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AP State Syllabus AP Board 9th Class Biology Solutions Chapter 9 Adaptations in Different Ecosystems Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 9th Lesson Adaptations in Different Ecosystems

9th Class Biology 9th Lesson Adaptations in Different Ecosystems Textbook Questions and Answers

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Question 1.
What do you understand by adaptations in organisms and why do they adapt? (AS 1)
Answer:

• The ways and means that organisms adapt or develop over a certain period of time in different conditions for better survival are adaptation of organisms.
• Adaptation is a feature that is common in any population because it provides some improvement for better survival.

Question 2.
With the help of two examples, explain how these organisms have adapted themselves in the ecosystem? (AS 1)
Answer:

• Mangroves grow in a wet and salty place.
• They have evolved to have curious looking projections from their roots called pneu- matophores or knees.
• These pneumatophores develop from the lateral roots that are growing near the surface, and protrude upto 12 inches out of the soil.
• Pneumatophores aid the plants in maintaining adequate root respiration in a watery environment.
• We don’t find such structures in plants growing around us.
• Another example is in kaiabanda, the leaves are reduced to spines so that there is little transpiration loss and water is stored in the tissues of the stem (succulent stems)
• This helps the plant to live in conditions of water scarcity as we come across in deserts.
• With the above two examples, we can say that these organisms have adapted them-selves in the ecosystem.

Question 3.
Collect some aquatic plants- cut the leaves and stems. Observe them under microscope and record your observations like presence air /absence of air spaces etc. and answer the below. (AS 3)
a) Are there any other reasons for their floating?
Answer:
The bodies of aquatic plants are delicate with more than 80% of their weight consisting of water.

b) Draw a diagram of what you have observed under microscope.

Question 4.
What special adaptations can be seen in the following organisms? (AS 1)
a) mangrove trees
b) camel
c) fish
d) dolphins
e) planktons

a) Mangrove trees :
Answer:

1. Mangroves grow in a wet and salty place near the sea shore.
2. From their roots arise pneumatophores or knees.
3. These pneumatophores develop from the lateral roots that are growing near the surface and protrude upto 12 inches out of the soil.
4. Pneumatophores. aid the plants in maintaining adequate root respiration in a watery environment.

b) Camel:
Answer:

1. In camel hump stores fat fordater use.
2. Long eyelashes protects eye from sand.
3. Nostrils closes voluntarily to protect from blowing sand.
4. Long legs keeps the body away from hot ground.

c) Fish :
Answer:’

1. The body is covered by scales.
2. Fishes bear specialised structures to swim like fins.
3. Fishes have floaters in their body (special structures of their digestive canals) to be able to inhabit particu¬lar levels in the water body.
4. Fishes respire with gills.

d) Dolphins:
Answer:
Dolphins have adapted to their environment in the following ways :
Fins shape – A dolphins tail goes up and down to help it dive up to get the air. The shape of their fins also help to propel them through the water.

To help dolphins save oxygen while they dive under water, their heart beat slower during a dive and their blood is diverted from other parts of their body to their heart, lungs and brain. They also save oxygen via muscles, which have a protein called myo-globin which in turn stores oxygen.

They have a blubber or fat which provides insulation helping the dolphin stay warmer under cold water.

They have a body covering of skin. The upper most layer of skin produces an oil which forms a film that cover the dolphin’s body.

Being mammal dolphin breathe with lungs rather than gills. So they breathe from a blow hole which closes before the dolphin goes into the water. The long nose helps the dolphin to fight sharks and their teeth help them to catch fish.

They have well developed echo location by which they locate other animals and also communicate with each other.

e) Planktons:
Answer:
Microscopic photosynthetic organisms like planktons have droplets of oil in their cells that keeps them float.

Question 5.
If an animal of euphotic zone has to survive in abyssal zone, what adaptations are required to survive there? (AS 1)
Answer:
Adaptations required to survive in abyssal zone are :

1. The animals should have wide mouths and huge curved teeth which prevent escape of any prey.
2. Absence of skeleton, flattened bodies are required.
3. Animals should have special structures that produce light on their bellies, around their eyes, and at the sides of their bodies.
4. The animals should show bioluminescence in the dark waters.

Question 6.
Marine water fishes drink more water than fresh water fishes. Do you agree? Justify.
Answer:

• Yes, marine water fishes drink more water than fresh water fishes.
• Because several marine fishes have a lower internal salt concentration than that of the water they swim in.
• So they tend to dehydrate as water is lost by osmosis.
• To compensate, they drink large amount of water and excrete the salts both via their kidneys and through highly specialised cells in the gills.

Question 7.
Visit a nearby pond or lake. Record the organisms you have observed and their adaptations. (AS 4)
Answer:

• Nearby pond or lake consists of three zones namely littoral zone, limnetic zone, and profundal zone.
• In the topmost littoral zone, the edge of a water body is home to snails, insects, several crustaceans, fishes and amphibians, and the eggs and larvae of dragonflies.
• Predators present are tortoise, snakes, and ducks.
• Adaptations : Several organisms have well developed sight, usually have dull and greyish bodies, and are fast swimmers.
• Limnetic zones contains fresh water fish, crustaceans like daphnia, cyclops, and small shrimps are present.
• Floating- plants like water hyacinth, wolfia, pistia along with algae are present.
• Adaptations seen in the plants this zone are presence of air space, leaves covered with wax, etc.
• In the profundal zone scavengers and predators for example crustaceans, crabs, fishes like eels and snails, turtles are present.
• They adapt themselves by feeding on dead animals that settle down.

Question 8.
Draw a lake showing different zones. Why are they called so? (AS 5)
Answer:
Zones of Lake :

1) Littoral zone :
The zone close to shore. They reaches all the way. Plants living in this zone perform photosynthesis.

2) Limnetic zone :
Sunlit part of the lake surrounded by the littoral zone. This zone extends at a depth where sunlight penetrates.

3) Profundal zone :
It is much colder and denser than previous zones.

Question 9.
Collect information of one lake from internet and prepare a table of organisms adapted at different zones.
Answer:
Different zones in lakes and types of organisms present:
1) The littoral zone :
a) The topmost and warmest zone at the edge of a water body is home to snails, clams, insects, several crustaceans, fishes and amphibians and eggs and larvae of dragonflies.
b) Plants like mosses, water lily, vallisneria, hydrilla etc. are found along with several types of algae.
c) Predators of this zone are tortoise, snakes and ducks.

2) The limnetic zone :
a) This zone contains variety of fresh water fish with bright shiny scales.
b) Transparent or whitish bodied crustaceans like daphnia, cyclops, small shrimps are also found in this zone.
c) There are different types of floating plants like water hyacinth wolfia, pistia along with a variety of algae.

3) The profundal zone :
a) Mostly heterotrophs are present.
b) Scavengers and predators like crustaceans, crabs, fishes like eels and glossogobius (isika dondu), snails, turtles etc are present.
c) Many kinds of bacteria are also present in this zone that help in decomposition.

Question 10.
Write the effect of temperature on the organisms adapted in a lake and pond in a tabular form. (AS 1)
Answer:

• In deeper lakes during summer only the surface water is heated up while the deeper layer remain cold. During summer the ponds dry up.
• In tropical regions water gets heated up and evaporates in lakes. During average temperatures the water in the pond heated up and evaporates.
• The requirements necessary to the organisms like oxygen and nutrients gets decreased in the lake.
• The salinity of the water increases, concentration of oxygen decreases and availability of food decreases in pond during average temperatures.
• In the cold regions upper layers of the lake gets frozen during winter and lower layers does not.
• The entire pond gets frozen during winter.
• Aquatic animals in tropics undergo aestivation or hybernation to overcome extreme cold or hot seasons.

Question 11.
Amphibians are wonderful creatures on the earth. How do you appreciate their adaptation? (AS 6)
Answer:

• Amphibian body has small waist, no neck. Streamlined body shape helps in swimming.
• Skin is thin and moist allows gaseous exchange in cutaneous respiration.
• Front legs used to keep the front portion of the body off the ground.
• Hind legs able to jump great distances and change direction quickly.
• Eyes are positioned on top of head gives the frog a wide angled visual field.
• Mouth is very large and broad can able to catch and eat large prey.
• Tongue attached at front of mouth enables it stick the prey when caught.
• Frogs start their lives as aquatic tadpoles with gills to breathe. As tadpole grows into frogs lungs replace the gills and allows frog to breathe on land.

Question 12.
Some animals and plants survive only in certain conditions. Nowadays human activities cause damage to these conditions. What do you think about this? (AS 7)
Answer:

• Human activities are causing lot of damage to biodiversity.
• Human activities such as deforestation, overgrazing, conversion of forest land to agricultural land, hunting and indiscriminate killing of animals for their products, and pollution can endanger the plant and animal species.
• If proper care is not taken plants and animals may disappear totally from the surface of the earth.

Question 13.
In the chapter on ecosystem, we had studied about the mangrove ecosystems. What kind of abiotic conditions did you study in them? (AS 1)
Answer:
Kinds of abiotic conditions in mangrove ecosystems are soil, pH, oxygen, nutrients, winds and currents, light, temperature, humidity, tides, salinity.

Question 15.
Are there any rivers meeting in the Bay of Bengal in the Coringa ecosystem? Collect information and make a note on them.
Answer:

• Coringa mangrove is situated South of Kakinada Bay and is about 150 km South of Visakhapatnam.
• Coringa is named after the river Coringa.
• Coringa mangroves receive fresh water from Coringa and Gaderu rivers, distributors of Gautami, Godavari rivers, and neritic waters from Kakinada Bay.
• Numerous creeks and canal traverse this coringa ecosystem.

Question 17.
The Murrel (Korramatta) and Rohu are fishes found in rivers. Will they be able to live in the coringa ecosystem ? Give reasons for your answer.
Answer:

• Yes, Murrel and Rohu be able to live in the coringa ecosystem.
• Because coringa ecosystem gets fresh water from rivers coringa, Gaderu and distributories of Gautami, Godavari rivers.
• If the salinity of the water in the coringa ecosystem increases, the water enters the body of fresh water fishes.
• The water can be excreted in the form of urine, but to maintain a suitable salt bal¬ance fresh water fish need to reabsorb salt through the kidneys and salt collecting cells in gills.

Question 18.
How the frogs got protected themselves from cold and heat?
Answer:

• Frogs are cold blooded animals so they can’t tolerate extreme cold or heat conditions.
• They protect themselves from extreme cold conditions by a process called hibernation (winter sleep) and from extreme heat conditions by Aestivation (summer sleep)
• During these processes they burrow deep in the ground and remain motionless until the conditions are favourable.
• During this period the rate of metabolic activities slow down and the animal goes into a nearly unconscious sleepy condition.

Question 19.
How do you appreciate the processing protection pebble plants from the enemies?
Answer:

• Pebble plants are also called living stones.
• They protect themselves from their enemies by adapting themselves to their habitat.
• They survive by living partly underground.
• They avoid being eaten by blending in with surrounding rocks.
• Leaves of these plants are not green as in almost all higher plants, but various shades of cream, grey and brown, patterned with darker windowed areas, dots and red lines.
• The markings on the top surface disguise the plant in its surroundings (camouflage)
• Thus, they adopt wonderfully to their habitats and protect themselves from their enemies.

9th Class Biology 9th Lesson Adaptations in Different Ecosystems InText Questions and Answers

9th Class Biology Textbook Page No. 131

Question 1.
What is a habitat?
Answer:
Habitat is the immediate environment occupied by an organism or the living place of an organism.

Question 2.
Is a tree habitat only for a crow?
Answer:’
No. Tree is a habitat for variety of birds and insects.

Question 3.
In what way an ecosystem is different from habitat?
Answer:
In ecosystem biotic and abiotic components are present. Habitat is the place where organisms live in an ecosystem.

9th Class Biology Textbook Page No. 134

Question 4.
You may know animals that live in water. Do you find in them any suitable characters adapted to live in water? Write a note on them in your notebook.
Answer:

• Structural adaptations in the bodies like presence of special air spaces.
• Such air spaces help them to swim and float in water.
• The aquatic organisms bear specialized structures to swim like flippers as in turtles and fins in fishes.
• Fishes, dolphins have floaters in their body to be able to inhabit particular levels in the water body.

9th Class Biology Textbook Page No. 135

Question 5.
In what way flexible stem is useful to the aquatic plants?
Answer:

• In aquatic plants flexible stem contains a parenchymatous tissue known as arenchyma.
• Arenchyma consists of number of air filled spaces.
• These air spaces help the plant to float on water.

9th Class Biology Textbook Page No. 137

Question 6.
Observe the table and answer the following questions.
Answer:

a) How many zones can you see in the figure basis of light penetration? Name them.
Answer:
Three zones are present. They are eu- photic zone, bathyal zone and abyssal zone.

b) What types of abiotic conditions do you find as per the given table?
Answer:
Light, temperature and depth.

c) What will effect adaptation to marine life other than the conditions shown in the table and figure?
Answer:
Salinity, oxygen, rainfall, regular windflow, soil, pH, nutrients, humid-ity, oceanic currents effect adaptation to marine life.

d) What happens to the temperature and pressure as depth increases?
Answer:
As depth increases temperature decreases and pressure increases.

e) Which zone has more animals? Guess why.
Answer:

1. Bathyal zone has more animals. Because the conditions in this zone are suitable for the organisms to grow.
2. Red and brown kelps are the primary producers. They provide food to other organisms in that zone.

