribute to Telugu culture, language and customs of the land. The film was watched repeatedly soon after its release because people identified every character of the film with someone they knew in their immediate vicinity and the audience still do the same now.

The dialogues written by Pingali Nagendra Rao (as well the lyrics) were the same that the people were hearing or using in their conversations every day – if not, those became a part of Telugu life thereafter. Sasirekha’s, nay Ghatothkacha’s Manadi Sodara Prema ……………. became immortalized as much as Suryakantam’s antha alamalame kada which has become a way of life in greeting people.
As for songs, Aha naa pelli anta still reverberates in marriages and Vivaaha bhojanambu is yet another must.

Answer each of the following questions in three to four sentences. 4 × 3 = 12 M
Question 5.
flow do you feel when you identify yourself with one of the characters in a movie ?
Answer:
In any movie some characters possess positive qualities while some have negative qualities. Characters with positive qualities cause a good impression and have a lasting impact on the audience. They show their love, affection, sympathy, compassion, etc. towards characters with positive qualities. So I would like to identify myself with a character possessing positive qualities so that I would feel happy, satisfied and proud.

Question 6.
How did the dialogues in the movie ‘Maya Bazaar’ influence the audience ?
Answer:
The dialogues in the movie ‘Maya Bazaar’ have strong influence on the Telugu audience. Many of the dialogues are still being heard or used in their conversations every day. They became a part of Telugu people’s life.

Question 7.
How did the ‘Maya Bazaar’ reflect our lifestyle ?
Answer:
The movie ‘Maya Bazaar’ reflected our lifestyle greatly. It is a tribute to Telugu culture, language and customs of the land. Telugu people identified every character of the film with someone they knew in their immediate vicinity. The dialogues and songs in the movie became a part of Telugu life. Many of the dialogues became immortalized. As for songs, ‘Aha naa pelli anta’ and ‘Vivaha Bhojanambu’ still reverberate in marriage.

Question 8.
Pick Three TRUE statements according to the passage.
A) Maya Bazaar is based on the complete story of the Maha Bharatha.
B) Pingali Nagendra Rao wrote dialogues and songs for the movie.
C) The language used in the language of every day.
D) The characters in the movie represent the royal class of the society.
E) Every artist contributed to the perfection of Maya Bazaar.
Answer:
True statements: B, C, E

Question No. (9 – 12): Study the table carefully and answer the questions given. 4 × 2 = 8 M
Musical Instruments

Question 9.
What is the table mainly about ? How many categories and sub-categories are mentioned ?
Answer:
The table is mainly about Musical Instruments. Three categories and three sub-categories are mentioned in the table.

Question 10.
How is a clarinet different from a mouth organ and trumpet ?
Answer:
All the three instruments belong to the same class, i.e., wind instruments. The clarinet is a member of the woodwind family whereas the trumpet is the member of the brass family. Mouth organ belongs to other sub-class.

Question 11.
How are the stringed instruments different from the wind instruments ?
Answer:
In wind instruments, air is made to vibrate inside a wooden tube whereas stringed instruments are played by plucking the strings or drawing a bow across the strings.

Question 12.
If you are asked to choose one ofthe musical instruments, which instrument would you choose ? Why?
Answer:
I would choose the flute because it is an amazing instrument. It is one of the smallest instruments available. Listening to the flute music has many benefits. It relaxes the nervous system. It reduces negative effect of anxiety, stress and depression as it soothes the nerves. It aids in better sleep. It obviously makes our mind clearer and helps us to think better.

Question No. (13) : Read the passage given below focussing on the parts that are underlined. Answer any 4 of the questions as directed and write them in the answer booklet. 4 × 2 = 8 M

(i) A tiger was hunting in the forest. He saw a fox.
The fox noticed this. “Now the tiger will kill me,” the fox thought. “But I should do something.” Despite the danger, the fox said, “How dare you kill me !”
On hearing this, the tiger was surprised.
(ii) “Why can’t I kill you?” he asked.
The fox said,”
(iii) God has appointed me as the king of the forest. I have to rule my kingdom.
(iv) You can’t kill me. God will punish you.”
“Is it true?” the tiger asked doubtfully.
“Let’s go through the forest. I’ll show you how everyone is afraid of me,” said the fox.
The tiger agreed. The fox was walking through the forest followed by the tiger.
(v) The animals saw the tiger. They ran away.
“Have you seen how the animals are scared of me ?” the fox said. The tiger thought
(vi) the fox was the most powerful animal. “You’re right,” said the tiger. “You are the king of the forest.”
i) Combine the two sentences using’while’.
ii) Rewrite the sentence beginning with ‘He asked why …………
iii) Rewrite the sentence beginning with “I”
iv) Combine the two sentences using ‘if’.
v) Combine the sentences using ‘no sooner ………….. than’.
vi) Rewrite the sentence beginning with ‘no other’.
Answer:
i) While the tiger was hunting in the forest, he saw a fox. (Or)
While hunting in the forest, the tiger saw a fox. (Or)
The tiger saw a fox while he was hunting in the forest.

ii) He asked why he could not kill him.
He asked why he could not kill the fox.

iii) I was appointed as the king of the forest by God. (Or)
I was appointed by God as the king of the forest.

iv) If you kill me, God will punish you.

v) No sooner had the animals seen the tiger than they ran away. (Or)
No sooner did the animals see the tiger than they ran away.

vi) No other animal was as powerful as the fox. (Or)
No other animal was so powerful as the fox.

Question No. (14) :
In the text ‘My Childhood’, we have come across how the new teacher insulted Kalam and how Laxmana Sastry changed his mind. Finally, the new teacher realised and apologized for he had done.

Now write a possible conversation between the new teacher and Laxamana Sastry in ten to twelve exchanges. 10 M
You may use the following ideas:
* The new teacher couldn’t digest a Muslim boy sitting beside a Brahmin boy.
* He believed in customs related social status.
* Laxmana Sastry understood the pain of the two boys due to separation.
* Laxmana Sastry believed in equality and brotherhood.
* The new teacher realized and understood the feelings of the two boys.
(OR)
Nowadays family relationships are becoming weak. People have become so selfish that they always think about money and assets. Even members of the family do not show affection to-wards elders.
Prepare a speech script in this context in ten to twelve sentences.
You may use the following ideas.
* Importance of maintaining family relationship.
* Money not as important as family.
* Need to show love and affection towards elders.
* Maintaining good relationship among the members of the family.
Answer:
Possible conversation between Lakshmana Sastry and the teacher:
Lakshmana Sastry : Good morning, sir.
New teacher : Good morning, what can I do for you, sir ?
Lakshmana Sastry : Are you the new teacher ?
Teacher : Yes, I am. I joined this school yesterday only.
Lakshmana Sastry : Ramanatha Sastry is my son.
Teacher : Is there anything wrong with him ?
Lakshmana Sastry : Nothing with him, but I want to talk with you about the yesterday’s incident.
Teacher: What happened yesterday ?
Lakshmana Sastry : You separated Kalam from my son. You asked Kalam to go and sit on the back bench. Both of them are very sad. They are close friends. Why did you do this ?
Teacher : But, you are Brahmins and that boy is a Muslim.
Lakshmana Sastry : Stop talking. You should not spread the poison of social inequality and communal intolerance in the minds of innocent children.
Teacher : But, sir ……………..
Lakshmana Sastry : You must either apologize or quit the school and the island.
Teacher : Yes, sir. I am very sorry. My profound apologies to you, sir. The thing that happened yesterday was an unfortunate one.
Lakshmana Sastry : That’s good. You shouldn’t repeat it in the future.
Teacher : OK., sir

(OR)

Good morning respected Headmaster, beloved teachers and my dear fellow students.

Today 1, xxxx, would like to share my views on the family relations which are becoming weaker nowadays. There are several reasons for this unfortunate development. The most important one, of course, is the change in our lifestyles, values and goals.

Families that spend a lot of time together tend to be closer. And families where the members rarely come face to face with one another have hardly anything that ties them together. In the olden days parents and children used to live and work together. In many families, children pursued the same career as their father. They learned the trade or craft from their parents arid elder siblings. This allowed them to spend more time with their family and that made their relationships stronger.

Now we are living in the age of individualism. Everyone needs their space. This has made us more self centered than our parents or grandparents. This also reflects in our relationships. We no longer believe in making adjustments or compromises. We want everything our way. This attitude creates strain in our relationships.

There is yet another reason that affects family bonds. People are increasingly moving to another city or country in search of better education, jobs or living standards. They leave their family behind. While it is possible to keep in touch through modern communication means, an email or phone call is not equal a face-to-face interaction. The only way to forge better ties with our near and dear ones is to spend more time with them.

Nowadays people have become so selfish that they always think about money and assets. Even members of the family do not show affection towards elders. It pains to say that youngsters todays have neither time nor respect for the elders. Our busy lifestyle gives us no time to spend with them. They feel lonely and neglected. Our elders gave us their precious time, love, care, money, education, etc. If they had not done so, what would have been our situation ? So it is our responsibility to give them back their love and concern.

We must reslize that our biggest strength is our family. Relationships are like potted plants. We need to nourish them. If we don’t give them the love and affection they need, they will wilt. If we need better family relationships, we must be willing to adjust and compromise. We must realize that the only way to win love is to give love. We must remember that the strength of a family, like the strength of an army, is in its loyalty to each other.

Thank you for giving me this opportunity to share my feelings with you.

Question No. (15) :
On the occasion of International Women’s Day, your school is organizing Mandal Level Sports. Meet for girls of secondary level.

Imagine you are the girls’ representative of your class and draft a notice. Also request the girls of your class to participate in the Sports Meet. 5M
Answer:

KAMALA NIKETAN, CHATRAI. NOTICE MANDAL LEVEL SPORTS MEET We are pleased to inform that ‘Mandal Level Sports Meet’ for girls of secondary level is going to be organised in our school on the occasion of International Women’s Day. The Sports Meet will be held on 06 – 03 – 20xx. The girls are requested to participate in the programme and take active part in Sports Meet and make it a grand success. The girls who are interested to participate in the Sports Meet are asked to give their names to their class teachers. Sd/- Karuna Head Girl Date : 03.03.20xx

Question No. (16) :
Read the following information about Sachin Tendulkar, the famous Indian cricketer.


Now write a short profile of Sachin Tendulkar in a paragraph based on the above information. 5 M
Answer:
Sachin Tendulkar is an Indian former international cricketer who captained the Indian national team. He is regarded as one of the greatest batsmen in the history of cricket. He was born on 24 April, 1973 at Dadar in Muiribai. His father named Ramesh Tendulkar was a famous Marathi poet and novalist. His mother Rajni worked in the insurance industry.

On 11 December 1988 Sachin represented Mumbai team in Ranji Trophy and scored 100 runs. He was the youngest Indian cricketer to score a century on debut in first-class cricket. He made his Test Debut against Pakistan on 15 November 1989 and One Day International Debut aganist Pakistan on 18 December 1989. He played 200 Test matches and scored 15921 runs. His highest score in Test matches was 248. He got 46 wickets in Test matches. He played 463 ODI and scored 18426 runs. His highest score in ODIs was 200. He got 154 wickets in ODIs.

Sachin Tendulkar received many prestigious awards. He received the Aijuna Award in 1994, the Padma Sri Award in 1999 and the Padma Vibhushan Award in 2008 by the government of India. He was also honoured by the government of India with the India’s highest civilian award the Bharat Ratna in 2014.

Part – B (20 Marks)

Instructions :

1. Answer the questions on the question paper itself and attach it to the answer booklet of Part – A.
2. Avoid overwriting.

Questions 17 – 21 : Read the following poem.

Over the mountains, Over the plains, Over the rivers, Here come the trains.	Thousands of freight cars All rushing on Through day and darkness, Through dusk and dawn.
Carrying passengers, Carrying mail, Bringing their precious loads In without fail.	Over the mountains, Over the plains, Over the rivers, Here come the trains.

Now answer the questions. Each question has four choices. Choose the correct answer and write (A), (B), (C) or (D) in the brackets given. 5 × 1 = 5 M

Question 17.
The poem is mainly about ……………….
A) travelling in a train
B) the movement of trains
C) the speed of trains
D) the facilities in the train
Answer:
B) the movement of trains

Question 18.
The word ‘freight’ means ………………
A) goods carried from one place to another
B) passengers travelling in a train
C) everything in the train
D) important things in the train
Answer:
A) goods carried from one place to another

Question 19.
‘Through dusk and dawn’ means ……………….
A) in the morning
B) in the evening
C) in daytime
D) all the time
Answer:
D) all the time

Question 20.
The word ‘precious’ means ………………….
A) valuable
B) heavy
C) important
D) comfortable
Answer:
D) comfortable

Question 21.
The mood of the poem is ………………..
A) funny
B) serious
C) gloomy
D) enjoyable
Answer:
D) enjoyable

Questions (22 – 26): In the following passage, five sentences are numbered and each of them has an error. Correct them and rewrite them in the given.space. 5 × 1 = 5 M

An old tiger ran through the rain looking for shelter. (22) He was wet and cold so his cave was far away. (23) While hurried to his shelter he saw an old hut. (24) With a sigh of relief the tiger crawled in the thatched roof and lay down by the door. (25) Except for the sound of the rain was quiet all. Before he could nod off, however, he heard something heavy being dragged inside the hut. (26) This was followed on the voice of a woman.
Answer:
22. He was wet and cold and his cave was far away.
23. While hurrying to his shelter he saw an old hut.
24. With a sigh of relief the tiger crawled under / beneath the thatched roof and lay down by the door.
25. Except for the sound of the rain all was quiet.
26. This was followed by the voice of a woman.

Questions (27 – 31) : Complete the passage choosing the right words from those given below. Each blank is numbered and each blank has four choices (A), (B), (C) and (D). Choose the correct answer and write (A), (B), (C) or (D) in the brackets given. 5 × 1 = 5M

The Olympic Games ……………….. (27) in Olympia in 776 B.C. They took place from time to time ……………… (28) they stopped. At first, they lasted only one day and there was only one race. Later there were ………………. (29) races and contests and the Games lasted several days. People all over Greece ……………….. (30).

In 1894, a French man, Baron De Coubertin thought of starting the Games again and in 1896 the first Modern Olympics took place in Athens in Greece. The Greeks did not do very well, …………. (31) they won a very long race called Marathon.

Question 27.
A) begin
B) begins
C) has begun
D) began
Answer:
D) began

Question 28.
A) until
B) through
C) on
D) to
Answer:
A) until

Question 29.
A) more
B) the more
C) much
D) much more
Answer:
A) more

Question 30.
A) took up
B) took out
C) took part
D)took for
Answer:
C) took part

Question 31.
A) and
B) but
C) since
D) because
Answer:
B) but

Questions (32 – 36) : Read the following passage with focus on the underlined parts. Answer them as directed in the space given. 5 × 1 = 5M

I gave him the can of wine. He poured himself a mug and handed me the can. He drank all of it (32) at one go. He then arranged the belt that was (33) attatched to the trunk carefully on his forehead. So, this was the picture: my father carrying my luggage on his back and me following him with a (34) tiny bag in my hand. We were walking up a narrow hilly road, and neither of us uttered a word as if we were strangers who spoke different languages. I did not know what was going on in his mind. From time, to time it crossed my mind that it was improper for me to let father carry the luggage. I wanted to tell him that I would like to carry the trunk myself, but my guilt and shame did not allow me to do so. This self-consciousness had probably to do with my education, the white-collar job that I had, or quite simply my (35) proud. Somehow, I had the feeling that if I carried the luggage, my father and my people, in fact the (36) hole world would laugh at me and I would be belittled.

Questions :
32. Give the meaning of the underlined expression.
Answer:
all at once/all at one time/continuously/non-stop

33. Give the correct spelling of the underlined word.
Answer:
attached

34. Give the word that is opposite in meaning.
Answer:
huge / very big / large / vast

35. Replace the underlined word with its correct form.
Answer:
pride

36. Replace the underlined word with the suitable word which is pronounced similarly.
Answer:
whole

Timed practice with TS 10th Class Physical Science Model Papers Set 9 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 9 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section – I.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Suggest a new use with a spherical mirror.
Answer:

1. Concave mirrors are used in long focal length camera lenses.
2. In ‘Super snooper’ to focus and magnify sound from a great distance onto a microphone pick up.
3. At one end of some gas LASERS to focus the emerging beam of light.
4. In the telescopes to focus the very weak radio waves onto the suspended antenna.

Question 2.
What happens, if the eye lens lose its ability for accommodation ?
Answer:

1. If the eye lens lose its ability for accommodation, the vision becomes blurred.
2. A person who lose thsi ability of eye lens may suffer from eye defects like myopia, hypermetropia and presbyopia.