9th Class Biology Textbook Page No. 139

Question 7.
Does Pulikat lake of Nellore come under fresh water ecosystem or not? Why?
Answer:

• Pulikat lake of Nellore comes under marine or salt water ecosystem.
• Because the salinity of water in the lake is 3.5%.
• Main salts present in the Pulikat lake are sodium and potassium.

9th Class Biology Textbook Page No. 140

Question 8.
‘Think, how webbed feet helps ducks?
Answer:

1. Webbed feet of birds help them to adapt conditions on land as well as in water.
2. Webbed feet have enabled them to be good swimmers.

Question 9.
Why cranes have long legs and long beaks?
Answer:

1. Cranes have long, thin legs wander through the mud shallows searching for insects.
2. Long beak help them in searching of insects in the mud.

9th Class Biology Textbook Page No. 141

Question 10.
How are marine ecosystems different from fresh water ones?
Answer:

1. The saliny of water in marine ecosystem is 3.5% whereas it is 1.8% in fresh water.
2. Marine ecosystems are huge and they make up about three-fourths of the earths surface.
3. The number of organisms present in marine ecosystems are more when compared to fresh water ecosystem.

Question 11.
Write two types of adaptations you find in marine ecosystems, different from fresh water ecosystems.
Answer:

• Many marine animals have blubber fur insulation from the cold and some fish have an antifreeze like substance in their blood to keep it flowing.
• Marine animals must regulate the interaction of fresh water and salt water in their bodies.
• Specially developed kidneys, gills and body functions help to maintain salt concentrations across members through osmosis.

Question 12.
What are the similarities in adaptation on the basis of light penetration in the two aquatic ecosystems?
Answer:

• In both the aquatic ecosystems, light penetrates upto a depth of zoom only.
• The light intensity is sufficient to perform photosynthesis.
• In the low light intensities below 200 mts depth is sufficent to perform photosynthesis by some kelps.
• Due to the lack of light in abyssal and profundal zones, usually scavengers and predators exists.

9th Class Biology Textbook Page No. 142

Question 13.
Which zone do you think, when compared to marine ecosystems, is absent in fresh water ecosystem?
Answer:
Benthic zone is absent in fresh water ecosystem when compared to marine ecosystem.

Question 14.
What would be a major factor leading to different types of adaptations in marine, fresh water ecosystems?
Answer:
Light would be a major factor leading to different types of adaptation in marine, fresh water ecosystems.

Question 15.
Do all plants shed their leaves at same time in a year throughout the world?
Answer:

1. No. Some plants in temperate regions shed their leaves before the winter starts.
2. In tropical regions some plants shed their leaves before the start of summer.

9th Class Biology Textbook Page No. 143

Question 16.
Are thorny leaves also an adaptation to temperature?
Answer:

1. No. They are not adaptation to temperature.
2. They are adaptation to protect themselves from the animals who eat them.

Question 17.
If the trees have broad leaves at the time of snow fall season what will happen?
Answer:
If the trees have broad leaves at the time of snow fall season, the branches of tree can break due to the weight of snow gathered on each leaf and branch during snow fall.

Question 18.
Why polar bear has thick fur on its body?
Answer:

1. Polar bear has thick fur coat or hair covering on their bodies.
2. The fur act as insulator preventing heat loss from its body.

Question 19.
In what way thick skin helps the seal to protect from cold weather?
Answer:

1. In the thick layer of skin, fat is deposited in seals.
2. The thick layer of fat deposited under their act as insulators preventing heat loss from its body.
3. The fat not only insulates the body but helps in producing heat and energy.

9th Class Biology Textbook Page No. 132

Question 20.
Can you give some examples of fleshy leaf plants?
Answer:
Yes. Bryophyllum, Aloe, and Agave are the examples for fleshy leaved plants.

Question 21.
Why xerophytic plants do not have broad leaves?
Answer:
To prevent the excessive loss of water through respiration xerophytic plants do not have broad leaves.

Question 22.
You may see Kittanara, a xeric plant, grown as fence around crop fields in some areas in our state. Actually those places are not desert. How can they grow there?
Answer:
They grow there because this plant shows adaptations in that places.

9th Class Biology Textbook Page No. 133

Question 23.
Do all animals living in desert conditions show adaptations?
Answer:
Yes, all animals living in desert conditions show adaptations.

Question 24.
Why some animals have scales on their body?
Answer:

1. Scales mainly protect the animals from environment.
2. In desert animals scales allow them to retain moisture by preventing the evaporation of water through the skin.
3. This allows the animal to become dehydrated and animal requires small amount of water to survive.

Question 25.
Why the animals that lives in burrows usually comeout during night time only?
Answer:
To protect themselves from extreme hot conditions, animals that live in burrows usually comeout during night time only.

9th Class Biology Textbook Page No. 139

Question 26.
Which organism among jelly fishes and decomposers present in euphotic zone?
Answer:
Jelly fishes are present in euphotic zone.

Question 27.
What kinds of adaptations can be seen in the organisms of the euphotic zone?
Answer:

1. The organisms living in this zone are mostly floaten and swimmen.
2. Animals in this zone usually have shiny bodies reflecting light away to merge with shiny water surface are transparent.
3. These usually have sharp vision.

Question 28.
What kind of adaptations can be seen in the organisms of abyssal zone?
Answer:

• The larger animals in abyssal zone have wide mouths and huge curved teeth which prevent escape of any prey.
• Absence of skeleton, flattened bodies are some other characteristics observed.
• Some animals also have special structures that produce light on their bellies, around their eyes and at the sides of their bodies.
• Some animals shows bioluminiscence in the dark waters.

Question 29.
What differences can you find in the animals of bathyal zone when compared to animals of euphotic and abyssal zones?
Answer:

• Most of the plants found in this zone are the red and brown kelps, sponges, corals even animals with tubular bodies like squids and large animals like whales, etc.
• Some of the animals in the bathyal zone have a flat body like the ray fishes.
• Big eyes sensitive to very dimlight may present in bathyal zone animals.

Question 30.
How organisms of different zones of marine ecosystem are adapted?
Answer:

• The animals of euphotic zone are mostly floaters and swimmers.
• Animals in this zone usually have shiny bodies reflecting light away to merge with shiny water surface.
• Animals of euphotic zone have very sharp vision.
• Some of the animals in bathyal zone have a flat body like the ray fishes.
• The animals may have big eyes sensitive to very dim light in bathyal zone.
• Absence of skeleton, flattened bodies are some adaptations found in animals of abyssal zone.
• Some animals in abyssal zone may have special structures that produce light on – their bellies, around their eyes and at the sides of their bodies.
• Some animals in abyssal zone shows bioluminiscence in the dark waters.

9th Class Biology Textbook Page No. 141

Question 31.
Organisms of the oceans have a lesser salt content in their bodies than the seawater around 3.5%. The fluid could drain out of the body of the organisms into the sea. This could be dangerous and fatal to the organism. How do they survive under such conditions?
Answer:

• Several marine species have a lower internal salt concentration than that of the water they swim in. So they tend to dehydrate as water is lost by osmosis.
• To compensate, they drink large amounts of water and excrete the salts both via their kidneys and through highly specialised cells in the gills.

Question 32.
Can fish in estuarine ecosystem survive in river as well as in sea?
Answer:

• Yes, fish in estuarine ecosystem survive in river as well as in sea.
• Two of the main challenges of estuarine life are the variability in salinity and sedimentation.
• Many species of fish living in estuarine have various methods of control to the salt shifts.
• They regulate the salt concentrations using osmoregulaters.

9th Class Biology 9th Lesson Adaptations in Different Ecosystems Activities

Activity – 1

Question 1.
i) Take a Kalabanda (Aloevera) and a Balsam plant in two separate pots.
ii) Water each of them with two tablespoons of water.
iii) Do not water them for a week.
iv) Observe the condition of the plants after a week.

Observations :
a) Which plant showed growth?
Answer:
Kalabanda plant showed growth.

b) Which plant dried first? Why?
Answer:
Balsam plant dried first. Because Balsam plants are not watered regularly. They need water to grow.

Activity – 2

Question 2.
i) Collect an aquatic plant out of a water body (e.g. Duck weed, Hydrilla, Vallisneria etc.) ii) Carry it back home and plant it in a pot and water it.
Observations :
a) From the above activity we see that some plants dry up without water very quickly, while other can grow even with very little water.
b) Each of these plants are adapted to the conditions in their surroundings on the basis of need of water.

Activity – 3

Question 3.
You know some of the animals that reside in and around lake or pond. Make a list of those animals and the characteristics of their body.
List of animals and reside in and around lake or pond :
Insects : Dragonfly, Damsefly, Mayfix, Stonefly, Dobsofly, Caddisfly, Cranefly, Water bugs, Beetles, etc.

Crustaceans	Cray fish, Scuds, Shrimps
Molluscs	Snails
Annelids	Leeches
Fish	Blugill, Bass, Catfish, Sculpin, Minnow
Reptiles	Snakes, turtles
Amphibia	Frogs

Characteristics of the body of animais living in and around lake:

Animals	Characteristics
1) Mosquito	The body is segmented and it is a carrier of diseases.
2) Shrimps	‘ These are small, bottom dwelling crustaceans with a trans­lucent exoskeleton.
3) Snails	A soft bodied animal with a hard protective shell.
4) Swan	Swans are long necked water birds, webbed feet are present.
5) Crayfish	Fresh water crustaceans with four pairs of walking legs. Body is segmented with head and thorax united.
6) Dragonfly	it is a flying insect with a long abdomen. Body is elongated with two pairs of transparent wings.
7) Earthworm	It is a little animal with a long, soft body and no legs.
8) Fish	It lives in the water and breathe with gills.
9) Goldfish	It is a type of crap that makes a nice pet, kept in aquariums and swims with fins.
10) Toads	The skin is dry and leathery. Toads are amphibians with poison glands, short legs and snout like parotid glands. Drier skin. Webbed feet helps in walking and swimming.
11) Leech	The body is segmented. It sucks blood of other animals.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 8 Challenges in Improving Agricultural Products Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 8th Lesson Challenges in Improving Agricultural Products

9th Class Biology 8th Lesson Challenges in Improving Agricultural Products Textbook Questions and Answers

Improve Your Learning

Question 1.
Suggest some ways through which our country could increase the production of rice to meet at least global limits. (AS 1)
(OR)
Day by day population is increasing. But the cultivated land is very limited. To produce required quantity of food for the growing population, what are the poible solutions in your view?
Answer:

• Increasing area of cultivated land.
• Increasing production in the existing land.
• Developing high yielding rice varieties.
• Conserving the genetic diversity of rice so it can be used in the development of new varieties suited to different growing conditions.
• Developing rice crop management strategies that improve nutrient use efficiency.
• Management of crop protection and suitable irrigation methods.
• Organic manure to be used for rice crop.
• Alternating crops and mixed crop system to be followed.

Question 2.
How are biofertilizers more beneficial as compared to chemical fertilizers? (AS 1)
Answer:

• Biofertilizers add natural nutrients to soil.
• They increases soil organic matter and improves soil structures.
• Biofertilizers improves water holding capacity of the soil and reduces soil crusting problems.
• They reduces soil erosion from wind and water.
• Biofertilizers increases crop yield.
• Biofertilizers improves the percentage of humus and remained long time in the soil.

Question 3.
a) Find out the adverse effects of chemical fertilizers need for growing the high yielding varieties of crops. (AS 1)
Answer:

• Chemical fertilizers pollute lakes, rivers and streams.
• They destroy beneficial soil life including earthworms.
• By using chemical fertilizers, we can get high yielding for only 20 to 30 years.
• After that soil becomes reluctant to plant growth.
• Chemical fertilizers damage soil fertility.
• Make certain crops vulnerable to diseases.
• Prevent some plants from absorbing needed minerals.
• Food produced by using chemical fertilizers do not taste as good.

b) Can high yielding varieties be grown without them (Chemical Fertilizers) as well? How? (AS 1)
Answer:

• Yes, high yielding varieties be grown without chemical fertilizers.
• By using biofertilizers, instead of chemical fertilizers and synthetic pyrethroids we get higher yielding.

Question 4.
What threats to nature do chemical fertilizers, pesticides, insecticides and herbi-cides pose? (AS 6)
Answer:

• When we use insecticides to kill pests or weedicides/herbicides to destroy weeds, a large percentages of herbicides, chemical fertilizers, pesticides, insecticides remain in the soil.
• From the soil, these chemicals find their way into water sources.
• People spray these chemicals in fields are exposed to them and some of the chemi-cals enter their body.
• Insecticides destroy all insects in which some of them are useful in pollination.
• Extensive use of chemical fertilizers, pesticides, herbicides and weedicides makes the soil unsuitable to grow crops after sometime.