Question 3.
Write about electrolysis of NaCl.
Answer:

1. Fused NaCl is electrolysed with steel cathode and graphite anode.
2. The metal sodium (Na) will be deposited at cathode and chlorine gas liberates at the anode.
   At Cathode : 2 Na⁺ + 2e^– → 2Na
   At Anode : 2Cl^– → Cl₂ + 2e^–

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to 6 sentences.

Question 4.
The atomic number of elements P, Q, R, S and T are given below :

From the above table, answer the following questions :
a) Which element among these is a non metal ?
b) Which element belongs to 3rd period of periodic table?
c) Which is an inert gas ?
d) Which two elements are chemically similar ?
Answer:
The electronic configuration of the elements are as follows :

a) Element P is a non-metal since it has five valence electrons.
b) Element R belongs to third period since it has three shells.
c) Element Q is an inert gas element since it has complete octet.
d) Elements R and S are chemically similar since they have same number of valence electrons.

Question 5.
Write about ‘Hydrogen bond’.
Answer:

• Hydrogen bond is formed between molecules in which hydrogen atom is attached to an atom
  of an element with large electronegativity and very small size (F, O, N). Because of hydrogen bond the molecules associate themselves and hence possess higher B.P’s and M.P’s.
• The hydrogen bond formed between two molecules is called inter-molecular hydrogen bond.
• The hydrogen bond formed between different groups of the same molecule is called intra-molecular hydrogen bond.

Question 6.
Write the molecular formula of the first four compounds of the homologous series of aldehydes.
Answer:
Homologous series of aldehydes are Formaldehyde, Acetaldehyde, Propionaldehyde and Butylaldehyde.

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
What are the materials required for the experiment to show the chemical decomposition of water ? Write the procedure of the experiment. Name the products which we get in this reaction.
(OR)
What is meant by “water of crystallization” of a substance ? Describe an activity to show the water of crystallization.
Answer:
Aim : To Show Chemical decomposition of water
Materials required for chemical decomposition of water :
A plastic mug, two carbon rods, two corks, two test tubes, connecting wires,
9 V battery, water and some drops of acid.

Procedure :

1. Take a plastic mug, drill holes at the base.
2. Fit two one holed rubber stoppers in these holes:
3. Insert two carbon electrodes in these rubber stoppers.
4. Connect the electrodes to 9 V battery.
5. Fill the mug with water, so that the electrodes are immersed.
6. Add few drops of any acid.
7. Take two test tubes filled with water and insert them over the two carbon electrodes.
8. Switch on the circuit and leave the apparatus undisturbed for some time.
9. We notice the liberation of bubbles at both the electrodes. These bubbles displace the water in the test tubes.
10. After the test tubes are filled with gas, take them out and test the gases with a burning matchstick.
    The products which we get in this reaction: The gases are considered oxygen and hydrogen.

(OR)
Water of Crystallization : Water of crystallization is the fixed number of water molecules chemically attached to each formula unit of a salt in its crys¬talline form. The salts which contains water of crystallization are called hydrated salts.

Activity :

1. Take a few crystals of copper sulphate in a dry test tube.
2. Heat the crystals strongly by keeping the test tube over the flame of a burner for sometime.
3. On heating, the blue coloured copper sulphate crystals turns white and a powdery substance is formed.
4. We can also see tiny droplets of water in the test tube.
5. Now cool the test tube and add 2 or 3 drops of water on the white copper sulphate powder formed above.
6. The blue colour of copper sulphate crystals are reformed. They become blue again.
7. From the above activity we conclude that some water molecules are fixed in the blue coloured copper sulfate crystals.

Question 8.
How will you calculate the focal length of a biconvex lens that is used to correct the defect of Hypermetropia ? Explain it mathematically.
(OR)
Collect the information about the scientists who developed the atomic theories.
Answer:

1. Let us consider the object is at least distance of distinct vision (L).
2. Hypermetropia is corrected when the image is formed at the near point H.
3. This image acts like an object for the eye lens. Final image is formed at retina.
   Here object distance (u) = -25cm (least distance of distinct vision)
   image distance (v) = distance of near point (H) = -d
   Let ‘f’ be the focal length of a bi-convex lens
   Lens formula : \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
   = \(\frac{1}{-d}-\frac{1}{-25}\)
   = \(\frac{1}{-d}+\frac{1}{25}\)
   ⇒ \(\frac{1}{\mathrm{f}}=\frac{\mathrm{d}-25}{25 \mathrm{~d}}\) Hence f = \(\frac{25 d}{d-25}\)
4. From it if d > 25 cm, then f becomes +ve.
5. We need to use biconvex lens to correct the defect of hypermetropia.

(OR)

1. Atom is the smallest particle present in atom. Later it is proved proton (positive), electron (negative), neutron (neutral) are in atom.
2. Thomson proposed plum pudding model and atom as a whole electrically neutral.
3. Rutherford discovered nucleus and proposed that entire positive charge and mass of atom concentrate at nucleus and electrons are revolving around nucleus in different orbits.
4. Bohr proposed that electrons revolving around nucleus in stationary orbits. As long as electron revolve in a particular stationary orbit its energy does not change.
5. Quantum numbers finally explained the position of various sub-atomic particles that is electron, proton and neutron.

Question 9.
12 V battery is connected in a circuit and to this 4Ω, 12Ω resistors are cqnnected in parallel, 3Ω resistor is connected in series to this arrangement. Draw the electric circuit from this information and find the current in the circuit.
(OR)
A particle of charge 3.6 pC is projected into a uniform magnetic field of induction 5T with a speed of 2.5 × 10⁵ m/s. Calculate the radius and time period of the particle taking its mass to be 1 gm.
Answer:

Potential difference (V) = 12 V
Equivalent resistance between AB is R then,
\(\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{12}=\frac{3+1}{12}=\frac{4}{12}=\frac{1}{3}\)
R = 3 Ω
Equivalent resistance in the circuit
= R + 3 = 3 + 3 = 6Ω
Current in the circuit is I then I= \(\frac{V}{R}\)(∵ V = IR Ohm’s law)
= \(\frac{12}{6}\) = 2A.
∴ Current in the circuit (I) = 2A
(OR)
Mass of the particle : m = 1 gm = 10^-3 kg
Charge on the particle : q = 3.6
μc = 3.6 × 10^-6 C
Magnetic field induction : B = 5T
Speed of the particle : v = 2.5 × 10⁵ m/s
Radius of the circular part : r = \(\frac{\mathrm{mv}}{\mathrm{bq}}=\frac{10^{-3} \times 2.5 \times 10^5}{5 \times 3.6 \times 10^{-6}}\)
r = \(\frac{10^8}{7.2}\) ⇒ r = 1.3889 × 10⁷ ⇒ r = 1.4 × 10⁷ m
Time period of rotation: T = 2π\(\frac{\mathrm{m}}{\mathrm{Bq}}\) = 2π\(\frac{10^{-3}}{5 \times 3.6 \times 10^{-6}}\)
T = \(\frac{\pi}{9}\) × 10³ sec = T = \(\frac{\pi}{9}\) × 10³ sec

Part – B (10 × 1 = 10 Marks)

Instructions :

1. Answer ALL the questions.
2. Each question carries 1 mark.
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part – A.

Question 1.
In the case of …………….. mirror the magnification is always positive.
A) concave
B) plane
C) convex
D) all the above
Answer:
C) convex

Question 2.
A solution of potassium iodide reacts with lead nitrate to give ……………..
A) KNO₃
B) PbI₂
C) Both A and B
D) PbO
Answer:
C) Both A and B

Question 3.
Which of the following is produced by the action of chlorine on dry slaked lime?
A) Washing soda
B) Bleaching powder
C) Plaster of Paris
D) Baking soda
Answer:
B) Bleaching powder

Question 4.
Which of the following substances is used as an antichlor ?
A) CaOCl₂
B) Na₂S₂O₃
C) Na₂SO₄
D) CuSO₄
Answer:
B) Na₂S₂O₃

Question 5.
Which of the following precipitate is formed on mixing calcium hydroxide with carbon dioxide?
A) Red precipitate of calcium carbonate
B) Blue precipitate of calcium carbonate
C) White precipitate of calcium carbonate
D) Black precipitate of calcium carbonate
Answer:
C) White precipitate of calcium carbonate

Question 6.
The radius of curvature of a piano convex lens is doubled without changing the material of medium, it focal length will be
A) double
B) halved
C) remains same
D) insufficient data
Answer:
A) double

Question 7.
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
A) 1
B) 2
C) 1.5
D) -1.5
Answer:
B) 2

Question 8.
The angle of vision for a healthy human being is about
A) 60°
B) 30°
C) 45°
D) 90°
Answer:
A) 60°

Question 9.
Which of the following elements is electro-negative?
A) Sodium
B) Oxygen
C) Magnesium
D) Calcium
Answer:
B) Oxygen

Question 10.
The process of obtaining the pure metal from the impure metal is called ……………. of the metal.
A) Liquation
B) Refining
C) Poling
D) Distillation
Answer:
B) Refining

Timed practice with TS 10th Class Physical Science Model Papers Set 8 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 8 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section – I.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Write about various distances related to mirrors.
Answer:
The various distances related to mirrors are :

1. Focal Length (f) : The distance between vertex and focus is called focal length.
2. Radius of curvature (R) : The distance between vertex and centre of curvature is called radius of curvature.
3. Object distance : It is the distance between the object and pole of mirror and is denoted by ‘u’.
4. Image distance : It is the distance between the image and pole of the mirror and is denoted by ‘v’.

Question 2.
Why the least distance of distinct vision for children is 7 to 8 cm ?
Answer:

1. In this age the muscles around the eye are strong and flexible and can bear more strain.
2. So for children, the least distance of distinct vision is 7 to 8 cm.

Question 3.
Silicon is a metalloid. How do you support this ?
Answer:
Silicon exhibits following properties, so I conclude that it is a metalloid.

1. It is metallic lustre in nature.
2. It exists in several metallic and non-metallic compounds.
3. It having brittle nature.
4. All metalloids are usually occurs in combined states both metals and non-metals.

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to d sentences.

Question 4.
a) Which is more electro positive – Na or F ?
b) What is the number of electrons in the outer most orbit of Aluminium and Chlorine ?
c) Which forms cation easily – Mg or 0 ?
d) Which is better conductor of electricity – K or S ?
Answer:
a) Sodium (Z = 11) is more electro positive than F (Z = 9).
b) Al (Z = 13) : number of electrons present in outer most shell are = 3
Cl (Z = 17) : number of electrons present in outermost shell are = 7.
c) Magnesium (Z = 12) form cations easily than 0 (Z = 8).
d) Potassium (Z = 19) is better conductor of electricity than Sulphur (Z = 16).

Question 5.
Give electron dot formula for the following. ?
a) Magnesium chloride
b) Carbon dioxide
c) Carbon tetrachloride
d) Hydrogen bromide
Answer:

Question 6.
Which allotropes are used as good conductors ? Why ?
Answer:
1) Graphite and nanotubes are used as good conductors.
2) Both have layered structure. There is delocalised π electron system between layers. Because of this weaker system they are used as good conductors.

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
Explain the steps involved in balancing a chemical equation with an example.
(OR)
List out the material for the experiment to investigate whether all compounds containing Hydrogen are acids or not and write the experimental procedure.
Answer:
A chemical equation in which the number of atoms of different elements on the reactants side are same as those on product side is called a balanced equation.
Steps involved in balancing a chemical reaction : Let us consider the combustion reaction of Propane.

Step 1: Write the unbalanced equation using correct chemical formula for all substances.
C₃H₈ + O₂ → CO₂ + H₂O (Skeleton equation)

Step 2 : Compare number of atoms of each element on both sides.

Atom	No. of atoms in LH.S.	No. of atoms in RH.S.
C	3 ((in C3 H3)	1 (in CO2)
H	8 (in C3 H3)	2 (in H2O)
O	2 (in O2)	3 (inCO2, H2O)

Find the co efficients to balance the equation. In this case, there are 3 carbon atoms on the left side of the equation but only one on the right side. If we add a co-efficient of 3 to CO₂ on the right side the carbon atoms balance,
C₃H₈ + O₂ → 3 CO₂ + H₂O
Now, look at the number of hydrogen atoms. There are 8 hydrogen atoms on the left but only 2 on the right side. By adding a co-efficient of 4 to the H₂O on the right side, the hydrogen atoms get balanced.
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
Finally, look at the number of oxygen atoms. There are 2 on the left side but 10 on the right side. By adding a co-efficient of 5 to the 02 on the left side, the oxygen atoms get balanced.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Step 3 : Make sure the co-efficients are reduced to their smallest whole number values. The above equation is already with the co efficients in smallest whole numbers. There is no need to reduce its co efficients. Hence the final equation is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Step 4 : Check the answer. Count the numbers and kinds of atoms on both sides of the equation to make sure they are the same.
(OR)
Procedure :

1. Prepare glucose, alcohol, hydrochloric and sulphuric acid solution.
2. Fix two iron graphite rods on a rubber cork and place the cork in a 100 ml beaker.
3. Connect two different coloured electrical wires to graphite rods separately as shown in figure.
4. Connect free ends of the wire to 230 volts AC plug.
5. Complete the circuit as shown in the figure by connecting a bulb to one of the wires.
6. Now pour some dilute HCl in the beaker and switch on the current.

Observation : The bulb starts glowing.
Repetition : Repeat activity with dilute sulphuric acid, glucose and alcohol solutions separately.

Observation :

1. We will notice that the bulb glows only in acid solutions.
2. But the bulb not glows in glucose and alcohol solutions.

Result :

1. Glowing of bulb indicators that there is flow of electric current through the solution.
2. Acid solutions have ions and the movement of these ions in solution helps for flow of electric current through the solution.

Conclusion :

1. The positive ion (cation) present in HCl solution is H⁺.
2. This suggests that acids produced hydrogen ions H⁺ in solution, which are responsible for their acidic properties.
3. In glucose and alcohol solutions the bulb did not glow indicating the absence of H⁺ ions in these solutions.
4. The acidity of acids is attributed to the H⁺ ions produced by them in solutions.

Question 8.
Write the characteristics of the images formed by a convex lens having focal length of 25 cm, when an object is kept on the principle axis at a distance of 50 cm and 75 cm.
(OR)
Write postulates and limitations of Bohr Hydrogen atomic model.
Answer:
When the object is kept at 50 cm in front of the lens, the characteristics of image :

1. Image will form at 50 cm distance.
2. Size of image is equal to the size of the object.
3. Image is inverted.
4. Image is real.

When the object is kept at 75 cm in front of the lens, the characteristics of image:

1. Image will be formed between F and C (app. 37.5 cm)
2. Diminished image will be formed.
3. Image is inverted.
4. Image is real.

(OR)

Postulates :

1. Electrons revolve around nucleus in stationary circular orbits of fixed energies which are called energy levels.
2. As long as electron revolves in a stationary orbit, it neither loses nor gains energy.
3. These stationary orbits are denoted by the letters K, L, M, N …. or by the number n = 1, 2, 3, 4, ………….. , where, n stands for orbit number.

Limitations :

1. Bohr failed to explain atomic spectra of larger atoms which are heavier than hydrogen atom.
2. He also unable to explain the relative intensities of spectral lines, the existence of hyperfine lines and the Zeeman effect.

Question 9.
In a circuit, 60 V battery, three resistances R₁ = 10Ω, R₂ = 20Ω and R₃ = x Ω are connected in series. If 1 ampere current flows in the circuit, find the resistance R₃ by using Kirchhoff’s Loop law.
(OR)
A circular coil of radius 10 cm, 500 turns and resistance 2Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 sec. Estimate the magnitudes of the EMF and current induced in the coil. (Horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5T)
Answer:

According to Loop law,
60 – 10(I) – 20(I) – x(I) = 0
Substituting I = 1 Amp, in the above equation,
60 – 10 – 20 – x = 0
x = 30 Ω
∴ R₃ = 30 Ω
Another method :
R₁ = 10 Ω, R₂ = 2o Ω, R₃ = x Ω
As they are connected in series, the resultant
Resistance R = R₁ + R₂ + R₃
= 10 Ω + 20 Ω + 1 Ω
= 30 + x Ω
I = 1 Amp, V = 60 V
According to Ohm’s law.
V = IR
60 V = 1 × (30 + x Ω)
60 V = 30 + x Ω
x = 60 – 30
x = 30 Ω
∴ R₃ = 30 Ω.
(OR)
Earth’s magnetic field ‘B’= 3.0 × 10^-5 T
Area of the coil = πr² = π × 10^-2 m².
(radius = 10 cm = 10^-1m)
Initial flux through the coil
Q_(initial) = BA cos θ
= (3.0 × 10^-5) × (π × 10^-2) × cos 180° = 3π × 10^-7 wb
(since the plane of coil is perpendicular to the magnetic field, the angle is zero)
Final flux after rotation
Q_(initial) = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 180°
= – 3π × 10^-7wb
Total flux = (3π × 10^-7) – (- 3π × 10^-7)
= 6π × 10^-7 wb
The value of estimated emf in
ε = N.\(\frac{\Delta \phi}{\Delta t}\) = 500 × \(\frac{6 \pi \times 10^{-7}}{0.25}\) = 3.8 × 10^-3V
The value of estimated current is
I = \(\frac{3: 8 \times 10^{-3}}{2 \Omega}\) = 1.9 × 10^-3A.

Part – B (10 × 1 = 10 Marks)

Instructions :

1. Answer ALL the questions.
2. Each question carries 1 mark.
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part – A.

Question 1.
When parallel rays are Incident on a concave mirror, on reflection they meet at the
A) focus
B) pole
C) centre of curvature
D) none of these
Answer:
A) focus

Question 2.
If some amount of energy is released in a chemical reaction, then It Is called ……….. reaction.
A) exothermic
B) endothermic
C) oxidation
D) reduction
Answer:
A) exothermic

Question 3.
Why is universal indicator a better one than litmus paper ?
A) Litmus paper can only be used for acids.
B) Litmus paper can only be used for alkalies.
C) Universal indicator goes green if some thing is neutral.
D) Universal indicator is useful for all ranges of pH of the solution.
Answer:
D) Universal indicator is useful for all ranges of pH of the solution.

Question 4.
Tooth decay starts when the pH of the mouth is
A) 5.5
B) greater than 5.5
C) lower than 5.5
D) all of these
Answer:
C) lower than 5.5

Question 5.
Which of the following is the chemical formula for hydrated copper sulphate ?
A) CuSO₄.6H₂O
B) CuSO₄.7H₂O
C) CuSO₄.2H₂O
D) CUSO₄.5H₂O
Answer:
D) CUSO₄.5H₂O

Question 6.
If the convex lens is placed in water its focal length is ……………….
A) Increases
B) Decreases
C) remains constant
D) either increase or decrease
Answer:
A) Increases

Question 7.
If the refracted ray from a convex lens is travelling parallel to the principal axis, then image distance is
A) Equal to object distance.
B) Infinity
C) Equal to focal length of the lens
D) Equal to radius of curvature of the lens
Answer:
B) Infinity

Question 8.
In which type of defect in vision that people cannot see objects at long distances but can see nearby objects clearly ?
A) hypermetropia
B) presbyopia
C) myopia
D) all
Answer:
B) presbyopia

Question 9.
Which of the following indicatesThe correct order of variation in atomic size?
A) Be < C < F < Ne
B) Be > C < F < Ne
C) Be > C > F > Ne
D) F < Ne < Be < C
Answer:
B) Be > C < F < Ne

Question 10.
The reducing agent in thermite process is
A) Al
B) Mg
C) Fe
D) Si
Answer:
A) Al

Timed practice with TS 10th Class Physical Science Model Papers Set 7 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 7 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section – I.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Which objects at your home act as spherical mirrors ?
Answer:

1. The outer surfaces of stainless steel utensils, spoons, plates etc., act as convex mirrors.
2. The inner surfaces of the above said articles act as concave mirrors.
3. Any bulged surface with high polish acts as a convex mirror.

Question 2.
Doctor advised to use 2D lens. What is its focal length ?
Answer:
Given that the power of lens P = 2D
We know that, P = \(\frac{100}{\mathrm{f}(\mathrm{cm})}\)
2 = \(\frac{100}{\mathrm{f}}\); 2f = 100
f = \(\frac{100}{2}\) = 50 cm
∴ The focal length of the lens = 50 cm

Question 3.
What is 22 carat gold ? Why it is preferred for making jewellery ?
Answer:

1. Pure gold, known as 24 carat gold is very soft.
2. So it is not. suitable for making jewellery.
3. So they use 22 carat gold in which pure gold is alloyed with 2 parts of either silver or copper for making gold jewellery.

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to 6 sentences.

Question 4.
How many elements are there in 1^st period ? Why ?
Answer:

1. The first period contains only two elements.
2. The first period starts with k-shell, which is the first main shell.
3. It contains only one sub-shell that contains two types of electronic configurations.
4. So there are only two elements that are having electronic configuration as 1s¹ and 1s².

Question 5.
“Bond angle of ammonia reduced to 107°48’ from 109°28′” is said Ramya. Is she correct ? Justify your answer.
Answer:
Yes. She is correct.

Justification :

1. In NH₃ there are three bond pairs (N – H) and one lone pair of electrons (N) is present around the central atom of nitrogen.
2. So, the lone pair electron shows repulsion on bond pairs.
3. Hence, to minimize the repulsion force bond angle changes from 109°28’ to 107°48’.
4. At the same time the shape also changes from tetrahedral to pyramidal.

Question 6.
What is 22 carat gold ? Why it is preferred for making jewellery ?
Answer:

1. Pure gold, known as 24 carat gold is very soft.
2. So it is not suitable for making jewellery.
3. So they use 22 carat gold in which pure gold is alloyed with 2 parts of either silver or copper for making gold jewellery.

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
Balance the following chemical equations.
i) Zn_(s) + AgNO_3(aq) → Zn (NO₃)_2(aq) + Ag_(s)
ii) Fe₂O_3(s) + C_(s) → Fe_(s) + CO2(g)
iii) Ag + H₂S + O₂ → Ag₂S + H₂O
iv) Cu_(s) + O_2(g) → CUO_(s)
(OR)
List out the materials for the experiment “when” Hydrochloric acid reacts with NaHCO₃ and evolves CO₂”. Write the experiment procedure.
Answer:
i) Zn_(s) + AgNO_3(aq) → Zn(NO₃)_2(aq) + Ag_(s)
Step – I: Zn + 2 AgNO₃ → Zn (NO₃)₂ + Ag
Step – II: Zn_(s) + 2 AgNO_3(aq) → Zn (NO₃)_2(aq) + 2Ag_(s)

ii) Fe₂O_3(s) + C_(s) → Fe_(s) + CO_2(g)
Step – I: Fe₂O₃ + C → 2 Fe + CO₂
Step – II: 2Fe₂O₃ + C → 2 Fe + 3 CO₂
Step – III: 2Fe₂O_3(s) + 3C_(s) → 4 Fe_(s) + 3 CO_2(g)

iii) Ag_(s) + H₂S_(s) + O_2(g) → Ag₂S_(s) + H₂O_(l)
Step – I: 2 Ag+ H₂S + O₂ → Ag₂S + 2 H₂O
Step – II: 2 Ag + 2H₂S + O₂ → Ag₂S + 2 H₂O
Step – III: 2 Ag + 2H₂S + O₂ → 2Ag₂S + 2 H₂O
Step – IV: 4 Ag_(s) + 2H₂S_(g) + O_2(g) → 2Ag₂S_(s) + 2 H₂O_(l)

iv) CU_(s) + O_2(g) → CUO_(s)
Step – I: Cu + O₂ → 2CuO
Step – II: 2Cu_(s) + O₂2(g) → 2CuO_(s)
(OR)
Material required : Stand; two hold rubber cork, two test tubes, thistle funnel, delivery tube, Ca(OH)₂, Na₂CO₃, HCl.

Procedure :

1. Take two test tubes, label them as A and B.
2. Take about 0.5 gm of sodium carbonate – (Na₂CO₃) in test tube A and about 0.5 gm of sodium hydrogen carbonate (NaHCO₃) in test tube B.
3. Add about 2ml of dilute HCl to both the test tubes.
4. Pass the gas produced in each case through lime water and record your observations.
5. We observe that the lime water turns to milky white.
6. The reactions are
   Na₂CO_3(s) + 2HCl_(aq) → 2NaCl_(aq) + H₂O_(l) + CO_2(g)
   NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(g)
   Reactions when the gas is passed through lime water.
   Ca(OH)_2(aq) + CO_2(s) → CaCO₃ ↓+ H₂O(l)
   (white precipitate)
   On passing excess carbon dioxide, the following reaction takes place.
   CaCO_3(s) + H₂O_(l) + CO_2(s) → Ca(HCO₃)_2(aq) (Soluble in water)
   Thus from the above activity we can conclude that the reaction of metal carbonates and hydrogen carbonates with acids give a corresponding salt, carbon dioxide and water.

Question 8.
Refractive index of a lens is 1.5. When an object is placed at 30 cm, image is formed at 20 cm. Find its focal length. Which lens is it ? If the radii of curvature are equal then what is its value ?
(OR)
1s² 2s² 2p³ 3s² 3p⁶ 3d¹⁰ 4s¹ is the electronic configuration of Cu (Z = 29) which rule is violated while writing this configuration ? What might be the reason for writing this configuration ?
Answer:
Object distance u = – 30 cm (Infront of the lens)
Image distance v = 20 cm
From len’s formula = \(\frac{1}{\mathrm{f}}\) = \(\frac{1}{\mathrm{v}}\) – \(\frac{1}{\mathrm{u}}\) = \(\frac{1}{20}\) + \(\frac{1}{30}\)
⇒ f = 12 cm
The value of the focal length of the lens we get is positive so the lens is convex type.
For convex lens, Len’s maker’s formula is
\(\frac{1}{\mathrm{f}}\) = (1 – n) (\(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}\))
R₁ = R₂ = R
\(\frac{1}{12}\) = (1.5 – 1)(\(\frac{2}{R}\))
\(\frac{1}{12}\) = 0.5 × \(\frac{2}{R}\) ⇒ R = 12 cm
∴ Radius of curvature is 12 cm.
(OR)

1. The predicted electronic configuration of Cu (Z = 29) is 1s² 2s² 2p⁶ 3s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹.
2. But experimentally it has been showed that the actual configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.
3. Here Aufbau principle is violated. According to Aufbau principle, the electron occupies the orbital having least energy first.
4. That means in the ground state the electronic configuration can be built up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to the atomic number.
5. But the atoms with half filled or fully filled orbitals in their outermost orbit are more stable
   when compared with their neighbouring elements.
6. In order to get fully filled orbitals in the outermost orbit in copper the electrons in 4s and 3d re-distribute their energies to attain stability.
7. Hence the actual electronic configuration of copper is Cu (Z = 29) [Ar] 4s¹ 3d¹⁰

Question 9.
What are the factors affecting the resistance of an electric conductor ? Explain any two factors.
(OR)
Collect information about material required and procedure of making a simple electric motor from internet and make a simple motor on your own. (Or) Make a simple electric motor on your own with the help of internet information about simple electric motor for material; procedure.
Answer:
The factors affecting the resistance of an electrical conductor are

1. Nature of material
2. Temperature
3. Length of the conductor
4. Area of cross section of conductor

Explanation:

1. As the temperature increases the resistance increases and vice versa.
2. As the material changes resistance changes.
3. Resistance(R) is directly proportional to length(l) of the conductor (if T and A are kept constant) R ∝ l
4. Resistance(R) is inversely proportional to area of cross section (A) (if l and T are kept constant)
   R ∝ \(\frac{l}{\mathrm{~A}}\)

(OR)
Aim : Preparation of a simple electric motor.

Materials required : A wire of nearly 15 cm, 1.5V battery, iron nail, strong magnet paper clip.

Procedure :

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

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

Procedure:

1. Wind copper wire on the battery nearly 10 15 turns to make a coil.
2. Remove the coil and fix the ends as shown in the figure.
3. Scape the insulation completely on one end of the coil.
4. Scape the insulation on top, left and right of the other end. The bottom should be insulated.
5. Now complete the electric motor as shown in the figure.
6. We observe the rotation of coil.

Part – B (10 × 1 = 10 Marks)

Instructions :

1. Answer ALL the questions.
2. Each question carries 1 mark.
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part – A.

Question 1.
Which of the following image cannot be caught on the screen ?
A) real
B) lateral inversion
C) virtual
D) none of these
Answer:
C) virtual

Question 2.
The metal formed when zinc reacts with AgNO₃ is
A) Au
B) Ag
C) Al
D) Cu
Answer:
B) Ag

Question 3.
Which of the following is slightly soluble in water ?
A) Ca(OH)₂
B) KOH
C) Mg(OH)₂
D) Be(OH)₂
Answer:
D) Be(OH)₂

Question 4.
Which of the following acid is injected by stinging hair of leaves of nettle plant that causing burning pain?
A) Methanoic acid
B) Hydrochloric acid
C) Nitric acid
D) Sodium thiosulphate
Answer:
A) Methanoic acid

Question 5.
Which of the following acid is produced by human’s stomach ?
A) CH₃COOH
B) HCl
C) H₂SO₄
D) HNO₃
Answer:
B) HCl

Question 6.
Name of the following lens.

A) Concave lens
B) Convex lens
C) Plano concave
D) Plano convex
Answer:
D) Plano convex

Question 7.
denotes which lens ?
A) Convex
B) Concave
C) Plano convex
D) Plano concave
Answer:
B) Concave

Question 8.
The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
A) Presbyopia
B) Accommodation
C) Near sightedness
D) Far-sightedness
Answer:
B) Accommodation

Question 9.
What type of oxide would Ekaaluminium form ?
A) E₃O₂
B) E₂O₃
C) EO₃
D) EO
Answer:
B) E₂O₃

Question 10.
Steel is an alloy of …………..
A) Fe and Zn
B) Fe and C
C) Cr and Fe
D) Zn and Cr
Answer:
B) Fe and C

Timed practice with TS 10th Class Physical Science Model Papers Set 6 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 6 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section – I.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Identify types of mirrors without actually touching it. (move the mirror to and from)
Answer:

1. Plane mirror: It forms the image is of same size.
2. Concave mirror: Image is curved, bringing the mirror closer, magnifies the image. Moving it away, the image is inverted and reduced.
3. Convex mirror: Mirror image is always diminished but erect. The view point is wider.

Question 2.
Why do presbyopia is seen in aged people ?
Answer:

1. Presbyopia is generally seen in aged people.
2. This is due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens.
3. So that the ability of accomodation of the eye gradually, decreases with ageing.

Question 3.
Which ores of metals are needed roasting during extraction process ? Why ?
Answer:

1. The ores of such metals like zinc, iron, tin, lead and copper need roasting process.
2. Because they generally present as sulphides or Carbonates in nature! So first they converted into oxides by heating in excess air. This is nothing but roasting.
3. Then these, metal oxides are reduced into their metals.

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to 6 Sentences.

Question 4.
Explain Newland’s law of octaves with an example.
Answer:

1. Newlands arranged the elements in the ascending order of their atomic weights.
2. According to Newlands, if we start with hydrogen in the table, the eighth element is fluorine and next eighth element is chlorine and so on.
3. The properties of hydrogen, fluorine and chlorine are similar.
4. Every eighth element of a certain element have similar chemical properties in the table of seven groups.

Question 5.
Electronic configuration of a molecule ‘X’ is 1s² 2s² 2p¹. It should form XF molecule. But it forms . XF₃. HOW can it possible ? What the ‘X’ may be ?
Answer:
1. ‘X’ is nothing Boran.

2. It has to form one covalent bond with fluorine because of one un paired electron.

3. But it undergoes excitation and forms 3 unpaired electrons and the electronic configuration
of excited boran is 1s² 2s¹ 2p¹2p_y¹.

1. Each carbon is only joining to two other atoms rather than four or three.
2. Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridizes two of the orbitals.
3. They use the ‘s’ orbital (2s) and one of the 2p orbitals uncharged because they are made by an s-orbital and a p-orbital reorganizing themselves.
4. The new hybrid orbitals formed are called sp-hybrid orbitals.

Question 6.
What is sp hybridisation ? Explain.
Answer:

1. Each carbon is onlyjoining to two other atoms rather than four or three.
2. Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridizes two of the orbitals.
3. They use the ‘s’ orbital (2s) and one of the 2p orbitals uncharged because they are made by an s-orbital and a p-orbital reorganizing themselves.
4. The new hybrid orbitals formed are called sp-hybrid orbitals.

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
Write the required material and experimental procedure for the experiment, “Hydrochloric acid reacts with ‘Zn’ pieces and liberates H₂O.
(OR)
List out the materials required to test whether the solutions of given acids and bases contain ions or not. Explain the procedure of the experiment.
Answer:
Aim: To show the reaction of acids with metals.
Required Materials :
1. Test tube,
2. Delivery tube,
3. Glass trough,
4. Candle,
5. Soap water,
6. Dil. HCl,
7. Zinc granules.