Question 5.
What are the adverse effects of using high yielding varieties of seeds? (AS 1)
Answer:
The adverse effect of using high yielding seeds is – they use more nutrients from the soil. Thus the soil can lose its fertility if they are used continuously.

Question 6.
What are the essential measures that a farmer needs to take before sowing the seeds of a crop? (AS 1)
Answer:

• Preparation of soil is done before sowing the seeds.
• The soil is ploughed to loosen and break the solid pieces of soil.
• The field is watered before sowing.
• Seed treatments against/soil-borne diseases to reduce the incidence of diseases.

Question 7.
Suppose you had a farm in a drought striken area of your state, what crops would you grow and how? (AS 1)
Answer:

• Sorghum, Pearlmillet, Red gram, Green gram, Horse gram can be grown in drought striken area.
• We can grow these crops by rain water harvesting building check dams, drip irrigation methods, watershed management and soil and water conservation methods.

Question 8.
What measures will you take to save your field from seasonal outburst of insects? (AS 1)
Answer:
Nowadays farmers use insecticides and other chemicals to save their crop field.

• I prefer catching the insects manually and removed from the field.
• I also use predatory insects to remove insects from the field.
• I place lighted bulb (Deepapu Teralu) so that insects could cluster around it.
• Insecticides are sprayed at regular intervals.

Question 9.
What basis would you adopt to explain to a farmer using chemical fertilizers switch over to organic fertilizers? (AS 4)
Answer:

• Organic fertilizers replenish the soil, keeps soil easily broken up into small pieces.
• Organic fertilizers promotes beneficial soil life.
• Organic fertilizers increase crop yield.
• They maintain a natural balance in the soil.
• They protect certain crops from diseases.
• Benefit the environment by recycling agricultural wastes into energy for local community.

Question 10.
A farmer had been using a particular insecticide for a long time. What consequences will it have on – a) insect population b) soil ecosystem? (AS 1)
Answer:
a) Insect population :

1. Insect develop immunity to the insecticide used.
2. And it has any effect on the insect it targets. Hence the number of insects increases,

b) Soil ecosystem :

1. A large percentage of insecticide chemicals remain in the soil.
2. These chemicals kill the worms in the soil which are useful to soil.
3. Hence soil ecosystem destroys by increasing the concentration of salts in the soil.

Question 11.
Venkatapuram village is in drought prone area. Somaiah wants to cultivate sugar-cane in his fields. Is it beneficial or not? What questions will you ask him to convey your belief? (AS 7)
Answer:

• It is not beneficial for Somaiah to cultivate sugarcane crop.
• Sugarcane grown in places where rich water resources are present.
• “Where do you get water to cultivate sugarcane crop?” I ask this question to Somaiah.
• I advise him to grow crops which needs less water (Aruthadi Pantalu) in drought prone area.

Question 12.
Draw a block diagram of water resources in your village. (AS 5)
Answer:

Question 13.
Ramaiah has soil testing done in his field. The percentages of nutrients are 34-20-45. Is it suitable for cultivating sugarcane crop ? Which crops can be cultivate without using pesticides in Ramaiah’s field? (AS 2)
Answer:

• Ramaiah’s field is not suitable for cultivating sugarcane crop.
• Because sugarcane needs 90% of nitrogen in the soil but Ramaiah’s field has only 34% of nitrogen.
• Maize and groundnut can be cultivated in Ramaiah’s field.

Question 14.
Organic manure is helpful to biodiversity. How do you support this statement? (AS 6)
Answer:

• Biological research on soil and soil organisms has proven beneficial to the system of organic farming.
• Varieties of bacteria and fungi breakdown chemicals, plant matter and animal waste into productive soil nutrients.
• In turn the producer benefits by heal their yields and more suitable soil for future crops.

Question 15.
Make a list of major weeds in your area (You have already conducted the project) Find out the different weeds that grow along with different crops in your area. (AS 4)

Answer:
Cynodondacylon, Digitaria longifolia, Dacty loctenium colonum, Setaria glauca, Cyperus rotundus, Cyperus difformis, Eichornia crassipes, Salvinia mollusta, Alternathera sps. Celosia argentea, Leucas aspera, Portulaca oleracea, cleome sps. Solanum nigum, Argemone mexicana, Abutilon indicum, Euphorbia sps. Vernonia Cinnera, Eichnochloa colonum, Commelina bengalensis, Avenafatua, Eichnochloa Crusgalli, Eleusine indica, Euphorbia hirta, Achyranthus, despera, Eclipta prostrata.

Name of the Crop	Weeds that grown on crop
Paddy	Cynodon dactylon, Digitaria longi folia, Cyperus rotundus, Eichornia.
Groundnut	Leucas aspera, portulaca oleracea, Cleome sps, Abutilon indicum, Euphorbia cynodon dactylon, Commelina bengalensis, Cyperus roturdus.
Black gram	Cynodon dactylon, Cyperus rotundus, Abutilon indicum, Commalina bengalensis, Euphorbia hirta.
Maize	Euphorbia hirta, solanum nigrum, cyperus rofundus, cynodon dactylon.
Green gram	Eichnochloa colonum, cyperus rotundus cynodon dactylon, Argemone mexicana, Portulaca oleracea.

Question 16.
Spraying high dose of pesticides is hazardous to biodiversity and crop yielding. How can you support this statement?
Answer:

• When we use pesticides large percentage of it will remain in the soil. These kill the germs in the soil.
• From the soil pesticides find their way into water bodies affecting the aquatic animals.
• People who spray these pesticides in the fields are exposed to them and some of the chemicals enter their bodies causing health problems or some times the person dies.
• Pesticides destroy all the insects which are useful to the plants causing hazardous to biodiversity.

Question 17.
Natural pest controlling methods are useful to biodiversity. Comment it.
Answer:

• Some insects control the harmful insects and they are called friendly insects.
  E.g.: Spiders, Dragonfly, Krisopa etc.
• Trachoderma bacterium lives in the eggs of stemborer, tobacco caterpillar destroy the pests at the egg stage.
• Some bacteria like Bacillus Turengenisis destroy some pests.
• Mixed crops also control some pests and diseases.
• Hence natural pest control methods are useful to biodiversity because these methods destroys only the selected pests.

Question 18.
Observe the fields in your surroundings and collect the information from farmers about the process to remove weeds.
Answer:
Farmers use different methods to remove weeds. Some of them are
1) Manual method:
Many farmers still remove weeds by manually pulling them out of the field, making sure to include the roots that would otherwise allow them to resprout.

2) Stale seed bed method :
This method involves cultivating the soil, then leaving it follow for a week or so when the initial weeds sprout, the farmer lightly hoes them away before planting the desired crop.

3) Using Herbicides :
Selective herbicides kill certain targets while leaving the de¬sired crop relatively unharmed.

4) Biological control:
Vinegar kills the visible part of the weed. They will wrinkle and die next day.

5) Ploughing & Tilling :
Ploughing includes filling of soil, inter-cultural ploughing and summer ploughing. Ploughing up roots weeds causing them to die. Mechanical tilling can remove weeds around crop plants at various points in the growing process.
Crop rotation method also helps in controlling weeds.

9th Class Biology 8th Lesson Challenges in Improving Agricultural Products InText Questions and Answers

9th Class Biology Textbook Page No. 110

Question 1.
Rate of growth of population and food grain production.

1. In which decade population growth is higher?
Answer:
Population growth is higher in 1961-1971.

2. In which decade food grain production is higher?
Answer:
Food grain production is higher in 1981-1991.

3. What major differences did you find in the table?
Answer:
The major differences find in the table are :
i) Foodgrain production is not increasing according to population growth.
ii) Ratio of FP/PG is in irregular order.

4. Is food grain production increasing according to population growth?
Answer:
No, last two decades food grain production is not increasing according to population growth.

5. In which decades production of food grains didn’t satisfy the needs of population? What will happen if the production is not sufficient?
Answer:
In 1991-2001 production of food grains didn’t satisfy the needs of population. If the food production is not sufficient then it leads to food crisis.

6. The decade 1991-2001 shows that rate of food production was nearly half as compared to population. What can you infer from the decade when population growth was highest?
Answer:
The reasons for the highest population growth :
i) Wide spread diseases are controlled.
ii) Health care programmes were made available in rural areas.
iii) So death rate declined.
iv) Therefore population growth become inevitable.

Question 2.

1. Find out from the graph the months in which the most water evaporates from plants.
Answer:
The months in which the most water evaporates from plants are May and June.

2. Are these the same months in monsoon season when the rainfall is heavy?
Answer:
No, these are not same.

3. So how does the availability of more water effect the plant?
Answer:
The availability of more water effects the plant with more evaporation.

9th Class Biology Textbook Page No. 110

Question 3.
When the weather is hot and the stomata dose, what effect would this have on the absorption of carbon dioxide by the plant?
Answer:
If the stomata closed, then the absorption of carbon dioxide by the plant is less.

Question 4.
What effect would a change in the amount of carbon dioxide absorbed have on the growth of the plant?
Answer:
If the absorption of carbon dioxide by the plant is less then the growth of the plant decreases.

Question 5.
If the plant does not get water at this time, what effect would this have on its growth? Discuss in your class and find out reasons.
Answer:
If the plant does not get water at this time then the growth of the plant will stopped.

Question 6.
What are the main water sources in your village for agriculture? How farmers utilise them?
Answer:

• Canals, Bore wells, ponds are the main water sources in our villages.
• Farmers utilise water from these sources to cultivate crops.

Question 7.
Make a list of crops which require less amount of water.
Answer:
Cotton, Jute, Bajra, Maize, Coconut, Black gram, Green gram etc.

9th Class Biology Textbook Page No. 111

Question 8.
If a field is cultivated for many years, what would happen to the nutrient content of the soil?
Answer:
If a field is cultivated for many years, then the nutrient content of the soil is decreased.

Question 9.
How does the soil get back or replenish these nutrients?
Answer:
The soil get back or replenish of these nurients by adding organic manure or chemical fertilizers.

9th Class Biology Textbook Page No. 112

Question 10.
A farmer cultivated sugar cane in his land for the last five years. Another farmer culti¬vated sugarcane in the first year and soya bean in the second year and sugarcane in third year.
– In which case do you think has the land lost most of its nutrients?
Answer:
The land lost most of its nutrients in the case of first farmer.

Question 11.
Have you ever seen two types of crops in the same field?
Answer:
Yes, I have seen two types of crops in the same field.

Question 12.
Which crops are grown in this way?
Answer:
In the fruit growing fields like Lemon, Pomegranate, Papaya, etc. pulses like Red gram, Black gram, Green gram, etc. are grown in this way.

Question 13.
What are the uses of cultivating mixed crops?
Answer:
The uses of cultivating mixed crops are :
1) The soil becomes fertile.
2) The nutrients which are used by one crop will be regained by cultivating another crop.

9th Class Biology Textbook Page No. 113

Question 14.
Is betel (Tamalapaku) a mixed crop? How can you justify your answer?
Answer:
Yes. Betel is mixed crop. Sorghum grown along with betel.

9th Class Biology Textbook Page No. 121

Question 15.
If we don’t use these chemicals, how can we get a good crop? How can we increase production? Is there an answer to this question? What coidd it be?
Answer:
Suppose we can use some other methods that do not give rise to these problems. For example, they say we can make use of the natural food chains to control pests. There are many insects that eat other insects. They are called predatory insects. We can make use of these insects. There are also birds that eat insects. We can use these birds to get rid of insects.

Question 16.
If insects that pollinate crops are killed, what effect will this have on crop production?
Answer:
If insects that pollinate crops are killed, the crop production will decrease.

Question 17.
In recent times, why farmers touch the flowers with handkerchiefs in sunflower fields?
Answer:
Farmers touch the flowers with handkerchiefs in sunflower fields to control the insects.

9th Class Biology Textbook Page No. 122

Question 18.
Do you know why Jatropa in cotton fields and marigold in mirchi fields are cultivated?
Answer:
Some mixed crops controls some diseases and pests. That’s why Jatropa in cotton field, marigold in mirchi fields are cultivated.

9th Class Biology Textbook Page No. 110

Question 19.
In what way this kind of water supply is useful to the crop as well as the farmer?
Answer:
To prevent water wastage and economically helpful to the farmer.

Question 20.
Water Shed is a process to improve ground water level. In what way it is related to irrigation? Support with your answer.
Answer:
If ground water level will be increased then it will help to irrigation.

9th Class Biology Textbook Page No. 117

Question 21.
In what way vermi compost is better than chemical fertilisers?
Answer:
After using vermi compost, investment on chemical fertilizers and other pesticides became reduced and the quantity of their agricultural products increased.

9th Class Biology 8th Lesson Challenges in Improving Agricultural Products Activities

Activity -1

Question 1.
Observe the Transpiration :

1. Take a polythene bag. Cover the bag on leaves and tie it.
2. Do this experiment during day time and night time separately.
3. Note the difference in your notebook.

Observations :

1. If we tie a plastic bag over a leaf, we will be able to see how much water a plant releases in the air.
2. It is estimated that a plant use only 0.1 percent of the water it absorbs to form carbohydrate.
3. The rate of transpiration is high during day time when compared to night time.