Procedure:

1. Set the apparatus as shown in figure.
2. Take about 10 ml of dilute HCl in a test tube and add a few zinc granules to it.
3. We will observe the formation of gas bubbles on the surface of zinc granules.
4. Pass the gas being evolved through the soap water.
5. Gas filled bubbles are formed in the soap solution which rises into the air.
6. Bring a burning candle near the gas filled bubbles.
7. The gas present in a soap bubble burns with a ‘pop’ sound.

Result :

1. Only ‘hydrogen’ gas burns making a ‘pop’ sound.
2. So we will notice that gas evolved is H₂.

Chemical reaction:
Acid + Metal → Salt + Hydrogen
2HCl_(aq) + Zn_(s) → ZnCl_2(aq) + H_2(g)

Additional Experiment:

1. Repeat the above experiment with H₂SO₄ and HNO₃.
2. We will get the same observation of the HCl experiment.

Conclusion : From the above activities we can conclude that when acid reacts with metal, H₂ gas is evolved.
(OR)
Required materials: 100 ml beaker, rubber cork, graphite rods, two different coloured wires, bulb,
glucose, alcohol, hydrochloric and sulphuric acids.

Procedure:

1. Prepare glucose, alcohol, hydrochloric and sulphuric acid.solution.
2. Fix two iron graphite rods on a rubber cork and place the cork in a 100 ml beaker.
3. Connect two different coloured electrical wires to graphite rods separately as shown in figure.
4. Connect free ends of the wire to 230 volts AC plug.
5. Complete the circuit as shown in the figure by connecting a bulb to one of the wires.
6. Now pour some dilute HCl in the beaker and switch on the current.

Observation : The bulb starts glowing.

Repeatation : Repeat activity with dilute sulphuric acid, glucose and alcohol solutions separately.

Observation:

1. We will notice that the bulb glows only in acid solutions.
2. But the bulb not glows in glucose and alcohol solutions.

Result :

1. Glowing of bulb indicates that there is flow of electric current through the solution.
2. Acid solutions have ions and the movement of these ions in solution helps for flow of electric current through the solution.

Conclusion :

1. The positive ion (cation) present in HCl solution is H⁺.
2. This suggests that acids produced hydrogen ions H⁺ in solution, which are responsible for their acidic properties.
3. In glucose and alcohol solutions the bulb did not glow indicating the absence of H⁺ ions in these solutions.
4. The acidity of acids is attributed to the H⁺ ions produced by them in solutions.

Question 8.
Radii of biconvex lens are equal. Let us keep an object at one of the centres of curvature. Refractive index of lens is ‘n’. Assume lens in the air. Let us take R as the radius of the curvature.
a) How much is the focal length of the lens?
b) What is the image distance ?
c) Discuss the nature of the image.
(OR)
Make the s, p and d-orbital models.
Answer:
Radii of curvatures (R) of biconvex lens are equal, so, R₁ = R₂ = R
a) According to lens formula

Object is placed at centre of curvature.
So, object distance u = R
Let the image distance v = v
From above equation
\(\frac{1}{v}=\frac{2 n-2}{R}-\frac{1}{u}=\frac{2(n-1)}{R}-\frac{1}{R}\)
\(\frac{1}{v}=\frac{2 n-2-1}{R}-\frac{2 n-3}{R}\)
∴ Image distance = v = \(\frac{R}{2 n-3}\)

c) The nature of the image is inverted and v < u.

(OR)
Materials required: lo Iron spokes, 9 i’ound shaped wooden pieces, dumb-bell shaped beeds having hole along its length.

Procedure :

1. Cut the each iron spoke into three pieces in which one of them is longer than the other two.
2. Make a hole at the middle of the wooden piece and insert the long spoke into that hole.
3. Fix the other two-spokes to the longer spoke horizontally such that the three spokes are perpendicular to each other.
4. Now the three spokes represents x, y and z-axis.
5. Make nine of this types of models, one for s-orbital, three for p-orbitals and five for d-orbitals,
6. Fix the dumb-bell shaped beeds to the iron spokes as shown in the fig (a) – to make three (P_x P_y, P_z) P – orbitals.
7. Fix the dumb-bell shaped beeds to the iron spokes as shown in the fig (b) – to make five d-orbitals. (d_xy d_yz, d_zx, d_x2 – d_y2 and d_z2)

Question 9.
How do you verify that resistance of a conductor of uniform cross-section area is proportional to the length of the conductor at constant temperature.
(OR)
Collect information of experiments done by Faraday. (Or) What is the information do you collect of experiments done by Faraday ?
Answer:

1. To verify the relation between resistance and length of the conductor. Collect some wires or spokes of different lengths with same cross-section area of the same metal.
2. Construct a circuit with Battery, Ammeter, Switch (key) and connecting wires, keeping some space at the both ends.
3. Connect the selected wires or spokes at the ends to complete the circuit.
4. Connect the wires or spokes individually and record the current using ammeter.
5. If the current flowing in the circuit decreases with an increase in the length of the wire or spokes (Resistance increases), we can say that the resistance of the conductor is proportional to the length of the conductor.

(OR)
Experiment – 1

1. Connect the terminals of a coil to a sensitive galvanometer.
2. Normally, we would not expect any deflection of needle in the galvanometer because there is no EMF in the circuit.
3. 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.
4. 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.
5. 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.
6. This activity proves that the change in magnetic flux linked with a closed coil, produces current.
7. From this Faraday’s law of induction can be 1 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

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

Part – B (10 × 1 = 10 Marks)

Instructions :

1. Answer ALL the questions.
2. Each question carries 1 mark.
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and
   write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part – A.

Question 1.
Starting from a long distance, a flame Is moved towards a convex mirror. Then the image
A) Decreases in size and moves towards pole
B) Increase in size and moves towards pole
C) Decrease in size and moves away from pole
D) Increase in size and moves away from pole
Answer:
B) Increase in size and moves towards pole

Question 2.
In exothermic reactions heat energy is
A) absorbed
B) created
C) released
D) none of these
Answer:
C) released

Question 3.
The acid which enters the body by the sting of bee is
A) Acetic acid
B) Methanoic acid
C) Sulphuric acid
D) Fatty acid
Answer:
B) Methanoic acid

Question 4.
An aqueous solution of the salt is acidic which of the following acids and bases react to give this salt?
A) Strong acid and strong base
B) Strong acid and weak base
C) Weak acid and strong base
D) Weak acid and weak base
Answer:
B) Strong acid and weak base

Question 5.
Select a pair of basic salts among the following.
A) Sodium chloride and Sodium acetate
B) Sodium acetate and Sodium bicarbonate
C) Sodium carbonate and Sodium sulphate
D) Sodium nitrate and Sodium oxalate
Answer:
B) Sodium acetate and Sodium bicarbonate

Question 6.
Which of the following lens act as converging lenses ?
A) Bi-convex lens
B) Plano-convex lens
C) Concave convex
D) All the above
Answer:
D) All the above

Question 7.
Which one of the following is normally used for the preparation of lenses?
A) Water
B) Glass
C) Plastic
D) All of these
Answer:
B) Glass

Question 8.
A person with ……… can see distances objects usually but cannot see objects at near distances.
A) myopia
B) hypermetropia
C) presbyopia
D) none of these
Answer:
B) hypermetropia

Question 9.
1 pm = …………..
A) 10^-11
B) 10^-12
C) 10^-13
D) 10^-18
Answer:
B) 10^-12

Question 10.
Food cans are coated with tin and not with zinc because
A) zinc is less reactive than tin
B) zinc is costlier than tin
B) zinc is costlier than tin
D) zinc is more reactive than tin
Answer:
D) zinc is more reactive than tin

Timed practice with TS 10th Class Physical Science Model Papers Set 5 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 5 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section – I.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Write about various distances related to mirrors.
Answer:
The various distances related to mirrors are :

1. Focal Length (f) : The distance between vertex and focus is called focal length.
2. Radius of curvature (R) : The distance between vertex and centre of curvature is called radius of curvature.
3. Object distance : It is the distance between the object and pole of mirror and is denoted by ‘u’.
4. Image distance : It is the distance between the image and pole of the mirror and is denoted by ‘v’.

Question 2.
Why do we use lenses in spectacles to correct defects of vision?
Answer:

1. The process of adjusting focal length is called “accommodation”.
2. This process has to be done by eye itself.
3. Sometimes the eye may gradually lose its power of accommodation. In such condition the person cannot be able to see the object clearly and comfortably.
4. In this situation we have to use lenses in spectacles to correct defects of vision.

Question 3.
How do various metals in activity series react with dilute strong acids ?
Answer:

1. The metals from potassium to lead displace hydrogen from dilute strong acids with decreasing reactivity.
   a) The reaction of potassium is explosive.
   b) The reaction ofmagnesium is vigorous.
   c) The reaction of iron is steady.
   d) The reaction of lead is slow.
2. The metals from copper to gold do not displace H2 from strong dilute acids.

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to 6 sentences.

Question 4.
The electronic configuration of atom A is 2, 8, 6
a) What is the atomic number of element A ?
b) State whether the atomic size of element A is bigger or smaller than the atom having atomic number 14. Why ?
c) Which of the elements exhibit similarity in chemical properties as element A O(8), C(6), N(7), AV(18). Why ?
d) How the element is formed inert gas configuration ?
Answer:
The electronic configuration of atom – A is 2, 8, 6.
a) Atomic number of element‘A is 16. i.e,, Sulphur.

b) The atom which having atomic number -14 is Silicon (Si).
Atomic size of element decreases across period from left to right. So. the atomic size of element ‘A’ is smaller than the atom having atomic number 14.

c) Element oxygen O₈ – exhibit similarity in chemical properties as element A, because they belong to the same group.

d) Given element – A becomes inert gas i.e., Argon configuration by gaining ‘2’ electrons.

Question 5.
What is Hybridisation ? Explain the formation of BeF₂ molecule using hybridisation.
Answer:
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.

This process is called hybridisation.
BeF₂ does not exist. Because valency of Be is 2 and the valency of F is 3. It means there are no equal energy orbitals in outer shells of Be and F.

Question 6.
How many isomers can be drawn for pentane with molecular formula C₅H₁₂ ? What are they ? Draw their structures and mention their common names.
Answer:
Isomers of pentane are three.
There are :

1. Pentane
2. Isopentane
3. Neo pentane

Structures :

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
Balance the following chemical equation after writing the symbolic representation.
a) Magnesium_(s) + Hydrochloric acid_(aq) → Magnesium chloride_(aq) + Hydrogen_(g)
b) Zinc_(s) + Calcium chloride_(aq) → Zinc Chloride_(aq) + Calcium_(s)
(OR)
Raghu wants to test the pH of given samples by using universal indicator. What is the procedure that he follows?
Answer:
A) a) Magnesium_(s) + Hydrochloric acid_(aq) → Magnesium

1) Write the unbalanced equation using correct chemical formula for all sub-stances.
Mg_(s) + HCl_(aq) → MgCl_2(aq) + H_2(g)

2) Balancing the atoms of each element on both sides.
1 Mg_(s) + 2 HCl_(aq) → 1 MgCl_2(aq) + 1 H_2(g)

3) Write the co-efficient of smallest ratio.
1 Mg_(s) + 2 HCl_(aq) → 1 MgCl_2(aq) + 1 H_2(g)

4) The above equation is balanced.
Mg_(s) + HCl_(aq) → MgCl_2(aq) + H_2(g)

b) Zinc_(s) + Calcium chloride_(aq) → Zinc Chloride_(aq) + Calcium_(s)

1) Write the unbalanced equation using correct chemical formula for all substances.
Zn_(s) + CaCl_2(aq) → ZnCl_2(aq) + Ca_(s)

2) Balancing the atoms of each element on both sides.
1 Zn_(s) + 1 CaCl_2(aq) → 1 ZnCl_2(aq) + 1 Ca_(s)

3) Write the co-efficient of smallest ratio.
1 Zn_(s) + 1 CaCl_2(aq) → 1 ZnCl_2(aq) + 1 Ca_(s)

4) The above equation is balanced.
Zn_(s) + CaCl_2(aq) → ZnCl_2(aq) + Ca_(s)
(OR)
B) To test the pH of given samples, Raghu has to follow the below procedure.

1. He has to take five clean test tubes in a test tube stand and label A, B, C, D and E to these tubes.
2. He has to pour 5 ml of given samples in each test tube.
3. He has to add two drops of universal indicator to each test tube.
4. After changing the colour he has to compare with standard colour chart.
5. The chart gives approximate value of pH of the concerned sample.
6. The observations to be recorded and finally he may confirm the nature of the given samples.

Question 8.
Derive expression for lens maker’s formula. (Or) Prove \(\frac{1}{\mathrm{f}}\) = (n – 1) )(\(\frac{1}{R_1}-\frac{1}{R_2}\))
(OR)
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ is the electronic configuration of Cu (Z = 29) which rule is violated while writing this configuration ? What might be the reason for writing this configuration ?
Answer:
A) 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.
  Substituting the values in
  We have \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{\left(n_2-n_1\right)}{R}\) ……… (1)
• But the ray 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 the concave surface of lens is medium -1 and surrounding is medium -2.
  ∴ n₁ = n_b and n₂ = n_a
  Substituting values in
  \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\left(\mathrm{n}_2-\mathrm{n}_1\right)}{\mathrm{R}} \frac{\mathrm{n}_{\mathrm{a}}}{\mathrm{v}}-\frac{\mathrm{n}_{\mathrm{b}}}{\mathrm{x}}=\frac{\left(\mathrm{n}_{\mathrm{b}}-\mathrm{n}_{\mathrm{a}}\right)}{\mathrm{R}_2}\) …………. (2)
  Adding (1) and (2) and dividing both sides by n_a we have

If the surrounding medium is air we have ba = n (absolute refractive index)
∴ \(\frac{1}{\mathrm{f}}\) = (n – 1) )(\(\frac{1}{R_1}-\frac{1}{R_2}\))
This is called lens maker’s formula.

(OR)
B)

1. The predicted electronic configuration of Cu (Z = 29) is 1s² 2s² 2p⁶ 3s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹.
2. But experimentally it has been showed that the actual configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.
3. Here Aufbau principle is violated. According to Aufbau principle, the electron occupies the orbital having least energy first.
4. That means in the ground state the electronic configuration can be built up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to the atomic number.
5. But the atoms with half filled or fully filled orbitals in their outermost orbit are more stable when compared with their neighbouring elements.
6. In order to get fully filled orbitals in the outermost orbit in copper the electrons in 4s and 3d re-distribute their energies to attain stability.
7. Hence the actual electronic configuration of copper is Cu (Z = 29) [Ar] 4s¹3d¹⁰

Question 9.
A circuit is made with a copper wire as shown in the diagram. We know that conductor’s resis¬tance is directly proportional to its length. Calculate the equivalent resistance between points 1 and 2.
(OR)
Explain the working process of induction stove.
Answer:
A) Let the resistance of the wire be ‘R’ and length of the wire be ‘l’.
The shape of the circuit be square length of the side (l) = R
In a square diagonal is √2 times its length = √2l

Resistance towards diagonal is √2R
The circuit diagrams for the given arrangement is along PTR and QTS is in effective as no current flows through it.

PQ and PS are in series so, effective resistance are R₁ + R₂ = R + R = 2R.
QR and SR are in series so, effective resistance are R₁ + R₂ = R + R = 2R
Redrawn of the circuit again as resultant resistance between the points 1 and 2 is,

(OR)
B)

1. An induction stove works on the principle of electromagnetic induction.
2. A metal coil is kept just beneath the cooling surface. It carries alternating current (AC). So that AC produces an alternating magnetic field.
3. When you keep a metal pan with water on it, the varying magnetic field beneath it crosses the bottom surface of the pan and EMF is induced in it.
4. Since pan has a finite resistance, the flow of induced current in it produces heat and this heat is conducted to the water.
5. That’s why we call this stove as induction stove.

Part – B (10 × 1 = 10 Marks)

Instructions :

1. Answer ALL the questions.
2. Each question carries 1 mark.
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part – A.