Question 2.
b) Draw the route map of Jawahar and Lai Bahadoor canals of Nagarjuna sagar in Andhra Pradesh & Telangana map.
Answer:

I. 1) Take one example from each of millets, cereals, vegetables, and fruits.
2) First you have to list out the known characters of the above and then list out the characters that you want to change or modify in them.
3) But you need to give your own reasons – why do you want to make such changes in them?

II. Red and yellow equal to rellow.

1) If you want to make your own hybrid flower you need to do the following. But it is time consuming process and patience job too.
2) For this you need red and yellow colour chandrakantha plants.
3) Select 5 or 6 red flowers on a plant.
4) Remove all the other flowers of that plant.
5) Take each flower, remove stamens carefully.
6) Take yellow flower and rub with that flower gently on the stigma of selected red flower for pollination. Do this process in the evening only.
7) Tie a tag with a thread loosely to the pollinated flower to avoid confusion in iden¬tifying these flowers for seeds in the next few days.
8) Within a week days you will get black seeds.
9) Keep them another two weeks to dry and sow them in a pot.
10) Take care to grow the plants until they flower.

Observations:
1) The colour of the flowers will be orange.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 7 Animal Behaviour Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 7th Lesson Animal Behaviour

9th Class Biology 7th Lesson Animal Behaviour Textbook Questions and Answers

Improve Your Learning

Question 1.
What is the advantage of reflex action? (AS 1)
(a) It has to be learned
(b) It happens differently each time
(c) It does not have to be learned
(d) None of them
Answer:
(c) It does not have to be learned.

Question 2.
If a rat is given a mild electric shock when it goes to a certain part of its cage, it eventually avoids going there. This is because of- (AS 1)
(a) Imitation
(b) Conditioning
(c) Instinct
(d) Imprinting
Answer:
(b) Conditioning

Question 3.
Describe all types of behaviour discussed in the lesson with appropriate examples. (AS 1)
(OR)
Describe different types of animal behaviours discussed in the classroom with suitable examples.
Answer:
Scientists categorize animal behaviour into different categories like instinct, imprinting, conditioning, imitation.
1) Instinct:
A) Instinctual behaviours are behaviours that need not be learned.
B) They are inborn behaviours and can be complex like making nest by birds, choose mates and forming into groups for protection.
C) Reflexes are also a type of instinct behaviour.

2. Imprinting:
A) Chickens and ducklings are able to walk almost immediately after hatching from the egg.
B) Duckling can even swim after a few days.
C) They recognise their mother because of a behaviour type called imprinting.
D) Imprinting lets young animals recognise their mother from a young age. They can follow her for food and protection.

3. Conditioning:
A) Conditioning is a type of behaviour involving a response to a stimulus that is different from the natural one.
B) It is a type of learned behaviour.
C) If we take ringing of school bell as an example, student shows different types of conditioning to a school bell as per the time.

4. Imitation:
A) It is a type of behaviour where one animal copies another animal.
B) Kohler conducted experiments on imitation in chimpanzees.
C) One chimpanzee tried to take a fruit from a tree. But it failed in reaching the fruit.
D) Later chimpanzee used sticks to reach the fruit. This time it succeded in reaching the fruit.
E) Chimpanzee used sticks to spear juicy grubs to eat.
F) Other chimpanzees copy this behaviour.

Question 4.
Differentiate between (AS 1)
Answer:
a) Imitation and Imprinting.

Question 5.
How human behaviour is different from behaviour of other animals? Explain with an example. (AS 1)
Answer:

• Humans show many of the same types of behaviour as other animals.
• But human behaviour is often more complex because we are more intelligent and aware of ourselves.
• For example, hungry persons might want to start eating immediately when they sit down at the dining table.
• But the humans have learned that good manners mean they should wait until everyone is seated and ready to eat.
• But animals eat food whenever they find it.

Question 6.
Observe ants going on a line. Ask your teacher how they communicate and write a note on this. (AS 4)
Answer:

• Ants talk to each other using chemical signals called pheromones that they detect with antennae.
• Ants use their antennas to pick up smells.
• For example, when ants find food they have a pheromone trail from the food soure to the colony.
• Other ants follow this trial. As the other ants follow the trial, the pheromone scent becomes stronger.
• The paired antennas of ants provide information about the direction and intensity of scents.
• Since most ants live on the ground, they use the soil surface to leave pheromene trail that may be followed by other ants.
• Some ants produce sounds using gaster segments and their mandibles.
• Sounds may be used to communicate with colony members or with other species.

Question 7.
“Understanding of animal behaviour creates positive attitude towards animals”. How do you support this statement? Explain with suitable examples. (AS 6)
Answer:

• I support the above statement that understanding of animal behaviour creates positive attitude towards animals.
• Animals usually make sounds depending upon their needs. They show different facial expressions.
• For example, cattle make sounds whenever they need food and water.
• After giving food and water by the master they calmdown.
• When a crow dies, all the other crows come around making sounds ‘kaww, kaww’ shows their sadness, we have to understand nature of the crows in this situation.
• We have to identify the unity and integrity among the ants when they go in line in search of food.
• When dogs bark during nights, we should understand that they are doing that for our safe.
• We have to show positive attitude towards animals who are useful in our daily life. Live and let live should be our motto.

Question 8.
Look at this picture. How do you feel about sibbiling care nature of animals. Have you ever seen such kind of situations in your surroundings? Explain in your own words. (AS 7)

Answer:

• Sibbiling care nature of animals are animals with instinct behaviour.
• Every animal take care of their young ones until they are grown adults.
• I observed sibbiling care situations in my surroundings.
• The newly hatched chickens are taken by their mother to surrounding places to feed them.
• Whenever the chicks faces danger mother brings them under her wings.
• When the eagle tries to take away the chicks the mother attacks the eagle to save its chickens.
• Chicken feeds and protect young ones until they are able to collect their own food.
• A new born kitten is born blind. Its eyes normally do not open until it is 10 to 12 days.
• Finding the milk source is accomplished with help from mom, who encourages young kittens to feed a few minutes after birth.
• Cat often changes its living place by transfering young kitten. It does so to protect kittens from enemies.

9th Class Biology 7th Lesson Animal Behaviour Activities

Lab Activity

Question 1.
Behaviour of Cockroach : For this we need a choice box and calcium chloride.
Answer:
Making of Choice box :

• Take a box, and divide it into four chambers with the help of a card board.
• Make tiny holes in any two chambers of one side so that light can pass through these holes into the chambers.
• Let other two chambers as it is (dark).
• Now create humid environment with help of moist cotton wool in one of the lightened and one of the dark chambers.
• Create dry atmosphere with help of calcium chloride in one of the lightened and one of the dark chambers.

• So, the box has been divided into four chambers with different conditions i.e., light and dry, light and humid, dark and dry, dark and humid.
• Make four groups in class. Each group will put several cockroaches into a choice of chamber with four different conditions.
• Cover the box and leave the setup for 15-20 minutes.
• Count the number of cockroaches in each chamber.

Observations :

1. Cockroaches prefer dark and damp conditions.
2. The quarter of the choice chamber with these conditions contains most or all of the cockroaches.

Activity – 1

Question 2.
Observe the following behaviours of different animals. Identify their instinct, imprinting, conditioning or imitation.
a) Our pet dog barks only on strangers. If is not stopped, how would it behave?
Answer:
Conditioning.

b) Ants which usually go in a line reach sweet kept in tin. How do they know the way to reach the tin?
Answer:
Conditioning.

c) Mosquitoes, cockroaches come out of their places only when it is dark. How do they know the difference between light and dark?
Answer:
Instinct.

d) Bats and owl move and search for food during night only. How could they know what is a day what is a night?
Answer:
Instinct.

e) When you untie the neck of your bull at the time of ploughing, it moves towards plough without any instructions. In the same way, it moves towards tub at the time of feeding. How does the bull respond differently?
Answer:
Conditioning.

f) Birds collect material which is soft, strong to build its nest. How do they know the quality of material?
Answer:
Instinct.

g) Puppies, kitten fight each other when they saw a piece of cloth. They try to tare it off why?
Answer:
Imitation.

h) In a particular season some birds in our surroundings migrate from long distances. . How do they know their way?
Answer:
Instinct.

Activity – 2

Question 3.
Select any one of the animals in your surroundings. Observe it how it behaves in the following situation.
Answer:
1) Name of the animal:
Crow (corvus species)

2) Place where it lives :
They live in nests build on trees. Usually, they build nest where they feel safe from predators.

3) How it builds its place :
Crow builds its nest using tree branches, small sticks, hay etc.

4) Way of collecting food/prey:
Crows go around places where food is available. Crows are omnivorous and they eat almost everything.

5) External characters :
Crows are usually black in colour or black with little white plumage.

6) Expressions :
A) Crows make a wide variety of calls or vocalizations.
B) In many species the pattern and number of numerical vocalizations have been observed in response to events in the surroundings like arrival or departure of crows.
C) Crows show their happiness, Jadness, fear, threat by making sounds like ‘KOWWS’.

7) Group behaviour:
A) If one crow finds food it call others to join.
B) If one crow dies, all the other crows make ‘KOWWS’ continuously without interference.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 6 Sense Organs Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 6th Lesson Sense Organs

9th Class Biology 6th Lesson Sense Organs Textbook Questions and Answers

Improve Your Learning

I. Give reasons for :

Question 1.
We usually do not see bright colours in dim light.
Answer:

• Retina contains cells called rods and cones.
• Nearly 125 million tiny rods are present In our eye which contain pigment ‘rhodopsin’.
• Rods detect low intensity of light at night.
• Rods cannot make the fine distinctions of bright colours in dim light.

Question 2.
Removal of wax layer too often will raise incidence of ear infection.
Answer:

• External ear or pinna has wax producing ceruminous glands and oil producing sebaceous glands.
• These glands help to keep the ear canal lubricated prevent the dust and other par¬ticles from entering into the ear canal called Auditory Meatus.
• If we remove wax layer ear diseases like formation of pus, infection of eardrum etc. may be caused by bacterial and fungal infections.

Question 3.
During severe cough and cold we lose taste of food.
Answer:
When we suffering from severe cold and cough our nasal passages are completely blocked, food becomes tasteless as we are unable to smell the foods delicious aromas.

Question 4.
While cutting onions our tears start flowing.
Answer:

• The cells of onion contains amino acids like mithionine and enzyme cystine. They are kept separate in the onion cell.
• When we cut the onion, enzymes start mixing and produce propanethiol. Sulpher oxide, which is a volatile compound that starts moving towards our eyes.
• The gas that is emitted reacts with the water of our eyes and forms sulphuric acid.
• The sulphuric acid thus produced causes burning sensation in our eyes and this, in turn, leads to the tear glands secreting tears.
• Thus we end up with watery eyes every time we cut onions at home.

II. Find out the false statements and rewrite them as correct ones.

1. The rationale behind seeing is just the impression of the image in the retina.
Answer:
True

2. Ear functions only to hear.
Answer:
False
Correct statement: Ear helps in hearing and also in maintaining the equilibrium of our body.

3. Iris patterns are like fingerprints used in identifying individuals.
Answer:
True.

4. Saliva helps the taste buds in taste sensation.
Answer:
False
Correct statement: Taste receptors help the taste buds in taste sensation.

5. We are not able to adapt to sensations.
Answer:
False
Correct statement: We are able to adapt to sensations.

III. State the differences between the two.

1. Rods and cones.
Answer:

Rods	Cones
1) Rods are responsible for detecting the dim lights only with black and white.	1) Cones are responsible for detecting the bright lights with colour.
2) 125 million rods are present in retina.	2) Seven million cones are present in retina.
3) Rods contain the pigment rhodopsin.	3) Cones contain the pigment idopsin.
4) Defect in rods causes night blindness.	4) Defect in cones causes colour blindness.

2. Iris and pupil.
Answer:

Iris	Pupil
1) This is the coloured part of our eye around pupil.	1) It is the hole located in the centre of the eye.
2) The colour of the iris may be brown, blue, green or grey.	2) The colour of the pupil is black.
3) It does not dialate or contract according to the intensity of light.	3) It dialates or contracts according to the intensity of light.

3. Pinna and tympanum.
Answer:

Pinna	Tympanum
1) It is also known as external ear.	1) It is also known as ear drum.
2) Pinna is the visible part of the ear on either side of our head.	2) It is present in between external and middle ear.
3) It is a flap (ring) like structure.	3) It is in the shape of a cone.
4) Pinna is made up of cartilage.	4) It is a thin membranous layer.
5) It collects the sound waves.	5) Sound waves strikes the tympanum and vibrates.
6) It is the first part of external ear.	6) It is the last part of the external ear.

4. Nasal cavity and ear canal.
Answer:

Nasal cavity	Ear canal
1) It is the cavity present in external nostrils.	1) It is the canal in the external ear.
2) Nasal cavity opens into internal names.	2) At the end of ear canal ear drum is present.
3) It filters the air that entering into internal names.	3) It carries the sound waves from external ear to ear drum.

IV. How do the following processes occur?

Question 1.
When we see an object, a real inverted image is formed on the retina.
Answer:

• When we see an object the eye gathers light through a convex lens, focusses it and forms an image in the retina at the back of the eye.
• The lens turns the image left to right and upside down.

Question 2.
The sound waves, collected by the pinna are changed as vibrations.
Answer:

• External ear or pinna collects the sound waves.
• They enter into the auditory meatus or ear canal. Then they strike the tympanum.
• The vibrations from the tympanum reach the malleus, incus and stapes in the middle ear.
• They magnify the intensity of the sound vibrations and send them to the membrane of oval window in middle ear.

Question 3.
We move our hand away from a hot object.
Answer:

• Moving our hand from a hot object is an unconditional reflex.
• Unconditional reflexes are inherited and shown from birth.
• Unconditional reflexes are present in all individuals and are basically same.

Question 4.
A pungent odour, makes us close our nose.
Answer:

• The olfactory receptors sense pungent odour, that information is sent to brain in the form of nerve impulses through sensory nerve.
• The brain interprets the information and identifies it as pungent odour.
• It sends message to our hand to close the nose immediately.
• The brain also sends information in the form of nerve impulses through motor nerves as we are unable to smell the food’s delicious aromas.

V. Fill in the blanks with suitable words. Then give reasons why the words are suitable.

1. Choroid layer provides ………………… to the eye.
2. The relationship between the tongue and ……………………. is more.
3. Iris pattern is used for individual …………………… .
4. Area where optic nerve leaves the eye is called the ……………………..
5. The ear drum is the …………………….
Answer:

1. Protection
2. Nose
3. Identification
4. Blind spots
5. Vibrating membrane

VI. Choose the correct option.

1. This vitamin is essential for the health of eye.
a) Vitamin ‘A’
b) Vitamin ‘B’
c) Vitamin ‘C’
d) Vitamin ‘D’
Answer:
a) Vitamin ‘A’

2. Sensation is a complex pathway involving
a) Sense organs
b) Sense organs and nerve impulses
c) Sense organs, nerve impulses, brain
d)Brain and nerve impulses
Answer:
c) Sense organs, nerve impulses, brain

3. The sound waves if not focused by external pinna and ear cannal will result in
a) Hearing several types of sound loudly
b) Not hearing anything
c) Slight hearing
d) Not being able to make out the type and origin of sound
Answer:
b) Not hearing anything

4. The muscles of the eyeball of a person becomes non functional, the invariable effect would be
a) The person fails to close eyes
b) Fails to move eye and see colours clearly
c) Feels pain in the eye
d) The nerves reaching the muscles become non-functional.
Answer:
b) Fails to move eye and see colours clearly

5. The tongue of a person is exposed to a high salty taste then:
a) The person learns to taste salty things better
b) Loves tasting salty things
c) Hates tasting salty things
d) Fails to taste a less salty thing just after the exposure.
Answer:
d) Fails to taste a less salty thing just after the exposure.

VII. Draw and label the diagrams, showing the structure of the
1. Eye

2. Ear

3. Tongue

VIII. How would you pay concern towards disabled people who is lacking sensory organs?
Answer:

• 1 will show utmost sympathy forwards disabled people who is lacking sensory organs.
• I will give my full cooperation in leading normal lives.
• With my deeds I will bring confidence among the disabled people.
• If the disabled people are blind everyday I will show the way to school and from school to home.
• I will see that the disabled people get the government help in a proper way.
• I will give my support to deaf people by giving symbols and signals to understand things.
• I will join the disabled people who are in the school age in the school mend for them.

IX. How do you appreciate the functions of sensory organs which helps us to enjoy the beauty of the nature?
Answer:

• Sense organs help us to enjoy the beauty of the nature.
• We enjoy the beauty of nature with our eyes, the melodious music with our ears, the taste of food with our tongue and feel the cool breeze on our skin.
• All these situations show just how our senses pick up information and react to them.
• Our sense organs are not just parts of us because nothing that we experience in our life, from the most important to the most boring, would be possible without the complicated power of our sense organs.
• Nothing in the entire universe of scientific exploration can even come close to match-ing the ability of our brain to use information sensed by our eyes, ears, skin, tongue and nose to produce a rich sensory experience in a matter of milli seconds.

X. Form a group with five students in your class and collect eye diseases and its char-acteristics by talking with ophthalmic assistant.
Answer:

Eye disease or defect	Characteristics
1. Age related macular degeneration	It is an eye condition that leads to the deterioration of the centre of the retina called macula.
2. Astigmatism	It is an imperfection in the curvature of retina.
3. Cataract	It is the clouding of the eye lens causing vision problems.
4. Central retinal vein occlusion	It is a blockage of the main vein in the retina.
5. Colour blindness	Occurs when we are unable to see colours in a normal way.
6. Conjunctivitis	It is the swelling of the conjunctiva, the eye becomes red, burning sensation in the eye, releases water.
7. Corneal transplant	Scars, swelling or an irregular shape can cause the cornea to scatter or distart light resulting in glare or blurring vision.
8. Diabetic retinopathy	It is a common diabetic eye disease caused by changes in retinal blood vessels.
9. Dry eye or Xeropthalmia	It is a condition where the eyes do not produce enough tears or the right quality of tears to be healthy or comfortable and eye becomes dry.
10. Far sightedness (Hypermetropia)	It is a refractive error, which means the eye does not bend or refract light properly. Images are formed behind the retina.
11. Glaucoma	It is a disease that damages the eyes optic nerve. This leads to high pressure in the back of the eye.
12. Kerolitis	It is a condition where the cornea be­comes swollen or inflamed, making the eye red and painful effecting vision.
13.Macular edema	It is a swelling or thickening of the macula, the area of the retina responsible for central vision.
14. Near sightedness (Myopia)	It is a refractive error, which means the eye does not bend or refract light properly. Image is formed in front of the retina.
15. Optic neuritis	It is an inflammation of the eye’s optic nerve.
16. Retinopathy of prematurity	Retinopathy of prematurity is an eye dis­ease that occur in a small percentage of premature babies where abnormal blood vessels grow on the retina.
17. Scleritis	It is a painful swelling of the white part of the eye, which is also known as sclera.
18. Detached retina and Torn retina	A torn retina is when the retina tears in one or more places. A detached retina is when the retina is lifted off the wall of the eye.
19. Night blindness	Person suffering from night blindness cannot see things in dimlight or at nights.
20. Trachoma	It is an eye infection affecting both eyes, is the world’s leading cause of prevent­able blindness. It is caused by a bacterium called Chlamydia Trachomatis.

XI. What happens if our skin loses its sensory nature?
Answer:

• The skin contains numerous sensory receptors which receive information from the outside environment.
• The sensory receptors of the skin are concerned with at least five different senses: pain, heat, cold, touch and pressure.
• The five are usually grouped together as the single sense of touch in the classification of the five senses of the whole human body.
• If the skin loss its sensory nature we cannot experience the pain, heat, cold, touch and pressure.

XII. Sagar is not able to listen things properly. Guess what would happen to him. What suggestions would you like to give him?
Answer:

• Sagar may be exposed to too much loud noise. This condition is noise induced hearing loss.
• Some times loud noise can cause a ringing, hissing or roaring sound in the ears called Tinnitus.
• Hearing problems may also be caused by a virus or bacteria.
• Hearing impairment happens when there is a problem with one or more partsflof the ear.
• So, I suggests Sagar to identify the reason for not listening things properly.
• I also advise him to consult a specialist called audiologist in ear problems.

9th Class Biology 6th Lesson Sense Organs InText Questions and Answers

9th Class Biology Textbook Page No. 76

Question 1.
Do you think our sense organs work together? Why, why not?
Answer:

• Yes, our sense organs work together.
• Every single function of the body is managed and controlled by the brain, including our organs and senses.
• Otherwise, we would have different interpretations of a stimulus, resulting confusion.
• But to be clear it is not the organs working together it is the brain constantly receiving stimuli from different senses.
• The brain is responsible for assimilating information and filling the pieces together.

9th Class Biology Textbook Page No. 85

Question 2.
If we do not have our external ear what will happen to us?
Answer:

• If we have no external ears, sound waves may not be collected by it.
• So we cannot hear anything and it leads to deafness.

9th Class Biology Textbook Page No. 87

Question 3.
If you are suffering from cold do you smell things in the natural way?
Answer:

• No, we cannot smell things in the natural way.
• When we have cold, we will notice foods seem tasteless because your nasal pas¬sages are blocked.

Question 4.
Do you find any relation between smell and taste?
Answer:

• Like smell, taste is also a sense based on identifying chemicals in food and the texture of it.
• The sense of taste and smell have a close and cooperative working relationship.

9th Class Biology Textbook Page No. 90

Question 5.
How sensitive is our skin?
Answer:

• Different parts of our body have different sensitivity and the skin around the neck and finger tips is more sensitive than skin on the palm, knee and arm.
• This is because sensitivity of our skin depends on
  a) thickness of our skin.
  b) the number of sensory receptors.

9th Class Biology Textbook Page No. 82

Question 6.
What will happen if we have no eyelashes?
Answer:

• The purpose of eyelashes is to keep moisture like sweat from setting into our eyes.
• Eyelashes protect the eye from debris and they are sensitive to being touched.
• If lashes are absent moisture like sweat will not be setting into our eyes.
• There will not be protection from debris to our eyes.

Question 7.
Are tears good for us?
Answer:

• Yes, tears are good for us.
• Whenever unwanted substances come in contact with the conjunctiva the lachry¬mal glands are stimulated to produce tears to wash the substances out of the eye.

9th Class Biology Textbook Page No. 89

Question 8.
Why are we suggested not to take too cool or too hot food material?
Answer:
To cool or too hot food material causes damage the sense of smell. It also damage the taste buds. If we take too cool or too hot food materials we will loose sense of smell and taste.

Question 9.
If you are suffering from fever that time you are not able to enjoy the taste of food why?
Answer:

• During fever, the temperature of our body increases from the normal body temperature of 98.6°F to high temperature sometimes to 105°F.
• At this temperature, the function of the enzymes in our tastebuds stop since they can work efficiently only in the temperature range of 77 to 98.6°F.
• The cells in the tastebuds cannot send messages to the nerve centres in the brain.
• That is why during fever we are not able to enjoy the taste of food.

9th Class Biology 6th Lesson Sense Organs Activities

Activity – 1

Question 1.
Note down a few lines of any text in your book. Write about the stimuli and re¬sponses and the sensory and motor functions with respect to the sense organs in¬volved.
Answer:
(Stimuli from the environment around are received by our body through some sense organs. As we already know, they are the eyes, ears, nose, tongue and skin. Let’s try to understand the path of receiving a stimulus to expressing a response (sensation))

1. The sense organs involved in writing the para above are eyes and skin.
2. Asking to write few lines in textbook is stimulus and writing the lines is response.
3. The sensory nerve in the eye take the information about writing lines to brain in the form of nerve impulses.
4. The brain interprets the signals and send the message through motor nerves to write the lines.
5. The skin in the palm helps to hold the pen in writing the lines.

Activity – 2

Question 2.
Testing tastes with tongue.
Answer:

• Dissolve a pinch of sugar in a glass of water.
• Drink a little of this. It doesn’t taste good.
• Try this for different concentrations of sugar adding by proper quantification that is weighing and preparing solutions to find out taste.
• You could take l/4th teaspoonful sugar each time which would be nearly 2 grams.
• After adding three spoons of sugar the taste of water becomes sugary.

Activity – 3

Question 3.
1. Observe the external structure of your friend’s eye, draw the diagram and lable it.
Answer:

2. Observe the eye ball of your friend in normal light. Then focus a beam of torch light on youj^friend’s eye.
Answer:

• Our eye contains eyelids, eyelashes, eyebrows and lachrymal glands.
• In normal light my friend’s eye ball is normal as usual, but after focusing a beam of torch light he immediately closed his eyes.
• After closing the eyes for two minutes, the black portion of eye is somewhat big in size.
• After opening forcibly when we throw the beam of torch light, the size of the small dark portion decrease.
• The small black portion in the centre of the eye expands in dark whereas in bright light it decreases in size.

Activity – 4

Question 4.
Testing the visual system.
Answer:

• Hold the text at arm’s length, close your right eye, and fix your left eye straight on the figure.
  
• Keep your right eye closed and bring the book slowly closer.
• When it is about 8 to 10 inches away the gap disappears as it is on the blind spot of your left eye.
• But you will not see a ‘hole’ in your visual field.
• Instead, your visual system “fills in” the missing area with information from the blue line on either side.

Activity – 5

Question 5.
Observe the iris and its surrounds of your friend’s eye. Can you find the pupil? Observe the colours and patterns in the iris of your friend’s eyes. Is there any difference from one another?
Answer:

• I found pupil in my friend’s eye.
• The colour of iris in my friend’s eyes are different.
• The colour of iris are blue in some, green in some and grey or brown in some of my friend’s eyes.
• The colour of iris are different but their shape is round in all my friend’s eyes.