Question 1.
Which mirrors are used in many vehicles as rearview mirror and window glasses of a car?
A) convex
B) bi convex
C) bi-concave
D) concave
Answer:
A) convex

Question 2.
The formula of hypochlorous acid is
A) HCl
B) H₂SO₄
C) HOCl
D) H₂Cl₃
Answer:
C) HOCl

Question 3.
The reaction of an acid with a base to give a salt and water is known as
A) Neutralization
B) Reduction
C) Oxidation
D) Crystallisation
Answer:
A) Neutralization

Question 4.
Which one of the following is weak acid ?
A) HNO₃
B) CH₃COOH
C) HCl
D) H₂SO₄
Answer:
B) CH₃COOH

Question 5.
If the pH value of a solution is 12, then the character of solution of the solution is ( )
A) Base
B) Strong acid
C) Acid
D) Strong base
Answer:
D) Strong base

Question 6.
The focal length of the piano 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
A) \(\frac{3}{2}\)
B) \(\frac{5}{2}\)
C) \(\frac{2}{5}\)
D) \(\frac{2}{5}\)
Answer:
C) \(\frac{2}{5}\)

Question 7.
If the object is placed beyond the centre of curvature (C2) on the principal axis, then the image is formed ……………
A) between F₁ and C₁
B) beyond C₁
C) ‘C’
D) at infinity
Answer:
B) beyond C₁

Question 8.
Rainbow is due to
i) scattering of light
ii) refraction of light
iii) dispersion
iv) total internal reflection of light
A) i and iii
B) i and ii
C) iii and iv
D) i, ii and iv
Answer:
C) iii and iv

Question 9.
Elements belonging to the same period have
A) Same valency electrons
B) Same number of shells
C) Same atomic size
D) Same electron affinity
Answer:
B) Same number of shells

Question 10.
Magnetite is an ore of
A) Hg
B) Ag
C) Mg
D) Fe
Answer:
D) Fe

Students must rely on TS 10th Class Biology Model Papers Set 5 to gauge their understanding of exam patterns.

TS 10th Class Biology Model Paper Set 5 with Solutions

Time: 1: 30 hours
Max. Marks:40

Instructions:

1. Read the question paper carefully and understand.
2. Answer the questions under Part – A in the answer sheet provided.
3. Part – A contains three sections: section – I, II, and III.
4. There is an internal choice to the questions under section – III.
5. Part-B Answers should be written in the given brackets and attached to the Part – A answer sheet.
6. Write the answers following the instructions given in each section.

Part – B (Marks 30)
Section-I (3 x 2 = 6 M)

Instructions:

• This section contains 3 very short answer questions.
• Answer ALL the questions.
• Write answer to each question in 3 to 4 sentences.
• Each question carries 2 marks.

Question 1.
How do you identify pellagra?
Answer:

1. Patients of pellagra suffer from dermatitis, diarrhea, and dementia.
2. When their skin is exposed to sunlight their skin becomes pigmented, scaly, and cracked.

Question 2.
Can we say that combustion and respiration are almost same actions? What evidences do you have for this?
Answer:

1. in both these processes sugar is converted to carbon dioxide and water.
2. Both these processes require oxygen.
3. Both combustion and respiration release energy.

Question 3.
Why doctors advised to take low amounts of salt in food?
Answer:
If salt is not reduced in food the salt levels increase in blood and cause other problems in the body. Salt (sodium) levels will be more in accumulated water at the time of edema. Hence it is advised to take low amount of salt in food.

Section – II (3 x 3 = 9 M)

Instructions:

• This section contains 3 short answer questions.
• Answer ALL the questions.
• Write answer to each question in 5 to 6 sentences.
• Each question carries 3 marks.

Question 4.
Divide the following into groups based on their actions. (walking, blinking of eye, heartbeat, laughing, digestion of food, reading etc.)
Answer:

• The given actions are divided based on the stimulus and response of our body.
• Some are voluntary actions like E.g.: Walking, laughing, and reading.
• There are involuntary actions like E.g.: Blinking of eye, heartbeat, and digestion of food.
• Voluntary actions are under the control of the brain.
• Involuntary actions are under the influence of reflex-are.

Question 5.
How do traits get expressed according to Mendel?
Answer:

1. Mendel hypothesized that each character or trait is expressed due to a pair of factors or alleles’, as he named them.
2. Now we know that these are known as genes.
3. Gene is a segment of nucleic acid called DNA which is present in the nucleus of every cell.

Question 6.
How do we follow sustainable forestry practices?
Answer:
Sustainable forestry practices are critical for ensuring resources well into the future.
They are:

• Low-impact logging practices.
• Harvesting with natural regeneration in mind. ,
• Avoiding certain logging techniques, such as removing all the high-value trees.

Section-III (3 x 5 = 15 M)

Instructions:

• This section contains 3 essay-type questions.
• Answer ALL the questions.
• There is an internal choice for each question.
• Write answer to each question in 8 to 10 sentences.
• Each question carries 5 marks.

Question 7.
Do the malnutrition reason for diseases? Why? Write any of such diseases and its characters.
(OR)
Classify different types of blood vessels in humans. On what basis you classified the blood vessels?
Answer:

1. Yes, malnutrition is the reason for occurrence of diseases.
2. Malnutrition is eating of food that does not have one or more than one nutrient in required amount.
3. This results in scarcity of nutrients for the proper growth and health of the individual.
4. Kwashiorkor disease occurs in children due to the deficiency of proteins in the diet.

Characteristic features of kwashiorkor disease:

1. Body parts becomes swollen due to accumulation of water in the intercellular spaces.
2. Very poor muscle development, swollen legs, fluffy face, difficult to eat, diarrhea, dry skin are the symptoms of kwashiorkor disease.
3. The child becomes lethargic and shows little interest in its surroundings or in playing and learning.

(OR)
There are two different types of blood vessels in human body i) Arteries ii) Veins. They are classified on the basis of blood flow in between the body parts and heart. Arteries carry the blood from heart to the cells of body. Veins carry the blood from the cells of body to heart.

Blood vessels that bring blood to the heart:

1. Superior vena cava (Or) Precaval vein: It collects deoxygenated blood from the upper parts of the body.
2. Inferior vena cava (Or) Post cava vein: It brings deoxygenated blood from lower parts of the body to heart.
3. Pulmonary vein: It brings oxygenated blood from the lungs and opens into left auricle of the heart.

Blood vessels that carry blood away from heart:

1. Systemic aorta or artery: It arises from left ventricle and carries oxygenated blood to all the body organs except lungs.
2. Pulmonary aorta: It carries deoxygenated blood from right ventricle to lungs.

Question 8.
Describe female reproductive system with well labelled diagram.
(OR)
What are the changes that occur ¡ri Metaphase, Anaphase and Telophase of the mitosis cell division?

Answer:

1. The female reproductive system consists of ovaries, oviduct, uterus, and vagina.
2. Two ovaries where ova are formed are located deep in the lower abdomen of female body.
3. The ova develops in tiny cellular structures called follicles.
4. Which at first look like cellular bubbles in the ovary and are called graffian follicles.
5. As a follicle grows it develops a cavity-filled Female reproductive system with fluid.
6. When an ovum is matured, the follicle ruptures at the surface of the ovary and the tiny ovum is flushed out.
7. Generally, the ovum enters the widened funnel of an oviduct, a tube that extends from the neighbourhood of an ovary to the muscular thick-walled uterus.
8. This uterus which is sac-like called vana.
9. Ovaries secretes two types of hormones which help in ovulation and menstruation.
10. Women got the capacity to produce ova from the age of 10 – 12 years and it will continue upto the age of 45 – 50 years.

(OR)
Metaphase:

1. During metaphase chromosomes move to the spindle equator, centromeres attached to spindle fibers.
2. Centromeres split, separating the chromatids.

Anaphase:
In this phase fibres attached to centromeres contract, pulling chromatids towards poles.
Telophase:

1. In this phase, chromatids elongate become invisible.
2. Nuclear membranes form round daughter nuclei.
3. Cell membrane pinches in, to form daughter cells or new cell wall materials becomes laid down across spindle equator.
4. Nucleus divides into two and division of cytoplasm starts.

Question 9.
What do you think how the food is controlled while it passes in the oesophagus?
(Or)
Explain the movement of food in oesophagus through peristaltic movements along with diagram.
(OR)
Which pyramid is always upright? Why?
Answer:

1. Due to the contraction and relaxation of the muscles the food bolus in oesophagus, propels into the stomach. It is called peristalsis. It is an involuntary action.
2. Due to the peristaltic movements in the oesophagus the food moves forward.
3. The walls of the food pipe secrete a slippery substance called mucus. It lubricates and protects the oesophageal wall from damage.
4. The wall of the oesophagus comprises two types of mouth muscles. The inner layer comprises circular muscles and the outer layer comprises longitudinal muscles.
5. Contraction of the circular muscles results in narrowing of the oesophagus just behind the bolus. So the food is squeezed downwards.
6. Contraction of the longitudiñal muscles in front of the bolus, widens the tube, this results in shortening of that particular part of oesophagus.
7. The contraction and relaxation of these muscles create a wave-like motion, that propels the food bolus into stomach.
8. The total process is under the control of the autonomoùs nervous system.

(OR)

1. The pyramid of energy is always upright.
2. This is because only 10% of the energy present in atrophic level transfers to the other level.
3. If there are 1000 calories of net production at producer level, only 100 calories of secondary production would be expected at the herbivore level, only 10 calories at first carnivore level, and I calorie at top carnivore level.
4. The flow of energy from producers to consumers is unidirectional.
5. So energy at producer level is always greater than the energy at primary consumer level and so on.
6. Hence the pyramid of energy is always upright with a typical pyramid shape.
7. But in case of pyramid of numbers, number of producers may be less than consumers. In some ecosystems or number of primary consumers may be less than secondary consumers etc., causing inverted or partly upright.
8. In case of pyramid of biomass also the pyramid may be inverted as in aquatic ecosystem, where the biomass of producers is less than that of consumers.

Part-B (Marks 10)

Instructions:

• Write the answers to the questions under Part – B on the question paper itself and attach it to the answer book of Part – A.
• Each question carries 1 mark.
• Marks will not be awarded in any case of overwriting, rewritten or erased answers.
• Write the capital letter (A / B / C / D) showing the correct answer for the following questions in the brackets provided against them.

Question 1.
The four important factors of photosynthesis are ( )
A) light, CO₂, chlorophyll and temperature
B) light, chlorophyll, water and temperature
C) light, CO₂, water and chlorophyll
D) water, chlorophyll, temperature, and CO₂.
Answer:
C) light, CO₂, water and chlorophyll

Question 2.
Glucose → Pyruvate → CO₂ + Water + More energy ( )
A) Bacteria
B) Animals
C) Plants
D) B and C
Answer:
A) Bacteria

Question 3.
The interior lung is divided into millions of small chambers called ( )
A) Nephrons
B) Corpus Leuteum
C) Alveolus
D) Neuron
Answer:
B) Corpus Leuteum

Question 4.
Which of the following opinion is correct? ( )
A) Ravi said, xylem and phloem cells arranged one upon the other to form a tube-like structure.
B) John said that xylem and phloem are not separate tube-like structures.
C) Salma said, the xylem and phloem cells connect together to form a tube-like structure.
D) Han said, because of its shape they said to be tube-like structures.
Answer:
C) Salma said, the xylem and phloem cells connect together to form a tube-like structure.

Question 5.
Tears are ……………….. . ( )
A) Involuntary fluid
B) Secretion
C) Excretion
D) Water
Answer:
C) Excretion

Question 6.
Some animals can’t eat some parts of the plants. Because ( )
a) They do not have taste
b) Wastes are stored in them
c) They have thorns’
d) Animals don’t prefer them for food
A) a, b are false
B) c, d are true
C) c, d are false
D) a, b are true
Answer:
D) a, b are true

Question 7.
Read the given table. ( )
Organism Excretory system organs
Coelenterate Water vascular system.
Annetida Nephridla
Birds Kidney
Name the group organism that performs water vascular system other than coelenterate.
A) Protozoa
B) Porifera
C) Mollusca
D) Manmali
Answer:
B) Porifera

Question 8.
We can’t see anything soon after entering a dark room. Find the reason.
a) Cones in the eyes do not respond
b) The radius of pupil decreases
A) both a and b are true b explains ‘a’.
B) both a and b are true. b Does Flot explain ‘a’
C) a is true,b is false
D) b is false, a is false
Answer:
D) b is false, a is false

Question 9.
Find out the correct sentence ( )
a – Centromeres in metaphase split and separate the chromatids.
b – Nucleus divides into two in prophase
A) both a & b are true
B) a is true, b is false
C) a is false, b is true
D) both a & b are false
Answer:
B) a is true, b is false

Question 10.
The chromosomal number becomes half in certain types of cells in cell division. This type of cell division takes place in
( )
A) In testes
B) In ovaries
C) In both
D) In all body cells
Answer:
C) In both

Timed practice with TS 10th Class Physical Science Model Papers Set 4 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 4 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section – I.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Suggest a new use with a spherical mirror.
Answer:

1. Concave mirrors ate used in long focal length camera lenses.
2. In ‘Super snooper’ to focus and magnify sound from a great distance onto a microphone pickup.
3. At one end of some gas LASERS to focus the emerging beam of light.
4. In the telescopes to focus the very weak radio waves onto the suspended antenna.

Question 2.
How can we appreciate the working of “Iris” ?
Answer:

1. Iris consists of muscles which helps in controlling the amount of light entering the eye through pupil.
2. Increase of low light the iris makes the pupil to expand and allow more light to enter the eye.
3. In case of bright (or) excess light the iris makes the pupil to contract in order to decrease the amount of light entering the eye. The iris consists of muscles that expand and contract the pupil.

Question 3.
Give some examples for corrosion.
Answer:
Examples for corrosion:

1. The rusting of iron (Iron oxide)
2. Tarnishing of silver (Silver sulphide)
3. Development of green coating on copper (Copper carbonate) and bronze.

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to 6 sentences.

Question 4.
Some elements belonging to second period of periodic table, and their atomic radii are given below. Observe them and write answers.

a) Write the elements in the ascending order of their atomic radii.
b) Which of the 2nd period elements closer to the configuration of inert gas ?
c) Which is the outermost orbit of all these elements ?
d) Which elements atomic size is bigger. Beryllium or Carbon ? Why ?
Answer:
a) The ascending order of atomic sizes is O, N, C, B, Be and Li.
b) Lithium has closest inert gas configuration i.e., 1s2 2sl. Its nearest inert gas is helium.
c) The outermost orbit for all these elements is second orbit.
d) Beryllium has more atomic size than Carbon. Because when we move across a period the atomic number increases. So nuclear attraction of outermost orbit increases. So carbon has lesser size than Beryllium.

Question 5.
Elements X, Y have atomic numbers 9 and 12 respectively. Which one of these
a) forms negative ion ?
b) forms positive ion ?
Answer:
a)
i) An element whose atomic number 9 is fluorine.
Electronic configuration is 1s²2s²2p⁵.
ii) To attain octet configuration one electron is required. So, it possesses high electronegativity. By gaining one electron it becomes F^Θ(negative ion).
F + le^Θ → F^Θ
iii) Then the electronic configuration of F is 1s²2s²p⁶.

b) iv) An element whose atomic number 12 is magnesium. Electronic configuration is 1s²2s²2p⁶3s².
v) To attain octet configuration two electrons it has to loose. So, it possesses high electropositivity. By loosing two electrons it becomes Mg²⁺(Positive ion).
Mg – 2^e- → Mg²⁺
vi) Then the electronic configuration of Mg²⁺ is 1s²2s²2p⁶.

Question 6.
Draw the structures of the following, a) Ethanoic acid b) Propanol c) Propene
Answer:

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
What is the difference between dis-placement and double displacement reactions ? Write equations for these reactions.
(OR)
Describe how sodium hydroxide is obtained from common salt.
Answer:

1. Displacement reaction: The reaction in which an element displaces another element from a substance is called “displacement reaction.” It is represented as
   AB + C → AC + B
   Ex: Zn + 2HCl → ZnCl₂ + H₂
2. In displacement reaction one element displaces another element from its compound and takes its place therein.
3. Here in above equation hydrogen has displaced from acids by metals.
4. Double displacement: The reaction in which the constituent atoms of the reaction interchange by the decomposition of the two reactants, to produce new compounds are called double displacement reactions.
   They are generally represented as follows.

Ex:
1) MgCl₂ + 2KOH → Mg(OH)₂ + 2KCl
2) HCl + NaOH → NaCl + H₂O
3) Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
(OR)

1. When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide.
2. The process is called chloralkali process because of the products formed chlor for chlorine and alkali for sodium hydroxide
   2NaCl_(aq) + 2H₂O_(l)
   2NaOH_(aq) + Cl_2(g) + H_2(g)
3. Chlorine gas is liberated at the anode and hydrogen gas at the cathode.
4. Sodium hydroxide solution is formed near the cathode.
5. The three products produced in this process are all useful.

Question 8.
Draw ray diagrams of image formed by a convex lens at various distances.
(OR)
Explain Bohr – Sommerfeld model of an atom. What is the merit of this model ? What are its limitations?
Answer:
1) Object at infinity.

2) Object placed beyond the centre of curvature (2F).

3) Object placed at the centre of curvature.

4) Object placed between C and F.

5) Object at the focal point.

6) Object placed between F and P.

(OR)

1. In an attempt to account for the structure of line spectra, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.
2. While retaining the first of Bohr’s circular orbit as such, he added one elliptical orbit to Bohr’s second orbit, two elliptical orbits to Bohr’s third orbit etc.
3. Nucleus of the atom is one of the principal foci of these elliptical orbits because periodic motion under the influence of a central force will lead to elliptical orbits with the force situated at one of the foci.
   Merit : Bohr – Sommerfeld model successful in accounting for the fine line structure of hydrogen atomic spectra.

Limitations:

1. This model failed to account for the atomic spectra of atoms of more than one electron.
2. It did not explain Zeeman and Stark effects.

Question 9.
Derive an expression to measure emf of a battery.
(OR)
Give one application of Faraday’s laws of induction in daily life.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force’ to move unit pos¬itive charge from negative terminal to positive terminal of the battery.

1. 1. Let this chemical force be F_c.
   2. This chemical force does some work to move a negative charge ‘q’ from positive terminal to negative terminal against the electric force F_e Let this work be ‘W’.
   3. The work done by the chemical force to move 1 coulomb of charge from the terminal to negative terminal is given by \(\frac{w}{q}=\frac{F_c d}{q}\).
      Where ‘d’ is the distance between the terminals.
      If we assume F_c = F_e, then \(\frac{w}{q}=\frac{F_c d}{q}\)
      This \(\frac{w}{q}\) is known as emf (e) of the battery.
      ∴ emf(s) = \(\frac{w}{q}=\frac{F_c d}{q}\)
      This S.I. unit of emf is ‘volt’ and is measured using voltmeter.

(OR)
Now-a-days we can’t imagine the modern world with out electricity.
The generation of electricity utilizes the principle of electromagnetic induction. Transfer of generated power is possible with transformer the most important electrical device in our day to day life.
In bed lamps we are using step down transformer and to increase the voltage we are using step-up transformer.
Transformer used the principle of mutual induction i.e., when current flowing through a coil or magnetic flux linked with a coil changes an induced emf is produced in other coil.

Part – B (10 × 1 = 10 Marks)

Instructions:

1. Answer ALL the questions.
2. Each question carries 1 mark. .
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part-A.

Question 1.
The image formed by a convex mirror is always
A) real and magnified
B) real and diminished
C) virtual and diminished
D) virtual and magnified
Answer:
C) virtual and diminished

Question 2.
CaO reacts with water to form ………
A) CaCO₃
B) Ca(OH)₂
C) CaCl₂
D) Ca(NO₃)₂
Answer:
B) Ca(OH)₂

Question 3.
Identify the pair of pH values of strong acid and strong base in the following.
A) (6, 14)
B) (1, 8)
C) (7, 7)
D) (2, 14)
Answer:
D) (2, 14)

Question 4.
We are using tooth paste to clean our mouth and to avoid tooth decay. The nature the tooth paste is …………..
A) Acidic
B) Base
C) Neutral
D) Amphoteric
Answer:
C) Neutral

Question 5.
The acid forms in stomach is
A) HCl
B) H₂SO₄
C) CH₃COOH
D) HNO₃
Answer:
A) HCl

Question 6.
Which of the following lens act as converging lens ?
A) Biconvex lens
B) Plano convex lens
C) Concave convex
D) All the above
Answer:
D) All the above

Question 7.
What happens to the focal length of a convex lens when it is kept in water ?
A) decreased
B) we can’t tell
C) constant
D) increased
Answer:
D) increased

Question 8.
The sky looks blue on a clear, sunny day because of ……………….
A) dispersion of light
B) scattering of light
C) reflection of light
D) refraction of light
Answer:
B) scattering of light

Question 9.
Ionization energy is expressed in
A) KJ mol^-1
B) J/K mol^-1
C) JK mol^-1
D) K/J mol^-1
Answer:
A) KJ mol^-1

Question 10.
Rusting of iron takes place in
A) ordinary water
B) both ordinary and distilled water
C) distilled water
D) none of the above
Answer:
B) both ordinary and distilled water

Students must rely on TS 10th Class Biology Model Papers Set 4 to gauge their understanding of exam patterns.

TS 10th Class Biology Model Paper Set 4 with Solutions

Time:1:30 hours.
Max. Marks:40

Instructions:

1. Read the question paper carefully and understand.
2. Answer the questions under Part – A in the answer sheet provided.
3. Part- A contains three sections: sections – I, II and III.
4. There is an internal choice to the questions under section – III.
5. Part-B Answers should be written in the given brackets and attach to the Part – A answer sheet.
6. Write the answers following the instructions given in each section.

Part – A (Marks 30)
Section – I (3 x 2 = 6 M)

Instructions:

• This section contains 3 very short answer questions.
• Answer ALL the questions.
• Write answer to each question in 3 to 4 sentences.
• Each question carries 2 marks.

Question 1.
Name the enzymes present in the gastric juice. What are their functions?
Answer:
Pepsin and lipase are. the enzymes present in the gastric juice. Pepsin breaks down proteins into peptones. Lipase converts fats into fatty acids and glycerol.

Question 2.
What are respiratory substrates?
Answer:
Substances which are oxidised in the body during respiration to produce energy are called respiratory substrates.
E.g.: Glucose, fatty acids.

Question 3.
What is a heartbeat?
Answer:

1. The word heartbeat represents one contraction and one relaxation of heart.
2. The contraction phase is called systole and relaxation phase is called diastole.

Section – II (3 x 3 = 9 M)

Instructions:

• This section contains 3 short answer questions.
• Answer ALL the questions.
• Write answer to each question in 5 to 6 sentences.
• Each question carries 3 marks.

Question 4.
Write a brief note on the nervous system that regulates pupil of eye.
Answer:

1. When we enter a dark room we cannot see anything immediately. Slowly we are able to see the things around us in the room.
2. This is because of increase in diameter of pupil, which allows more light in.
3. When we come out of the room into broad day light the diameter of the pupil decreases allowing less light to enter into the eye.
4. Both these functions occur under the influence of the autonomous nervous system.

Question 5.
Who discovered DNA? Write a short note on it?
Answer:

1. Francis Crick and James Watson discovered DNA in 1953.
2. DNA molecule looks rather like spiral staircases having a shape known as double helix.
3. The framework of the staircase consists of alternate sugar and phosphate groups and the steps which join the framework together are the pairs of chemical compounds called bases.
4. They are adenine, guanine, thymine and cytosine.

Question 6.
What are the effects caused by the consumption of fossil fuels?
Answer:
Effects of overconsumption of fossils fuels.

1. Fossil fuels emit CO₂ and other volatile gases into the atmosphere.
2. Affects acid rains.
3. Increase atmospheric pollution.

Section-III (3 x 5 = 15 M)

Instructions:

• This section contains 3 essay-type questions.
• Answer ALL the questions.
• There is an internal choice for each question.
• Write answer to each question in 8 to 10 sentences.
• Each question carries 5 marks.

Question 7.
What is heterotrophic nutrition? Explain the feeding mechanism of heterotrophs.
(OR)
How scientists prove that the food is transported through the phloem?
Answer:
Organisms which cannot prepare their own food but depend on the other organisms to get their food. Such type of organisms are called heterotrophs. This type of nutrition is called heterotrophic nutrition.

Feeding mechanism:

1. The size of the food particles varies from microscopic organisms to large animals and plants.
2. Each organism is adapted to the particular environment.
3. The form of nutrition differs depending on the availability of food materials as well as how it is obtained by the organism. Some animals use pseudopodia. Ex: Amoeba. Some animals use tentacles. Ex: Hydra. Some animals cut the food into small pieces before they take it. Ex: Snails have sharp teeth-like structures on their tongue called radula which helps in making the food materials into pieces.
4. In herbivorous animals like deer, cow teeth help for grazing the grass and masticating the food.
5. Some animals like lions and tigers are equipped with organs to chase, capture, kill, cut and eat the prey.
6. Some animals feed on only liquid food. For example, scorpions and spiders kill the prey and inject digestive enzymes into the body of dead and convert it into liquid form, which the animal feed.
7. Honeybees have tube-like proboscis to suck nectar from flowers. Mosquitoes have special organs to pierce through skin and suck the blood. Animals like earthworms feed on soil containing decomposed organic material.

(OR)

1. Food such as sugar is synthesized in the green parts of plants, mainly in the leaves, but this food has to be transported to all living cells, especially to actively growing cells and the cells which stores food.
2. Phloem sieve tubes are extremely small and the analysis of their contents is not easy.
3. Biologists studied about food transportation in plants with the help of aphids.
4. When you see aphids clustering round the young stems of plants as they feed on the plant juices.
5. To obtain this juice an aphid pierces the plant tissues with its long needle-like organ “Proboscis”.
6. It can be shown when a feeding aphid is killed and the stem carefully sectioned.
7. The proboscis only penetrates up to a phloem sieve tube.
8. An aphid is killed while in the act of feeding and the body is then carefully cut away, leaving the hollow proboscis still inserted into the phloem.
9. It is found that because the contents of the phloem sieve tubes are under slight pressure.
10. The fluid slowly excludes from the cut end of the proboscis in the form of drops.
11. These drops are then collected and analyzed.
12. The fluid is found to contain sugars and amino acids.

Question 8.
Describe male reproductive system of human.
(OR)
Write a short note on structure of the ovule.
Answer:

1. In human males, the two testes are located in pocket-like structure outside the body wall called ‘scrotum.
2. Each testes has several lobules and Seminal vesicle each lobule contain several seminiferous Prostate gland tubules.
3. They are small, highly coiled tubes and 80 cm in length.
4. Vasefferentia, collect from epididymis. Here Urethra sperms are stored temporarily and moved – Epididymis into vas deference then to urethra of penis and expel out of the body.
5. One prostate, two Cowper glands which Scrotum are accessory glands in male reproductive system secretes a fluid called ‘semen’.
6. They provide nutriènts for sperm to keep alive and helps as a medium for the movement of sperms.
7. The sperm is a flagellated structure with long tail. This helps them to move towards the ovum.

8. The development of the male reproductive organs is regulated by the male sex hormone called ‘testosterone’.
9. Men produce sperm, from the age of about 13 or 14 years and can go on doing so most of their lives, although their power to do so decreases as they grow older.

(OR)

1. An ovule is an egg-shaped structure attached by a stalk to the inner side of the ovary.
2. An ovary may have one, two, several or even hundreds of ovules.
3. At the centre of each ovule is a microscopic embryo sac filled with food and water.
4. The embryo sac is composed of gametophyte cells.
5. The majority of flowering plants have an embryo sac consisting of seven cells and eight nuclei.
6. The structure which receives pollen is called stigma and the long tube-like structure which helps in the passage of compatible male sex cells is called ‘style.

Question 9.
What is peristaltic movement? Explain the food movement in alimentary canal comparing with the experiment of moving potatoes in cycle tube.
(OR)
What determines the terrestrial ecosystems on the Earth?
Answer:
Peristaltic movement is the contraction and relaxation of the muscles of the digestive system.
The movement of food through food pipe is known as peristaltic movement.

Food movement in alimentary canal:

1. The walls of the food pipe secrete a slippery substance called mucus. Mucus lubricates and protects the oesophageal walls from damage.
2. This helps the food bolus to slide down easily just as the oiled potatoes that move in the tube. Oil acted as lubricant to push the potatoes easily in the forward direction.
3. The wall of the oesophagus is made up of two kinds of smooth muscles. The inner layer consists of circular muscles and the outer layer of longitudinal muscles.
4. Contraction of the circular muscles results in narrowing of the oesophagus just behind the bolus.
5. So the food is squeezed downwards.
6. Contraction of the longitudinal muscles in front of the bolus widens the tube, this results in shortening of that particular part of the oesophagus.
7. Contraction and relaxation of these muscles bring in wave-like motion that propels the food bolus into the stomach by the action called as peristalsis.
8. This is involuntary and under the control of autonomous nervous system.

(OR)

1. The terrestrial ecosystems on the Earth are being determined largely by the variations in climatic conditions between the poles and equator.
2. The main climatic influences which determine these ecosystems are rainfall, temperature and availability of light from the Sun.
3. For instance, forests are usually associated with high rainfall, but the type is influenced by temperature and light.
4. The same applies to deserts which occur in regions where rainfall is extremely low.
5. Thus, the climatic conditions along the horizontal climatic regions determine the terrestrial ecosystems on the Earth.
6. If we move from the equatorial region to the polar region, we can come across tropical rain forests, savannah, deciduous forests, coniferous forests and then tundras respectively.
7. Similarly, the altitude of the place is also a determining factor.
8. Let’s climb a mountain such as Kilimanjaro in equatorial Africa. We can go through a comparable system of ecosystems, starting with tropical rainforest at the base and ending with perpetual snow and ice.

Part-B (Marks 10)

Instructions:

• Write the answers to the questions under Part – B on the question paper itself and attach It to the answer book of Part – A.
• Each question carries 1 mark. ,
• Marks will not be awarded in any case of overwriting, rewritten or erased answers.
• Write the capital letter (A / B / C / D) showing the correct answer for the following questions in the brackets provided against them.

Question 1.
In which part of the alimentary canal digested food is absorbed? ( ).
A) Stomach
B) Mouth
C) Large intestine
D) Small intestine
Answer:
B) Mouth

Question 2.
Oil is applied on the lower side of the leaf. The result is ( )
A) Photosynthesis does not take place.
B) Respiration does not take place.
C) Transpiration does not take place.
D) All the above.
Answer:
D) All the above.

Question 3.
From nasal cavity, the air goes into pharynx. After pharynx, the track is divided into passages. Those are ……………, …………….. ( )
1) Stomach, duodenum
2) Trachea, digestive canal
3) Larynx, epiglottis
A) 1 only
B) 1 and 2
C) 2 only
D) 1 and 3
Answer:
B) 1 and 2

Question 4.
B.P. means ( )
A) Atria Pressure
B) Lymph Pressure
C) Ventricular Pressure
D) Blood Pressure
Answer:
D) Blood Pressure

Question 5.
This organ also involved in urea formation. ( )
A) Lungs
B) liver
C) Stomach
D) Colon
Answer:
B) liver

Question 6.
Taking of ……………… rich diet results in more urea in urine. ( )
A) Carbohydrates
B) Proteins
C) Fats
D) Vitamins
Answer:
B) Proteins

Question 7.
Animal that excretes by diffusion ( )
A) Earthworm
B) Scorpion
C) Amoeba
D) Leech
Answer:
C) Amoeba

Question 8.
The correct answer is ( )
1) Abscisic acid (a) a) Ripening of fruit
2) Cytokinins (b) b) Closing of stomata
3) Gibberellins (e) c) Elongation of stem
A) 1 is correct
B) 3 is correct
C) 2 is correct
D) 1, 2, 3 are correct
Answer:
B) 3 is correct

Question 9.
ASHA stands for ( )
A) Accredited Social Health Association
B) Accredited Social Health Activist
C) AIDS Social Health Activist
D) AIDS Sexual Health Activist
Answer:
B) Accredited Social Health Activist

Question 10.
img. What will be in the box? ( )
A) Ovary
B) Graffian follicle
C) Cavity of ovule
D) Canal
Answer:
B) Graffian follicle

Timed practice with TS 10th Class Physical Science Model Papers Set 3 is crucial for improving speed and efficiency during exams.

TS 10th Class Physical Science Model Paper Set 3 with Solutions

Time: 1 Hour 30 minutes
Maximum Marks: 40

General Instructions:

1. Read the question paper and understand every question thoroughly and write answers in given 1.30 hrs. time.
2. 3 very short answer questions are there in section – I. Each question carries 2 marks. Answer all the questions. Write answer to each question in 3 to 4 sentences.
3. 3 short answer questions are there in section – II. Each question carries 3 marks. Answer all the questions. Write answer to each question in 5 to 6 sentences.
4. 3 essay type answer questions are there in section – III. Each question carries 5 marks. Answer all the questions. Write answer to each question in 8 to 10 sentences. Internal choice is given in this section.

Part – A (30 Marks)
Section – I (3 × 2 = 6 Marks)

Instructions :

1. 3 Very short answer questions are there in section -1.
2. Answer ALL the questions. Each question carries 2 marks.
3. Write answer to each question in 3 to 4 sentences.

Question 1.
Why concave and convex mirrors are called spherical mirrors ?
Answer:
Concave and convex mirrors are called spherical mirrors because :

1. The reflecting surfaces of concave and convex mirrors are considered to form a part of a hollow sphere.
2. Their reflecting surface is not a plane surface.