Activity – 6

Question 6.
Enter into a dark room from a very bright place. What happens? Sit in a dark room for sometime. Then go into a bright light room. What happens?
Answer:

• If we enter into a dark room from a very bright place first we cannot see anything in the room.
• Because the pupil is very small in size and very less amount of light enters into the eyes.
• As the time progresses the pupil becomes large in size and we will be able to see things properly.
• If we go into a bright light room from a dark room, at first we cannot see anything in light from because the pupil does not bear the intensity of light.
• The pupil gradually lessens its size and we are able to see things properly.

Activity – 7

Question 7.
Eye and illusions.
Answer:

1. Take two pieces of white papers with same size.
2. Draw the picture of a cage on one pa¬per and the parrot on the other.
3. Then insert a stick and attach the blank sides of the papers with gum, see the figure.
4. Let it dry then twist the stick rapidly.
5. When we twist the stick rapidly, we see the parrot in the cage. What we are experi¬encing is an illusion.

Activity – 8

Question 8.
Testing of sound.
Answer:

• Take a plastic or iron funnel.
• Stretch a piece of rubber balloon and cover the wide part of the funnel with it.
• Tie it with rubber band.
• Ask your friend to shout ‘Oh’ at the narrow opening of the funnel.
• Observe the movements of the rubber sheet while he is shouting.
• Observe the rice grains also.

Observations:

1. Due to the vibrations in the rubber balloon the rice grains move up and down when we shout ‘Oh’ at the narrow opening of the funnel.
2. When we put the narrow end at the opening of our eye we hear the sound of heart as lub dub, lub dub …………….

Activity – 9

Question 9.
Touching test.
Answer:

• Blindfold your friend and ask him/her to identify different things by smell like lemon, tea, coffee, potato, tomato, tamarind, spinach, curd, brinjal etc.
• Keep as many things but be careful in choosing them.
• They should not be in powdered form.
• Don’t allow your friend to touch them.

Observations:

1. Biologically, the sense of smell or olfaction, begins with chemical events in the nose.
2. Their odours interact with receptor proteins associated with specialized nerve cells.
3. These cells incidentally are the body’s only nerve cells that come in direct contact with the outside environment.
4. Receptors present at the base of the skin lining the inner walls of the nose are highly sensitive to odour chemicals.
5. These odour chemicals can be complex and varied.

Activity -10

Question 10.
Tongue test
Answer:

• Close the eyes of your friend with a piece of cloth.
• Give her/him a piece of ginger, garlic, tamarind, banana and jaggery one by one.
• Ask her/him to taste by just taking these one at a time on the tongue.
• Remember that your friend needs to rinse his/her mouth between each test.
• Could your friend tell the taste by just putting the substances on the tongue? Yes, my friend told the taste.
• Now repeat the above experiment by asking your friend to take a bite and press the food on the plate.
• As food enters our mouth, we bite and chew it and press it against the palate with our tongue.
• This releases chemicals in food that trigger off our taste buds to act and carry stimulus to the brain to be processed for recognition of taste.
• The same taste bud is capable of producing different signals corresponding to the different chemicals in food.

Activity -11

Question 11.
Observe your tongue by standing in front of the mirror by sticking your tongue out. See how many different kinds of structures you can see on your tongue. Compare with the given diagram.
Answer:

• We can clearly see flake like structures that are filiform papillae.
• The roundish structures are fungi¬form papillae.
• There are large roundish ones at the back of the tongue which are circumvallate papillae.
• On the sides of the tongue, the bump like structures are foliate papillae.
• Taste buds are present on all of these except the filiform papillae that are not the sites of taste sensation.

Activity -12

Question 12.
Smell test
Answer:

• Blindfold your friend and ask him/her to close his or her nose as well.
• Give a few cumin seeds to your friend and ask him/her to chew.
• Ask your friend to identify what you have gave.
• You could try this with a small piece of potato as well.
• My friend has identified the cumin seeds and piece of potato.

Activity – 13

Question 13.
1) Make bundles of three toothpicks.
2) See to it that their pointed ends are at the same level.
3) Now ask your friend to make an outline of one of her/his palm.
4) Ask your friend to close her/his eyes. Now starting from the tip of the thumb keep pricking lightly with your toothpick bundle all over the plam.
5) Ask your friend how many points she/he could identify each time.
6) Repeat this with some of your friends.
Answer the following questions.
a) Where do you find maximum sensation on the palm?
Answer:
In the centre of the palm we find maximum sensation.

b) Where do you find minimum sensation?
Answer:
We find minimum sensation on the beginning of the palm.

c) Are palm sense patterns same for all your friends?
Answer:
Yes, palm sense patterns are same.

Activity – 14

Question 14.
Press your thumb gently on the tip of a sharpened pencil. Later press it on the blunt end of the pencil. How do you feel? Why?
Answer:

• When we press our thumb gently on the tip of a sharpened pencil it makes us feel pain.
• The sensory receptors sense the sharpened pencil’s press and the message is sent to brain through sensory nerve.
• Then the brain interprets the message and sends signals to feel pain through motor nerves.
• When we press our thumb with the blunt end of the pencil we do not feel pain because the sensory receptors (touch receptors) sends the message to brain through sensory nerve.
• Brain interprets the message and sends signals to our thumb to feel soft through motor nerves.

 

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 1 Cell its Structure and Functions Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 1st Lesson Cell its Structure and Functions

9th Class Biology 1st Lesson Cell its Structure and Functions Textbook Questions and Answers

Improve Your Learning

Question 1.
Differentiate between a) Plant cell and animal cell b) Prokaryotic and Eucaryotic cells. (AS 1)
Answer:
a)

Plant Cell	Animal Cell
1. Cell wall present.	1. Cell wall absent.
2. Chloroplasts present.	2. Chloroplasts absent.
3. Plant cell can perform photosynthesis.	3. Animal cell cannot perform photosynthesis,
4. Vacuoles are large in size.	4. Vacuoles are small in size.
5. Centrioles are absent. They appears only at the time of cell division.	5. Centrioles present.

b)

Prokaryotic cell	Eucaryotic cell
1. Nuclear membrane is absent.	1. Nuclear membrane is present.
2. The membrane bound cell organelles absent.	2. Cell organelles are enclosed by membranes.
3. Except ribosomes other organelles are absent.	3. All cell organelles are present.
4. They has a tough cell wall.	4. Flexible, porus cell wall present in plants, plasma membrane present in animals.
5. E.g. : Cynobacteria, blue green algae.	5. E.g. : All higher plants and animals.

Question 2.
What happens if plasma membrane ruptures or breaks? (AS 2)
Answer:

• Cell membrane or plasma membrane is covering of the animal cell.
• It separate cytoplasm from the external environment.
• It defined the shape and size of the cell.
• It plays a crucial role in maintaining a balance of various substances inside the cell.
• It controls the exchange of substances between the cell and its external environment.
• If it ruptured or broke, then the above activities will stop, the cell will die.

Question 3.
Prepare a model of plant cell or animal cell with locally available materials. (AS 5)
Answer:

Question 4.
What would happen to the life of cell if there was no golgi complex? (AS 2)
Answer:

• The golgi apparatus packed various substances before they are transported to other parts of the cell.
• If there was no golgi complex in the cell the proteins and other substances are not altered and packed.
• Substances transport will not occur.
• Regeneration or repair of the membrane will not takes place.

Question 5.
What happen to the cell if nucleus is removed? Give two reasons to support your answer. (AS 1)
Answer:

• If nucleus is removed from a cell, there would be no control on the functions of a cell.
• Cells are not involved in the process of cell division.
• The cell will not live for more time.
• E.g.: Red blood cells, not having nucleus live less time than the other cells, which are having nucleus.

Question 6.
Lysosomes are known as suicidal bags of the cell. Why? (AS 1)
Answer:

• Lysosomes contained the destructive enzymes.
• Thus the enzymes normally do not come in contact with the rest of the cell.
• The materials that need to be destroyed are transported to the lysosomes.
• At times, the lysosomes burst and the enzymes are released to digest the cell.
• Hence, lysosomes are known as suicidal bags of the cell.

Question 7.
Why do plant cell possess large sized vacuole? (AS 1)
Answer:

• Vacuoles are fluid filled sac-like structures.
• In a newly formed plant cell, the vacuoles are small.
• As the cell becomes old, these vacuoles, fuse to form a single large vacuole.
• In mature plant cells, they might occupy almost the entire cell space.

Question 8.
Prepare a temporary mount of any leaf peel observe the stomata draw their picture. Write a short note on the same. (AS 5)
Answer:

• A fresh leaf of Rheo is taken.
• Making a slit in the pith material and keep the leaf inside the slit.
• To get the T.S a leaf, section cutting with a blade should be done.
• The thin section with brush and keep the section on the slide.
• Putting a drop of water, glycer¬ine on it.
• Staining the section with saffronin.
• Cover the section with a cover slip.
• By observing under the microscope of the leaf. We can see stomata in the lower epidermis.
• They are enclosed by two kidney shaped cells, called guard cells.
• In between two guard cells a pore formed stomata.

Question 9.
“Cell is the basic unit of life” – explain the statement. (AS1)
Answer:

• The fundamental organizational unit of life is the cell.
• All living organisms are composed of cells.
• In unicellular organisms, a single cell performs all the the functions.
• In multicellular organisms, a no. of cells together performs different functions.
• So, we can say that “Cell is the basic unit of life”.

Question 10.
How do you appreciate about the organisation of cell in the living body? (AS 6)
Answer:

• Cell is the basic unit in the structural organisation of all living organisms.
• Cell carry physiological functions like oxidise food materials to derive energy.
• Excrete the waste materials.
• Increase in number by dividing into two identical cells.
• Defend itself against the attack of foreign organisms.
• Try to adjust to the conditions in its surroundings.
• Function of an organism depends on the functions carried out by the cell.

Question 11.
If the organisation of cell is destroyed due to physical and chemical influence, what will happen? (AS 6)
Answer:

• If the organisation of cell is destroyed due to physical and chemical influence, the cell will die.
• Sometimes it also effects the functions of other cells nearby.

Question 12.
Read the chapter carefully collect the information about the functions of different cell organelles and make a table which contains serial number. Cell organelle, function. Don’t forget to write your specific findings below the table. (AS 4)
Answer:

Cell organelle	Functions
1. Nucleus	1. Regulates and controls all the functions of the cell. 2. Barrier of genetic information. 3. Determines the characteristics of the organism. 4. Cell division.
2. E.R	1. Transport of substances. 2. RER are the sites of protein manufactures. 3. SER helps in the manufactures of lipids.
3. Golgi Apparatus	1. Packing of various substances in the cell. 2. Secretion of proteins from the cell.
4. Lysosomes	1. Digestion of food materials 2. At the time of disease condition it digest the cell also.
5. Mitochondria	1. Generates and stores the energy.
6. Plastids	1. Chloroplasts trap the energy of sunlight during photosynthesis. 2. Chromoplasts are responsible for the colouring of fruits and flowers.
7. Vacuole	1. Storing of carbohydrates, amino acids, proteins, pigments and waste materials.

Question 13.
How could you appreciate the function of tiny cell in a large body of an organism? (AS 6)
Answer:

• Cell is the basic unit in the structural organisation of ail living organisms.
• It is the functional and structural unit of the organism.
• Functions essential for survival of the organism are carried out at the level of a cell only.
• Each cell acts as an individual unit.
• In each cell excretion, generation of energy, defending itself, adjust to the conditions, production of new cells etc. functions are carried out.
• So we must appreciate the function of a tiny cell in a large body of an organism.

Question 14.
Look at the following cartoon of a cell. Find out the functions of cell organelles. (AS 5)

Answer:

Cell organelle	Function
Nucleus	Nucleus regulates and controls all the functions of a cell and determines the characteristics of the organism.
Endoplasmic reticulum	1. It serve as channels for the transport of materials within the cell.
	2. It also functions as a cytoplasmic framework for various biochemical activities.
Golgi Apparatus	It package various substances. Proteins are altered slightly by golgi apparatus.
Lysosomes	It participates in intracellular digestion. It destroys the cell contents.
Mitochondria	It produces energy through cellular respiration.
Plastids	These are responsible for the colour of the plant cell.
A. Chloroplasts	These trap solar energy and convert this to chemical energy during photosynthesis.
B. Chromoplasts	These are responsible for the coloured fruits, flowers.
C. Leucoplasts	These are colourless, stores carbohydrates, oils and proteins.

Question 15.
Who and when was “The cell theory” proposed? When did they prepare it? What are its salient features? (AS 1)
Answer:
M.J. Schleiden and Theodar Schwann proposed “The cell theory”. They prepared it in 1838 – 39.

Statements of modern form of cell theory :

1. All the living organisms are made up of cells and their products.
2. All the cells are formed from pre-existing cells.
3. All the cells are made up of similar chemicals and show similar metabolic activities.
4. Functioning of an organism depends on the functions carried out and the interac-tion of different cells present in the organism.

Question 16.
When you observing the nucleus of cheek cell in laboratory, what precautions do you take?
Answer:
While observing the nucleus of cheek cell in laboratory the following precautions are to be taken.