Question 2.
How do you appreciate the working of “Retina”?
Answer:

1. It is the innermost delicate membrane having a large number of receptors called ‘rods’ and ‘cones’.
2. The rods identify the colour and the cones identify the intensity of light.
3. The retina is a part on which the image of an object is formed

Question 3.
Give an example of auto reduction of sulphide ores.
Answer:
In the extraction of Cu, from its sulphide ore, the ore is subjected to partial roasting in air to give its oxide.
2Cu₂S (S) + 3O₂ (g) → 2Cu₂O (S) + 2SO₂ (g)
When the supply of air is stopped and the temperature is raised, the rest of the sulphide reacts with oxide and forms the metal and SO₂.
2Cu₂O (S) + Cu₂S (S) 6Cu (S) + SO₂(g)

Section – II (3 × 3 = 9 Marks)

Instructions :

1. 3 Short answer questions are there in section – II.
2. Answer ALL the questions. Each question carries 3 marks.
3. Write answer to each question in 5 to 6 sentences.

Question 4.
Negative ion of an element has bigger size than its neutral atom. Explain with an example.
Answer:

1. When we take chlorine (Cl) atom, its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁵, which has 17 electrons and 17 protons.
2. The electronic configuration of chloride ion is 1s² 2s² 2p⁶ 3s² 3p⁶. Which has 18 electrons and 17 protons.
3. Therefore, the nuclear attraction is less in Cl^– ion when compared with chlorine atom.
4. So, the size of chlorine atom is less than Cl^– ion.

Question 5.
If the electronic configurations of atoms A and B are 1s², 2s², 2p⁶, 3s², 3p¹ and 1s², 2s², 2p⁴ respectively, then
a) Which atom forms negative ion ?
b) Which atom forms positive ion ?
c) What is the valency of atom A ?
d) What is the molecular formula of the compound formed by atoms A and B ?
Answer:
Given electronic configuration of atom A is 1s² 2s² 2p⁶ 3s² 3p¹ i.e., Alu-minium and B is 1s² 2s² 2p⁴i.e., Carbon.
a) The atom ‘B’ tends to form negative ion by gaining two electrons order to get nearest inert gas neon configuration is 1s² 2s² 2p⁶.
b) The atom ‘A’ tends to form positive ion by losing three electrons in order to get nearest inert gas. Neon configuration is 1s² 2s² 2p⁶.
c) Valency of atom‘A’ is ‘3’.
d) According to Criss-Cross method, the molecular formula of the corn-pound formed by atoms both A and B is A₂B₃ i.e., Al₂O₃.

Question 6.
Draw the electron dot structures of Ethanoic acid and Ethyne (Acetylene).
Answer:

Section – III (3 × 5 = 15 Marks)

Instructions :

1. 3 Essay answer questions are there in section.
2. Answer ALL the questions. Each question carries 5 marks.
3. Internal choice is given in this section.
4. Write answer to each question in 8 to 10 sentences.

Question 7.
Explain the steps involved in balancing a chemical equation with an example.
(OR)
Write the chemical equation of preparation of baking soda. What are the uses of baking soda ?
Answer:
A chemical equation in which the number of atoms of different elements on the reactants side are same as those on product side is called a balanced equation.
Steps involved in balancing a chemical reaction: Let us consider the combustion reaction of Propane.

Step 1 : Write the unbalanced equation using correct chemical formula for all substances.
C₃H₈ + O₂ → CO₂ + H₂O (Skeleton equation) .

Step 2 : Compare number of atoms of each element on both sides.

Atom	No. of atoms in LH.S.	No. of atoms in RH.S.
C	3((in C3H3)	1 (in CO2)
H	8 (in C3H3)	2 (in H2O)
O	2 (in O2)	3 (in CO2, H2O)

Find the co-efficients to balance the equation. In this case, there are 3 carbon atoms on the left side of the equation but only one on the right side. If we add a co-efficient of 3 to CO₂ on the right side the carbon atoms balance.
C₃H₈ + O₂ → 3 CO₂ + H₂O
Now, look at the number of hydrogen atoms. There are 8 hydrogen atoms On the left but only 2 on the right side. By adding a co-efficient of 4 to the H₂O on the right side, the hydrogen atoms get balanced.
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
Finally, look at the number of oxygen atoms. There are 2 on the left side but 10 on the right side. By adding a co-efficient of 5 to the 02 on the left side, the oxygen atoms get balanced.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Step 3 : Make sure the co-efficients are reduced to their smallest whole number values. The above equation is already with the co-efficients in smallest whole numbers. There is no need to reduce its co-efficients. Hence the final equation is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Step 4 : Check the answer. Count the numbers and kinds of atoms on both sides of the equation. to make sure they are the same.
(OR)
Preparation of baking soda:

1. The chemical name of baking soda is sodium hydrogen carbonate and the formula is NaHCO,
2. It is prepared as follows.
   NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃

Uses of baking soda:

1. Baking soda is added for faster cooking.
2. It is used as an ingredient in antacids.
3. It is also, used in soda – acid fire extinguishers.
4. It acts as a mild antiseptic.
5. Baking powder can be prepared as a mixture of baking soda and a mild edible acid such as tartaric acid.
6. Baking powder is used in preparation of bread or cake to make them soft and spongy.

Question 8.
Find the relation among refractive indices of media, object distance, imagfe distance and radius of curvature.
(OR)
Draw Moeller chart of filling order of atomic orbitals. (Or) Draw a diagram showing the increasing value of (n + 1) of orbitals.
Answer:
1) Consider a curved surface separating two media of refractive indices n₁ and n₂ respectively.

2) O = Point object on the principal axis.

3) The ray, which forms an angle a with principal axis, meets the curved surface at A.
θ₁ = The angle of incidence.

4) The ray bends and passes through the second medium along the line
AI = Refraction ray in 2^nd medium.

5) θ₂ = The angle of refraction.

6) I = The image is formed there.

7) γ = The angle made by the refracted ray AI with principal axis,
b = The angle between the normal and principal axis.

8) In the above figure
PO = the object distance = ‘u’
PI = image distance = ‘v’
PC = radius of curvature = ‘R’
n₁, n₂ are refractive indices of two media.
In a triangle ACO, θ₁ = α + β
And in a triangle ACI, β = θ₂ + γ
……. β – γ = θ₂
According to Snell’s law,
we know n₁ sin θ₁ = n₁ sin θ₂
Substituting the values of θ₁ and θ₂,
we get, n₁ sin (a + b) = n₂ sin (β – γ ) ………… (1)
If the angles α, β and γ very small,
sin (α + β) = α + β and sin (β – γ) = β – γ
… n₁(α + β) = n₂(β – γ)
… n₁α + n₁β = n₂β – n₂γ ………….. (2)
Since all angles are small,
tan α = α = \(\frac{\mathrm{AN}}{\mathrm{NO}}\); tan β = β = \(\frac{\mathrm{AN}}{\mathrm{NC}}\) ; tan γ = γ = \(\frac{\mathrm{AN}}{\mathrm{NI}}\)
Substitute these values in equation (2),
⇒ n₁\(\frac{\mathrm{AN}}{\mathrm{NO}}\) + n₁\(\frac{\mathrm{AN}}{\mathrm{NC}}\) = n₂\(\frac{\mathrm{AN}}{\mathrm{NC}}\) – n₁\(\frac{\mathrm{AN}}{\mathrm{NI}}\) ……….. (3)
As the rays move very close to principal axis, the point N coincides with pole of the interface (P).
∴ NI = PI, NO = PO, NC = PC
Substituting those values in eqation (3),
⇒ \(\frac{\mathrm{n}_1}{\mathrm{PO}}+\frac{\mathrm{n}_1}{\mathrm{PC}}=\frac{\mathrm{n}_2}{\mathrm{PC}}-\frac{\mathrm{n}_2}{\mathrm{PI}}\)
⇒ \(\frac{\mathrm{n}_1}{\mathrm{PO}}+\frac{\mathrm{n}_2}{\mathrm{PI}}=\frac{\left(\mathrm{n}_2-\mathrm{n}_1\right)}{\mathrm{PC}}\) ……….. (4)
This equation shows the relation among refractive indices of media, object distance, image distance and radius of curvature.
If we take sign convention
PO = -u ; PI = v; PC = R
Substituting these values in (4),
⇒ \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\left(\mathrm{n}_2-\mathrm{n}_1\right)}{\mathrm{R}}\) …………….. (5)
It can also be used for plane surfaces.
For plane surfaces :
f = R = infinity ⇒ \(\frac{1}{\mathrm{R}}\) = 0
Then (5) ⇒ \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}\) = 0 ⇒ \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}\)

(OR)

Question 9.
Sudhakar has taken a substance in the form of wire. He applied different voltages to the wire and measured electrical currents. For this he used Ammeter and Voltmeter. He tabulated five
measurements. Then plotted a graph as shown in the figure.

In the graph he measured voltages in volts (V) and current (I) in Amperes.
Basing on graph. Answer the following questions :
a) What type of material Sudhakar had selected for his experiment ?
b) What is the resistance of the substance ?
c) If potential difference is 20V at the ends of wire, how much electrical power is utilized by wire ?
d) What is the law associated with the above graph ?
(OR)
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid.
Answer:
a) The graph is a straight line passing through origin.
It is in the form of y = mx,
i.e., I = mV
Here, ‘m’ is slope of the graph.
Here, m = \(\frac{1}{\mathrm{R}}\) = \(\frac{1}{\text { Resistance }}\)
∴ I = (\(\frac{1}{\mathrm{R}}\))V
The substance is Ohm’s substance i.e., obeying Ohm’s law. So, it is a metal like iron spoke
(or) Copper, Aluminium etc.

b) The resistance can be known from graph is
V = IR ; R = \(\frac{V}{I}=\frac{10}{0 \times 2}=\frac{100}{2}\) = 50Ω
The reciprocal of slope of graph gives resistance.

c) The electrical power can be measured by taking the area of graph i.e., area enclosed between the straight line and X-axis.
Power (P) = Voltage × Current = VI
Area = Area of triangle = \(\frac{1}{2}\) × 4 = 4 Watts

d) 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.
(OR)
Procedure:

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

Part – B (10 × 1 = 10 Marks)

Instructions :

1. Answer ALL the questions.
2. Each question carries 1 mark.
3. In this section there are 4 options (A / B / C / D) to each question. Choose the appropriate answer and write the answer in the brackets given against the question. Part – B must be attached to the answer booklet of Part – A.

Question 1.
The mirror which has a wide field of view must be
A) concave
B) convex
C) plane
D) npne of these
Answer:
B) convex

Question 2.
A chemical equation should be balanced because the law …………… should be verified.
A) constant proportions
B) conservation of mass
C) law of equality
D) law of balance
Answer:
B) conservation of mass

Question 3.
The aqueous solution of …………….. conducts electricity.
A) Ethyl alcohol
B) Acetic acid
C) Acetone
D) Ether
Answer:
B) Acetic acid

Question 4.
The reaction of an acid with a base to give a salt and water is known as
A) Neutralization
B) Reduction
C) Oxidation
D) Crystallisation
Answer:
A) Neutralization

Question 5.
Which of the following aqueous solution is called brine ?
A) sodium chloride
B) potassium chloride
C) copper chloride
D) calcium chloride
Answer:
A) sodium chloride

Question 6.
A person is standing on the bank of a river. A fish inside water will see the person to be
A) taller
B) shorter
C) original height
D) depends on fish
Answer:
A) taller

Question 7.
The minimum distance between an object and its real image formed by a convex lens is
A) \(\frac{2}{3}\)f
B) 2f
C) \(\frac{5}{2}\)f
D) 4f
Answer:
A) \(\frac{2}{3}\)f

Question 8.
The persistence if vision ofr normal eye is
A) (\(\frac{1}{18}\))^th of a second
B) (\(\frac{1}{6}\))^th of a second
C) (\(\frac{1}{10}\))^th of a second
D) (\(\frac{1}{16}\))^th of a second
Answer:
D) (\(\frac{1}{16}\))^th of a second

Question 9.
Which of the following element belongs to 3 period and also a 17 (VILA) group?
A) Fluorine
B) Bromine
C) Oxygen
D) Chlorine
Answer:
D) Chlorine

Question 10.
Which of the following is the outlet through which fuel gases go out the furnace ?
A) Chimney
B) Flux
C) Hearth
D) Fire box
Answer:
A) Chimney

Students must rely on TS 10th Class Biology Model Papers Set 3 to gauge their understanding of exam patterns.

TS 10th Class Biology Model Paper Set 3 with Solutions

Time:1:30 hours
Max.Marks:40

Instructions:

1. Read the question paper carefully and understand.
2. Answer the questions under Part – A in the answer sheet provided.
3. Part – A contains three sections: section – I, II and III.
4. There is an internal choice to the questions under section – III.
5. Part B Answers should be written in the given brackets and attach to the Part – A answer sheet.
6. Write the answers following the instructions given in each section.

Part – A (Marks 30)
Section-I (3 x 2 = 6 M)

Instructions:

• This section contains 3 very short answer questions.
• Answer ALL the questions.
• Write answer to each question in 3 to 4 sentences.
• Each question carries 2 marks.

Question 1.
Why chloroplast are green in colour?
Answer:

1. Chloroplast are green in colour due to the presence of chlorophyll, a green-coloured pigment.
2. Chlorophyll traps the solar energy required for the photosynthesis.

Question 2.
How does exhalation takes place?
Answer:
The ribs come back to their original position the diaphragm relaxes and assumes its dome shape. The pressure increases on the lungs. The elastic tissue contracts and squeezes the air out through the nose to the external atmosphere.

Question 3.
The wall of left ventricle is thicker than the wall of the right ventricle? Give reason.
Answer:
As it pumps blood to more distant parts of the body (such as fingers and toes) to sustain high pressure of oxygenated blood the left ventricle is thicker than the right ventricle.

Section – II (3 x 3 = 9 M)

Instructions:

• This section contains 3 short answer questions.
• Answer ALL the questions.
• Write answer to each question In 5 to 6 sentences.
• Each question carries 3 marks.

Question 4.
Sunil was unable to see outside when he came out from the theatre. He doesn’t know how he come out with other audiences. What type of questions he would have got in his mind to know the reason?
Answer:

1. Why am I unable to see when I came out of the theatre?
2. How was I able see the things in the theatre though it was dark?
3. Is there any adjustment happening in the eyes? What is it?
4. After some time, how was I am able to see normally? What would be the reason?

Question 5.
Why are the small number of surviving tigers and lions a cause of worry from the points of view of genetics?
Answer:

1. Sometimes, species (either animal or plant) may die out. It may become extinct.
2. Once species is extinct its genes are lost forever. It can’t reemerge at all.
3. The small number of tigers and lions are a cause of worry from the point of view of genetics because if they all die out and become extinct, their genes will be lost forever. Overcoming generations will not be able to see tigers at all.

Question 6.
Why is dependence of man on nature greater than that of any other organism?
Answer:
Man’s dependence on the environment is greater than that of other organisms, because

1. Man develops curiosity for more comforts and security.
2. Man consumes large amounts of material and energy.
3. Man develops a new kind of social-economic environment which consists of things developed by man through tools and techniques.

Section-III (3 x 5 = 15 M)

Instructions:

• This section contains 3 essay-type questions.
• Answer ALL the questions.
• There is an internal choice for each question.
• Write answer to each question in 6 to 10 sentences.
• Each question carries 5 marks.

Question 7.
Which issues do you take into consideration to tell that plants play a key role in animals’ nutrition?
(OR)
When do you know about the human heart’s pumping method, the “circulatory system” which issue you will remember particularly? What’s the reason for that?
Answer:

1. Plants play a very important role in the nutrition of animals. Actually, plants are the producers where as animals are dependent on plants for their nutrition.
2. Many plants (or) plants’ parts are eatable as food. There are around 2000 plant species which are cultivated for food. Nutrients like carbohydrates, proteins, fats, vitamins and minerals are available from these plants.
3. Seeds of plants arc good sources of food for animals including humans because they contain many healthful fats.
4. Infact majority of the food consumed by human beings are seed based food. Edible seeds include cereals like wheat, rice, maize etc., and legumes like pea, groundnut and nuts.
5. Oil seeds are often pressed to produce rich oils like sunflower, groundnut, sesame et&
6. Seeds are typically high in unsaturated fats and considered as a health food.

(OR)
Particularly I remember about the importance of closed type of circulatory system and the double circulation in man.
In most of the animals, tubes called blood vessels are present and the heart pumps blood into the blood vessels. This type of circulatory system is called closed type of circulatory system. The heart that pumps blood to lungs for oxygenation is called pulmonary heart in man blood passes twice through the heart once between heart and lungs and second time between heart and body parts. Such a circulation is called double circulation and the heart is called double circuit heart.