Precautions:

1. Do not scrap the cheek too hard as it may injure the buccal mucosa.
2. Scrapped material should be spread uniformly on the slide.
3. Excess stains should be drained off.
4. There should be no air bubbles under the coverslip.

Question 17.
Draw the typical animal cell and lable its parts.
Answer:

9th Class Biology 1st Lesson Cell its Structure and Functions InText Questions and Answers

9th Class Biology Textbook Page No. 1& 2

Question 1.
Observe the following figures.

a) What common features do you see in both the cells?
Answer:
We can observe some common features in plant and animal cells. They both are having plasma membrane, mitochondria, golgi apparatus, endoplasmic reticulum, nucleus etc.

b) Which cell organelles are found exclusively in plant cell?
Answer:
Chloroplasts and big vacuoles are the cell organelles exclusively found in plant cell.

9th Class Biology Textbook Page No. 3

Question 2.
What is the role of the cell wall in plant cells?
Answer:
It exerts an inward wall pressure to resist the outward pressure exerted by the cell sap.

9th Class Biology 1st Lesson Cell its Structure and Functions Activities

Activity – 1

Question 1.
How do you observe cell membrane in a peel of Rheo leaf under microscope? Draw the diagram of it. Write your observations.
Answer:
Take Rheo leaf, tear the leaf in single stroke take a small piece of leaf peel with light coloured (transparent) portion. Put it on slide and put a drop of water on it. Cover it with cover slip and observe the light portion of leaf under the microscope.

Observations:

1. Cells are arranged in rows.
2. Cell membrane is clearly seen.
3. Nucleus is present in the cell.

Lab Activity

Question 2.
To observe the nucleus in cheek cells.
Answer:
Aim :
To observe the nucleus in cheek cells.

Material:
A tooth pick or ice-cream spoon or spatula, glass slide, coverslip, watch glass, needle, blotting paper, 1% methylene blue, normal saline, glycerine, microscope etc.

Procedure:

1. Wash your mouth and scrap a little of the internal living of your cheek inside your mouth with a clean tooth pick or spatula or ice-cream spoon.
2. Place the scrap in a watch glass containing a very small quantity of normal saline.
3. Then place the material on a glass slide.
4. Put a drop of methylene blue and wait for a couple of minutes.
5. Wipe off the extra stain with a fine cloth of blotting paper.
6. Put a drop of glycerine over it.
7. Place a coverslip. Tap the coverslip with the blunt end of needle so as to spread the cells.

Precautions :

1. Do not scrap the cheek too hard as it may injure the buccal mucosa.
2. Scrapped material should be spread uni¬formly on the slide.
3. Excess stains should be drained off.
4. There should be no air bubbles under the coverslip.

Observations :

1. The shape of the cells are circular in shape.
2. These cells are not similar to the structure in onion peel cell.
3. Near the centre of the cell there is a darkly coloured oval dot like structure present.

Activity – 2

Question 3.
How do you observe mitochondria in onion peel ? Observe and make a sketch of mitochondria.
Answer:
Observing mitochondria :

1. Make a fresh solution of Janus Green-B in a beaker.
2. Mix 200 mg Janus Green-B in 100 ml of water.
3. Take a watch glass pour some solution. Put the onion peel in this solution and keep it about half an hour.
4. Keep a piece of onion peel on the slide and wash thoroughly with water. Mitochondria in onion peel ceil
5. Cover the slide with a cover slip and observe it under microscope at high magnification.

Observations :
Green oval or cylindrical grains scattered in the cytoplasm. They are mitochondria.

Activity – 3

Question 4.
Observe a chSoroplast in Rheo leaf under microscope ? Draw the diagram of it and write your observation.
Answer:
Observing chloroplast:

1. Take the peel of Rheo leaf and mount it in water on a slide.
2. Observe it under high power microscope.

Observations :

1. Small green granules called chloroplasts are present in the cells of Rheo leaf.
2. Chloroplasts mainly contain green substance called chlorophyll.

Activity – 4

Question 5.
How do you observe chloroplast in Algae under microscope? Draw the diagram and write your findings.
Answer:
Observing chloroplast:

1. Collect some algae from pond and separate out thin filaments of them.
2. Place a few filaments on slide. Observe it under microscope.

Observations:

1. In algae the chloroplasts are found as ladders, stars, spirals or reticulate.
2. The primary function of chloroplasts is to trap the energy of sunlight and transform it to chemical energy in photosynthesis.

Activity – 5

Question 6.
How do you observe under microscope the vacuoles of succulent plant like cactus?
Write small note on them.
Answer:
Observing vacuoles :

1. Take the leaf or stem of any succulent plant like cactus.
2. Take thin cross section of stem of cactus in a watch glass containing water.
3. Stain it with dilute safranine solution.
4. Observe it under low and high power microscope.

Observations :

1. The large empty spaces visible in the cell are vacuoles.
2. These are fluid filled sac like structures.

 

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 3 Animal Tissues Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 3rd Lesson Animal Tissues

9th Class Biology 3rd Lesson Animal Tissues Textbook Questions and Answers

Improve Your Learning

Question 1.
What do you understand by the term tissue? (AS 1)
Answer:
Tissue is a group of cells similar in structure and function.
Eg : Nerve tissue, epithelial tissue, muscle tissue etc.

Question 2.
Show the difference between the three types of muscle fibres with diagrams. (AS 1)
Answer:

Question 3.
What is the specific function of the cardiac muscle? (AS 1)
Answer:
Specific function of the cardiac muscle :

1. Cardiac muscle present in the heart.
2. It is responsible for pumping of blood.

Question 4.
Differentiate between striated and unstriated muscles on the basis of their shape and site/location in the body. (AS 1)
Answer:

Striated muscle	Unstriated muscle
Shape : Cells in striated muscle are long cylindrical and unbranched.	Cells in unstriated muscle are long with pointed ends.
Site / Location : These are located in limbs and attached to skeleton.	These are located in Alimentary canal, blood vessels, Iris of the eye, in uterus and in the bronchi of lungs.

Question 5.
Draw a neatly labelled diagram of a neuron. (AS 3)
Answer:

Question 6.
Name the following. (AS 1)
a) Tissue that forms the inner lining of our mouth.
b) Tissue that connects muscle to bone in humans.
c) Tissue that transports food in animals.
d) Tissue that stores fat in our body.
e) Connective tissue present in the brain.
Answer:
a) Epithelial cells
b) Tendon
c) Connective tissue/blood
d) Adipose tissue
e) Areolar tissue

Question 7.
Identify the types of tissue in the following : Linings of the organs, skin, bone, internal lining of kidney tubule. (AS 1)
Answer:

Linings of the organs	Epithelial tissue
Skin	Stratified squamous epithelium(epithelial tissue)
Bone	Connective tissue
Internal lining of Kidney tubule	Cuboidal epithelial tissue.

Question 8.
If the platelets are not present in the blood, what happens? (AS 2)
Answer:

• If the platelets are not present in the blood, blood loss may be more from the injury.
• Whenever a blood vessel is injured, at the site of injury formation of a blood clot will not takes place.
• The wound will not be sealed by the clot.

Question 9.
If you touch at elbow, you get a shock like feeling. Why? (AS 7)
Answer:

• In human beings ulnar nerve runs from the shoulder to the hand.
• The ulnar nerve comes close to the surface near the elbow.
• Due to the superficial location it is not protected by muscle, fat or other soft tissues.
• Thinner skin layer around bone at elbow makes ulnar nerve more receptive for any small stimuli.
• That is the reason for getting a shock like feeling if we touch at elbow.

Question 10.
Why the blood is called a connective tissue?
Answer:
Connective tissue :
A loosely spaced tissue mainly carrying different materials to different parts of the body as well as rendering support, making connection between organs is called connective tissue.

Blood is considered as connective tissue because of the following reasons.

1. Blood connects different organs of our body together by carrying oxygen, nutrients, hormones and other signaling molecules and removing the waste.
2. It has all the three components of connective tissue i.e., cells, fibers and matrix.
3. Similar to other connective tissues, blood is rich in fibres like collagen fibers and blood clotting fibres.
4. Blood originates from the mesodermal layer of the embryo from which all other connective tissues also originate.

Question 11.
Write the procedure to identify your blood group with the help of kit. (As 3)
Answer:
Aim :
Identification of blood group.

Apparatus :
Blood identification kit, glass slide, wax pencil, disposable needle.

Materials used :
Cotton, 70/6 alcohol, toothpicks.

Procedure:

1. Take one porcelain plate, clean and dry it.
2. With a wax pencil, draw three circles on the plate to divide the surface into three parts and draw three circles.
3. Place one drop of the corresponding antiserum near the edge of the circles.
4. Clean the fingertip with an alcohol and let it dry.
5. Press on the bottom of the fingertip with the thumb and quickly prick the fingertip with the help, of a needle.
6. Quickly, let one drop of blood get into each circle, but not touching the anti-serum.
7. Apply gently pressure to the wound with cotton ball.
8. Use a toothpick to mix the blood and anti-serum and stir gently.
9. Watch to see if any of the samples show agglutination.

Result and Inference :
By using the following table determine the blood group.

Anti – A	Anti – B	Type
Yes	No	A
No	Yes	B
Yes	Yes	AB
No	No	O

If agglutination occurs in anti – RhD serum, the Rh factor is positive, and if it does not . the Rh factor is negative.

Note :
1. See the needle is sterile.
2. Usually choose left ring finger.
3. Don’t use same needle to other body.

Question 12.
Ramu felt weak. Ramu’s father took him to hospital. The doctor advised a blood test. The report says that he does not have the required levels of haemoglobin. What are its ill effects?
Answer:
Ill effects of haemoglobin :

• Blood is red in colour due to the presence of red coloured protein called haemoglobin.
• Haemoglobin helps in the transport of oxygen and carbon dioxide.
• Low haemoglobin is the main cause for anemia.
• If there is not enough haemoglobin in blood. The oxygen supply to various parts will be less, which causes shortness of breath.
• Low haemoglobin levels many aggravate extant heart problems.
• People with low haemoglobin levels get very tired as their cells do not get enough oxygen to perform their activities.

Question 13.
How blood test is useful to diagnose the disease? Explain with daily life situation. (AS 1)
Answer:

• Blood test is useful to diagnose diseases such as malaria, typhoid, cancer, HIV/AIDS, diabetes, anemia coronary heart disease, abnormalities in the functioning of kidney, liver, thyroid, etc.
• Abnormal red blood cell levels might be a sign of anemia. Dehydration, bleeding, and other disorder.
• Complete blood count with differential can measure the amounts of different types of white blood cells in our body.
• Abnormal white blood cell levels might be a sign of infection, blood cancer or an immune system disorder.
• Abnormal platelet levels might be a sign of a bleeding disorder or thrombotic disorder.
• Abnormal haemoglobin levels might be a sign of anemia, sickle cell anemia or thalassemia.
• Abnormal glucose levels in the blood might be a sign of diabetes.
• Abnormal calcium levels in the blood might suggest kidney problems, bone disease, thyroid disease, cancer, or malnutrition.
• Abnormal electrolyte levels might be a sign of dehydration, kidney disease, liver disease, heart failure or high B.P.
• Abnormal Blood Urea Nitrogen (BUN) and creatinine levels might suggest a kidney disease.
• High levels of enzymes like Troponin and creatine kinase is a sign of Heart attack.
• Abnormal cholestrol or triglyceride levels might be a sign of increased risk of coronary heart disease.
• Abnormal coagulation pannel test results might suggest risk of bleeding or developing clots in blood vessels.
• Existence of microorganisms or their antibodies in the blood suggest occurence of corresponding disease.
  E.g. : Plasmodium – Malaria, HIV – AIDS etc.

Question 14.
Collect the old blood reports of your friends/relatives and prepare a project report on the contents of the blood.
Answer:
On collection and observation of old blood reports I came to know that the contents of blood should present in definite proportions such as.

Content of blood	Lower and upper limits
WBC	5.0 – 10.0 103 cells / ul
RBC	3.5 -5.5  106 cells/ul
HgB	Men 12 -16 g/dL; Women 9.9 – 13 g/dL
PLT (Platelet count)	1.0-3.0 105 cells/ ul
Neutrophil	40 – 75%
Lymphocytes	20 – 45%
Eosinophil	1 – 6%
Basophil	0-1%
Monocyte	0-3%

9th Class Biology 3rd Lesson Animal Tissues InText Questions and Answers

9th Class Biology Textbook Page No. 29

Question 1.
Why do old people shiver in winter when compared to youngsters? Is there any insulator like substance to prevent the escape of heat energy during winter?
Answer:

• Old people shivers in winter when compared to youngsters.
• They didn’t have enough fat storages below the skin.
• Fat storing adipose tissue is found below the skin and between internal organs.
• The cells of this tissue are filled with fat globules.
• Storage of fat also acts as insulator.

Question 2.
Which tissue gives definite shape to body of vertebrae?
Answer:

• Bone is one type of connective tissue.
• It forms the frame work that supports the body.
• It is a major component of the skeletal system of several vertebrae.