Oxygenated blood enters from lungs into the left atrium and flows into left ventricle. The left ventricle pumps blood to all other body parts. The deoxygenated blood from the body parts is collected into the right atrium and flows into right ventricle. The right ventricle sends it to lungs for oxygenation.

Question 8.
Write a short note on the propagation of Rhizopus.
(OR)
Write a short note on fission. Give examples.
Answer:

1. Rhizopus propagates by means of spores.
2. It consists of fine thread projects called hyphae and thin knob-like structure called sporangia.
3. Each sporangium produces hundreds of microscopic reproductive units called spores.
4. When the spore case bursts, the spores spread into air.
5. These airborne spores land on food or soil.
6. Under favourable conditions like damp and warm conditions, they germinate and produce new individuals.

(OR)

1. Fission is the asexual method of reproduction.
2. Single-celled organisms such as paramecium and bacteria reproduce by splitting into two or more off springs.
3. If the organisms split into two, it is called binary fission.
4. When the organism splits into more cells to form more off springs it is called multiple fission.
5. This is often the only mode of reproduction in these organisms.

Question 9.
Read the following passage:
As the process of digestion in the stomach nears completion, the contractions of the stomach decrease. This prompts the muscles called as pyloric sphincter at the opening of the stomach and the first part of the small intestine or duodenum to relax. This opens the pathway into the duodenum releasing the partially digested food (chyme) in small quantities into the dude num.

Peristalsis involves the contraction of the muscles behind the food and the relaxation of the muscles in front of the food giving rise to a thrust that pushes the food forward through the digestive canal. A wave of contraction followed by relaxation in muscles help in forward movement of food.
Write answers for the following questions.
i) What is the use of the peristalsis movement?
ii) How are the peristalsis movements caused by muscles in the oesophagus?
iii) What is the role of pyloric sphincter?
iv) What is the use of duodenum?
(OR)
Write a short note on food chain and food web.
Answer:
i) The wave-like movements of the muscles in the digestive tract are called peristalsis movements. They help for the movement of food in the alimentary canal. They may occur in a regular (or) reverse manner. The reverse peristalsis, in man, is a protective mechanism to expel unwanted substances.

ii) Peristalsis is a wave-like movement caused by the contraction and relaxation of the muscles of the oesophagus. The smooth muscles of oesophagus have circular muscles in the inner layer and longitudinal muscles in the outer layer. The contraction of circular muscles behind the food and contraction of longitudinal muscles in front of the food creates a wave-like motion and food is squeezed in the tube.

iii) Tle relaxation of pyloric sphincter opens the pathway from stomach to duodenum releasing small amounts of partially digested chyme into the duodenum from time to time. Its contraction closes the pathway.

iv) Duodenum is a small tube-like connection between stomach and small intestine. The partially digested chyme enters into the duodenum from the stomach. liver and pancreas opens into the duodenum through their ducts. Bile juice from liver and pancreatic juice from pancreas enter into it and helps for the digestion of proteins and fats.

(OR)

1. A food chain is a pathway along which food is transferred from one trophic level to another trophic level beginning with producers.
2. It shows who eats what in a particular habitat.
3. The arrows, between each item in the chain always point from the food to the feeder.
4. For example Grass → Rabbit → Snake → Hawk.
5. The elaborate interconnected feeding relationships in an ecosystem is said to be food web.
6. Many of the food chains in an ecosystem are cross-linked to form food web.
7. The food chain and food web help us to understand the food relations among living things.

Part – B (Marks 10)

Instructions:

• Write the answers to the questions under Part – B on the question paper itself and attach it to the answer book of Part – A.
• Each question carries 1 mark.
• Marks will not be awarded in any case of overwriting, rewritten or erased answers.
• Write the capital letter (A / B / C / D) showing the correct answer for the following questions in the brackets provided against them.

Question 1.
Observe the picture. If iodine is used on the leaf ………………… .? ( )

A) formation of deep blue colour at the non-light falling region.
B) formation of deep blue colour at the light falling region.
C) formation of deep blue colour on both light-falling and non-light-falling regions.
D) no colour formation on both regions.
Answer:
B) formation of deep blue colour at the light falling region.

Question 2.
What is the purpose of the experiment? ( )

A) To test the presence of oxygen in respiration
B) To test the evolution of CO₂ in respiration
C) To test the evolution of heat in respiration
D) To test anabolic activity
Answer:
B) To test the evolution of CO₂ in respiration

Question 3.
Maximum rate of respiration takes place at ( )
A) 0°C
B) 45°C
C) 100°C
D) 60°C
Answer:
D) 60°C

Question 4.
The correct order! sequence of different phases of human cardiac cycle. ( )
1) Ventricle systole
2) Auricle systole
3) Ventricle diastole
4) Auricle diastole
A) 1, 2, 3, 4
B) 2, 1, 4, 3
C) 3, 1, 2,4
D) 4, 3, 2, 1
Answer:
A) 1, 2, 3, 4

Question 5.
Sequence of urine formation in nephron is ( )
A) Glomerular filtration → tubular re- absorption → tubular secretion
B) Tubular reabsorption → tubular secretion → glomerular filtration
C) Tubular secretion → glomerular filtration → tubular reabsorption
D) Tubular reabsorption → concentration of urine → tubular secretion
Answer:
A) Glomerular filtration → tubular re- absorption → tubular secretion

Question 6.
Excretory organs are first appear in ( )
A) Nematodes
B) Platyhelmenthis
C) Aves
D) Reptiles
Answer:
B) Platyhelmenthis

Question 7.
Match the following. ( )

List – A	List – B
1. Flame cells	a) Nematada
2. “Water vascular system”	b) Arthropoda
3. Green glands	c) Starfish

A) 1 – c, 2 – b, 3 – a
B) 1 – b, 2 – c, 3 – a
C) 1 – a, 2 – b, 3 – c
D) 1 – a, 2 – c, 3 – b
Answer:
D) 1 – a, 2 – c, 3 – b

Question 8.
Electrical impulses travel in a neuron from ( )
A) Dendrite → axon → axon end → cell body
B) Cell body → axon → dendrite → axon end
C) Dendrite → cell body → axon → axon terminal
D) Axon terminal → axon → cell body → dendrite
Answer:
D) Axon terminal → axon → cell body → dendrite

Question 9.
Find out the correct sentence. ( )
A) Double fertilization occurs in Angiosperms
B) Double fertilization occurs in Gymnosperms
C) Asexual reproduction occurs in all flowering plants
D) Single fertilization occurs in flowering plants.
Answer:
A) Double fertilization occurs in Angiosperms

Question 10.
The scientist who expelled his research views by thinking, though he has poor eyesight is ( )
A) Theodor Boyen
B) Wilhelm Roux
C) Waither Flemming
D) August Weismann
Answer:
D) August Weismann

Students must rely on TS 10th Class Biology Model Papers Set 2 to gauge their understanding of exam patterns.

TS 10th Class Biology Model Paper Set 2 with Solutions

Time: 1:30 hours.
Max. Marks: 40

Instructions:

1. Read the question paper carefully and understand.
2. Answer the questions under Part – A in the answer sheet provided.
3. Part – A contains three sections: section – I, II, and III.
4. There is an internal choice to the questions under section – III.
5. Part-B Answers should be written in the given brackets and attach to the Part – A answer sheet.
6. Write the answers following the instructions given in each section.

Part – A (Marks 30)
Section-I (3 x 2 = 6 M)

Instructions:

• This section contains 3 very short answer questions.
• Answer ALL the questions.
• Write answer to each question in 3 to 4 sentences.
• Each question carries 2 marks.

Question 1.
What is RUBP?
Answer:
RUBP is the intermediary substance fortìed in the dark reaction. Ribulose 1,5 Bisphosphate is the full form of RUBP.

Question 2.
How does respiration takes place in plants where roots are present in wet places?
Answer:
The plants which have their roots in very wet places have much larger air spaces, connect the stems with the roots, making diffusion from upper parts.

Question 3.
What are the factors that transpiration depends on?
Answer:
Several factors such as temperature, humidity of wind, velocity, soil water content etc., determine the opening and closing of stomata and the rate of transpiration.

Section – II (3 x 3 = 9 M)

Instructions:

• This section contains 3 short answer questions.
• Answer ALL the questions.
• Write answer to each question in 5 to 6 sentences.
• Each question carries 3 marks.

Question 4.
Write a note on ‘feedback mechanism’.
Answer:

1. The timing and the amount of hormones released.by various glands are controlled by the feedback mechanism which is in built in our body.
2. The excess deficiency of hormones have harmful effect on our body. For example, the deficiency of insulin results in a disease called diabetes, whereas excess of insulin in the body can lead to coma.
3. For example, if the sugar level in the blood rises too much, they are detected by the cells of pancreas which respond by producing and secreting more insulin into blood. And as the blood sugar falls to a certain level, the secretion of insulin is reduced automatically.

Question 5.
What are the differences between monohybrid and dihybrid cross?
Answer:

Monohybrid cross	Dihybrid cross
1) In monohybrid cross only one pair of contrasting characters are taken into consideration.	1) In dihybrid cross two pairs of contrasting characters are taken into considerations.
2) Phenotypic ratio of F2 generation individuals is 3: 1.	2) Phenotypic ratio of F2 generation individuals is 9 : 3 : 3 :1.
3) In this cross the genotypic ratio of F2, generation individuals is 1: 2:1.	3) In this cross the genotypic ratio of F2 generation individuals is 1: 2 : 2: 4: 1: 2: 1: 2: 1.

Question 6.
Why is scarcity of water there in our country in spite of good monsoon?
Answer:
It has taken place due to the following reasons.

1. Failure to sustain water availability underground which has resulted largely from the loss of vegetation cover.
2. Diversion for high water-demanding crops.
3. Pollution from industrial effluents and urban wastes.

Section – III (3 x 5 = 15 M)

Instructions:

• This section contains 3 essay-type questions.
• Answer ALL the questions.
• There is an internal choice for each question.
• Write answer to each question in 8 to 10 sentences.
• Each question carries 5 marks.

Question 7.
Describe what disaster occurs on Earth, if photosynthesis life process stops.
(OR)
What is Lymph? What is the importance of lymphatic system in man?
Answer:

1. If photosynthesis by plants stops, food will not be available to all other living organisms. Because all these organisms cannot prepare their own food material. Hence heterotrophs dies due to lack of food.
2. Carbon dioxide is required for the process of photosynthesis by plants.
3. If it stops the concentration of CO₂ increases iñ the air that leads to increase in the temperatures of the Earth.
4. If the temperature increases it causes global warming which causes melting of ice.
5. This results in increase in sea levels and water inundates the low-lying areas near the sea coast. So all the low-lying areas drown leading to the death of many organisms.
6. Photosynthesis releases oxygen into atmosphere. If photosynthesis does not occur the oxygen may not be available for the living organisms. This leads to death of living organisms.

(OR)

1. In Latin Lymphs means water.
2. Lymph is the substance that contains blood without solid particles. Tissue fluid is the substance which contains lymph present in the tissues.
3. Lymph is the vital link between blood and tissues by which essential substances pass from blood to cells and excretory products from cells to blood.
4. The lymphatic system is a parallel system to various system which collects tissue fluid from tissues and transports it to the venous system.
5. The muscles which are attached to the skeleton (skeletal muscles) act as pumps when they contract and help in pushing the lymph flowing in lymphatic vessels and the blood flowing in veins towards the heart.
6. The valves that are present in the lymphatic vessels and veins stop the reverse flow of blood.

Question 8.
Explain the formation of fruits and seeds in plants.
(OR)
What are the advantages of artificial vegetative propagation?
Answer:

1. After fertilization, the fertilized egg present in the ovule divides several times to form embryo.
2. Mean while, the wall of the ovules becomes hard and they develop into seeds.
3. Along these developments, the wall of the ovary also starts to swell.
4. As a result of which ovary develops into fruit with seeds inside it.
5. The other parts of flowers like petals, sepals stamens, stigma, and style dry up and fall off.
6. The seed is the reproductive unit of the plant.

(OR)

1. The new plant produced by artificial propagation will be exactly like the parent plants. Any desirable features of the parent plant will be replicated in the new plant.
2. The fruit trees grow from cutting and grafting start bear fruits much earlier only after a few growing seasons.
3. Many plants can be grown from just one parent plant by artificial propagation.
4. We can also get seedless plants by vegetative propagation.
5. The plants grown in this method seed less attention in their early years than the plants grown from seeds.
6. Rare hybrid plants (ornamental plants) with attractive flowers and leaves can be produced in large numbers.
7. It is very useful method of reproduction in plants that rarely produce flowers.

Question 9.
What happens if there is no epiglottis in human beings?
(OR)
How do you understand the interdependency of animals and plants? How do you appreciate that?
Answer:

1. From the pharynx, there are two passageways, one to the lungs and other to stomach by oesophagus.
2. Air goes into the windpipe and food goes into food pipe.
3. A flap-like valve, the epiglottis protects the windpipe from asserting entry of food.
4. The epiglottis is the valve partly closed when we swallow food and it opens more widely when we take breath and air enters the lungs.
5. If epiglottis is absent then both air and food passes through pharynx at a time and cause chocking.

(OR)

1. In nature, all plants and animals are always interdependent on each other.
2. Plants synthesize food for all the animals. Hence they are the “Universal food providers”.
3. Herbivores depends on plants for obtaining food. They become food for secondary and tertiary consumers. Hence, carnivores are indirectly dependent on plants.
4. Plants provide food, shelter, and protection to animals. In turn, they get. nutrients and pollination from the organisms.
5. Their interdependence brings the sustainable balance in the ecosystem. This relationship is really appreciable.

Part – B (Marks 10)

Instructions:

• Write the answers to the questions under Part – B on the question paper It self and attach it to the answer book of Part – A.
• Each question carries 1 mark
• Marks will not be awarded in any case of overwriting. rewritten or erased answers.
• Write the capital letter (A / B / C / D) showing the correct answer for the following questions in the brackets provided against them.

Question 1.
Fill the following flow chart with correct organs A & B. ( )

A) Pyloric stomach, Pancreas
B) Large intestine, Oesophagus
C) Cardiac stomach, Pancreas
D) Large Intestine, Pancreas
Answer:
D) Large Intestine, Pancreas

Question 2.
The energy released in respiration is obtained from ( )
A) Synthesis of proteins
B) Synthesis of carbohydrates
C) Cell division
D) Oxidation of glucose
Answer:
D) Oxidation of glucose

Question 3.
Observe the given apparatus and write the name of the experiment. ( )

A) Photosynthesis
B) Mechanism of Respiration
C) Mechanism of anaerobic respiration
D) O₂, Heat
Answer:
C) Mechanism of anaerobic respiration

Question 4.

List-1	List-2
1. Contraction of auricles	a) 0.11-0.14 sec
2. Contraction of ventricles	b) 0.8 sec
3. Cardiac cycle	c) 0.27- 0.35 sec

A) 1 – a, 2 – b, 3 – c
B) 1 – a, 2 – c, 3 – b
C) 1 -c, 2 – a, 3 – b
D) 1 – b, 2 – a, 3 – c
Answer:
D) 1 – b, 2 – a, 3 – c

Question 5.
Read the contents of the urine as per the information given in table ( )

Glucose	65	mg/dl	50-80
Chlorides	128	m.mol/L	120-130
Urea	35	gm/day	20-30

The volume of urea per day present in the urine
A) 20 – 30 gm
B) 20-30 mg
C) 50-80 gm
D) 20-30 m.mol/L
Answer:
A) 20 – 30 gm

Question 6.
From Hevea braziliensis ……………….. (i) and …………………… (ii) from biodiesel will be extracted. ( )
A) (i) – Latex (ii) – Resin
B) (i) – Resin (ii) – Cinchona
C) (i) – Rubber (ii) – Jatropha
D) (i) – Jatropa (ii) – Latex
Answer:
C) (i) – Rubber (ii) – Jatropha

Question 7.
a) Synthesis of waste products in plants is very slow. ( )
b) Plants do not have specific excretory organs.
A) both a and b are true
B) a is true, b is false
C) b is true, a is false
D) both a and b are false.
Answer:
A) both a and b are true

Question 8.
The system that can changes both inside and outside the body ( )
A) Digestive system
B) Endocrine system
C) Nervous system
D) Circulatory system
Answer:
C) Nervous system

Question 9.
External fertilization is seen in ( )
A) Fish and snakes
B) Crocodiles and lizards
C) Man
D) Frogs and fish
Answer:
D) Frogs and fish

Question 10.
The cells in embryo sac are ( )
i) Antipodals
ii) Synergids
iii) Polar nuclei
iv) Female gametophyte
A) iv only
B) iii, iv only
C) i, ii only
D) All
Answer:
D) All