9th Class Biology Textbook Page No. 34

Question 3.
During winter, body shivers. Why?
Answer:

• When the body is exposed to cold air, we shiver.
• During shivering muscles contract and relax and produce large amount of heat.
• This keeps the body heat.
• It is one type of defensive mechanism of the body.

9th Class Biology Textbook Page No. 30

Question 4.
Blood is a type of connective tissue. Why is it called connective tissue?
Answer:
Blood is considered as connective tissue because of the following reasons.

1. Blood connects different organs of our body together by carrying oxygen, nutrients, hormones, and other signaling molecules and removing the waste.
2. It has all the three components of connective tissue i.e., cells, fibers, and matrix.
3. Similar to other connective tissues, blood is rich in fibres like collagen fibers and blood clotting fibres.
4. Blood originates from the mesodermal layer of the embryo from which ail other connective tissues also originate.

9th Class Biology 3rd Lesson Animal Tissues Activities

Lab Activity – 1

Question 1.
Aim:
Identification of tissue in collected sample.

Apparatus:
Microscope, slide, dilute HCl, forceps, brush.

Procedure:

1. Collect a small piece of chicken with bone from your nearby chicken centres or market.
2. Put it in dilute HCl for two hours.
3. Take the skin part of chicken piece.
4. Place the material with forceps or brush on the slide
5. Then keep the another slide on it and press both the slides gently.
6. Place a cover slip tap on it and observe under microscope.
7. Draw the diagram of what you have observed under microscope in your notebook.
8. Compare your diagram with the given picture.

Answer these questions.
1. Are all the cells similar?
Answer:
Yes. All the cells are similar.

2. How are they arranged?
Answer:
They are arranged in layers. Each cell is round and nucleated. Observed diagram

3. Are these cells tightly packed and formed as continuous sheath?
Answer:
Yes. The cells are tightly packed and formed as continuous sheath.

4. Is there any intercellular space?
Answer:
No. There is no intercellular space.

5. Think, why these cells look like continuous sheath.
Answer:
These cells are look like continuous sheath because there is no intercellular space and the cells are tightly packed.

6. Does this tissue covering protect inside and outside of the animal body?
Answer:
Yes. This tissue covering protect inside and outside of the animal body.

Question 2.
Aim:
Identification of tissue in collected sample.

Apparatus:
Microscope, slide, blood sample, syringe, cotton.

Procedure:

1. Take a sterilized syringe needle.
2. Collect one drop of blood from finger tip by pricking with syringe needle.
3. Take a slide. Keep the finger on the slide to collect one drop of blood.
4. Put another slide on it gently and press both :
5. Observe under microscope.
6. Draw the diagram of what you observe L microscope in your notebook. Compare diagram with the given picture.
   In this procedure we can identify red blood

Question 3.
Aim:
Identification of tissue in collected sample

Apparatus:
Microscope, slide, dilute HCl, vinegar, forceps.

Procedure:

1. Take a piece of muscle of chicken.
2. Put in diluted HCl or vinegar and leave it for two hours.
3. Next morning collect the piece of muscle on a slide with forceps.
4. Press gently with another slide, put few drops of water and place a cover slip on it.
5. Observe under microscope. Observed diagram
6. Draw the diagram what you have observed under microscope in your notebook. Compare your diagram with the above picture.

Answer these questions.

1. How are the cells arranged?
Answer:
Cells are arranged in layers one above the other.

2. Do you find any difference between skin cells and muscle cells?
Answer:
Muscle cells are long and nucleated.

3. If you want to observe the bone tissue in the chicken bone, settle it in vinegar or diluted HCl over night. Then only the bone becomes soft. Take a piece from it by using knife. Do you find any relation among these tissues?
Answer:
Usually muscle tissue is attached to bones.

4. Is this tissue useful for movements in our body?
Answer:
Yes. This tissue is useful for movements in our body.

Activity – 1

Question 4.
1. Collect the substance lining of mouth by using wooden spoon and observe this under microscope.
2. Draw the diagram that you observed in the microscope, in your notebook.
a) How are the cells arranged?
Answer:
Cells are extremely thin and flat and form a delicate lining.

b) Are there any intercellular spaces?
Answer:
No. Intercellular spaces are absent.

c) Think, why are the epithelial cells in skin are arranged in the form of layers?
Answer:

Because skin has to protect our body from cold, heat etc.

d) If you drink hot tea or chilled cool drink, how would you feel?
Answer:
Inner layers of our mouth cannot bear hot tea or chilled cool drink. We immediately spill hot or cold substances from our mouth.

e) If your skin burns or wounded, which tissue would effected ?
Answer:
Epithelial tissue.

Activity – 2

Question 5.
1. Take a permanent slide of cuboidal epithelium from your laboratory slide box and observe under microscope.
2. Draw the picture in your notebook.
3. How are the cells arranged?
Answer:
The cells are compactly arranged without intercellular spaces.

4. Conclusion :
These are the cuboidal epithelial cells which form the lining of organs or tubules or other parts and provide mainly mechanical support.

Activity – 3

Question 6.
1. Take a permanent slide of columnar epithelium from the slide box and observe under microscope.
2. Draw the figure that you observed under microscope. Observed diagram
3. How are the cells? Do you find any hair like projections on the outer surface of epithelial cells?

Answer:
a) The cells are long, compactly arranged without intercellular spaces.
b) Hair like projections are present on the outer surface of these cells.
c) These type of epithelial cells are present in the small intestine.

Activity – 4

Question 7.
1. Invite a scientist or doctor to your place.
2. Record an interview about blood structure and its functions.
3. It is important to make a questionnaire in order to conduct an interview.
4. After completion of interview, prepare a booklet about blood and display it on bulletin board or classroom library.

Booklet about blood.

1. Blood is a fluid connective tissue.
2. There are different types of cells in blood and each one has a different function.
3. All the cells in the blood cells float freely in the plasma.
4. Extracellular space is filled with fluid called plasma. There are no fibres in the blood.
5. Normal adult human beings have about 5 litres of blood. A chief component in plasma is water.
6. Besides water it also has several nutrients such as glucose, aminoacids, proteins, vitamins and hormones.
7. Plasma also contain factors responsible for blood clotting. Heparine helps to prevent blood clotting in blood vessels.
8. Cells present in blood are corpuscles. They are three types l.RBC, 2. WBC, 3. Blood platelets.
9. Red blood cells are also known as erythrocytes. They are red in colour due to the presence of haemoglobin.
10. haemoglobin helps in transport of oxygen and carbon dioxide.
11. When we are in mother’s womb, RBC are formed in the liver and spleen. After birth RBC are generated from the bone marrow of long bone.
12. RBC live for 120 days.
13. The second type of cells present in blood are white blood cells, which do not have haemoglobin. Hence they are called leucocytes.
14. There are two types of cells in WBC – granulocytes and agranulocytes.
15. There are three types of cells in the granulocytes – Neutrophils, Basophils and Esinophils.
16. These cells attack and destroy the microorganisms that enter the blood.
17. There are two types of agranulocytes – lymphocytes and monocytes.
18. Lymphocytes secret anti – bodies to guard against foreign material that enter into blood. So they are called microscopic policemen.
19. Monocytes move like amoeba and along with granulocytes. The foreign materials are destroyed inside these cells. They are called as ‘scavengers’.
20. Blood platelets are a separate group of cells which do not have a nucleus. They help in blood clotting.

Lab Activity – 2

Question 8.
Aim:
Identification of blood group.

Apparatus:
Blood identification kit, glass slide, wax pencil, disposable needle, cotton, tooth picks, 70% alcohol.

Kit components:

Components	Quantity (100 tests)
1. anti-A sera	5 ml
2. anti-B sera	5 ml
3. anti-RhD sera	5 ml
4. porcelaine white plate	2
5. wax pencil	1
6. needle (24G)	100
7. instructional mannual	1

Procedure:

1. Take one porcelain plate, clean and dry it. The plate must be very clean so that it does not interfere with the reaction.
2. With a wax pencil, draw three circles on the plate to divide the surface into three parts and draw three circles, one in each part as shown in figure.
3. Place one drop of the corresponding antiserum near the edge but within each of the circles as shown in figure.
4. Choose a left ring finger clean it with alcohol in a cotton ball and let it dry. Keep the cotton ball nearby, as it is needed again. Dangle the hand down to increase the amount of blood in the fingers.
5. Press on the bottom of the finger tip with the thumb of the same hand and quickly prick the finger tip with the help of a needle.
6. Quickly, let one drop of blood get into each circle but not touching the anti-sera.
7. After putting three drops of blood, apply gentle pressure to the wound with cotton ball.
8. Use a toothpick to mix the blood and antiserum and stir gently. Do it for each of the circles using a fresh toothpick every time.
9. Watch to see if any of the samples show agglutination.

Result and inference :
Determine the blood type depending on the result. Following table can be used to determine the blood type :

Anti – A	Anti – B	Type
Yes	No	A
No	Yes	B
Yes	Yes	AB
No	No	0

If agglutination occurs in anti-RhD serum, the Rh factor is positive; and if it does not, the Rh factor is negative.

Result should be noted in the given table :

Name	Blood group
Ramu	O
Gopal	B
Krishna	AB
Apparao	A
Gupta	B

Activity – 5

Question 9.
Collect three types of muscle slides (Striated muscles, Non-striated muscles, Cardiac muscles) from slide box. Then observe these under microscope. Write your findings in the following table.

Activity – 6

Question 10.
Collect the slide of nerve cells from the slide box. Observe it under microscope. Write your findings.
Answer:

1. We can identify three distinct parts in nerve cells.
2. They are
   1. Cell body or cyton,
   2. Axon and
   3. Dendrites
3. Cell body or cyton has a large nucleus and cytoplasm. The cytoplasm contains granular structure called Nissal’s granules.
4. Projections arising from the cell body are called dendrites. They are sharp, branched, more in number.
5. One projection of the cyton is somewhat longer than remaining projections. This is called axon.
6. Nerve cell is covered with myeline sheath. Nodes of Ranvier are present in myelin sheath.

 

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 2 Plant Tissues Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 2nd Lesson Plant Tissues

9th Class Biology 2nd Lesson Plant Tissues Textbook Questions and Answers

Improve Your Learning

Question 1.
Define the terms (AS 1)
A) Tissue
B) Meristematic tissue
C) Dermal tissue.
Answer:
A) Tissue :
Tissue is a group of cells similar in structure and performing similar functions.

B) Meristematic tissue :
Meristematic tissue is the dividing tissue present in the growing regions of the plant.

C) Dermal tissue :
Tissues that form outer coverings are called as dermal tissues. It gives protection to the plant.

Question 2.
Differentiate the following. (AS 1)
A) Meristematic tissue and Ground tissue :

Meristematic tissue	Ground tissue
1. Cells divide continuously.	1. Cells do not divide and cannot derived from the meristem.
2. It is a simple tissue.	2. It is a simple or complex tissue.
3. Cells are living.	3. Cells may be living or dead.
4. Dense cytoplasm is present in the cell.	4. Thin cytoplasm is present in the cell.

B) Apical Meristem and Lateral Meristem :

Apical Meristem	Lateral Meristem
1. Apical meristem is found in growing tips of root stem and apices of leaves.	1. Lateral meristems are present around the edges in a lateral manner.
2. It brings about growth in length of stems and roots.	2. It rise the growth in diameter of the stem and root.

C) Parenchyma and Collenchyma :

Parenchyma	Collenchyma
1. The cells of parenchyma are soft, thin walled and loosely packed.	1. The cells of collenchyma are thick walled and compactly arranged.
2. They store food, bears chloroplasts and contain larger cavities.	2. They give mechanical strength in young tissue of stem.
3. Cell wall is primary made up of cellulose.	3. Cell wall has deposition of extra cellulose.
4. Cells are oval, round and rectangular in shape.	4. Cells are elongated, round and spherical in shape.

D) Sclerenchyma and Parenchyma :

Sclerenchyma	Parenchyma
1. It is a dead tissue.	1. It is a living tissue.
2. Cells are thick walled.	2. Cells are thin walled.
3. Inter cellular spaces are absent.	3. Inter cellular spaces are present.
4. It provides mechanical strength.	4. It stores food, bears, air cavities and chloroplasts.

E) Xylem and Phloem :

Xylem	Phloem
1. Xylem conducts water and mineral salts from the roots to the leaves.	1. Phloem conducts food from the leaves to the growing parts of plants.
2. Xylem consists of trachieds, vessels, fibres and parenchyma.	2. Phloem consists of sieve cells, sieve tubes, companion cells, fibres and parenchyma.
3. Only xylem parenchyma is living.	3. Only phloem fibres are nondiving.

F) Epidermis and Bark :

Epidermis	Bark
1. It is the outermost layer of roots,	1. Several layered bark tissue is present
stems and leaves.	above the epidermis.
2. Epidermis is a living tissues.	2. Bark is a dead tissue.

Question 3.
Name the following. (AS 1)
A) Growing tissue, which causes growth in the length of the plant.
Answer:
Apical Meristem

B) Growing tissue, which causes growth in the girth (diameter) of the plant.
Answer:
Lateral Meristem.

C) Large air cavities in the aquatic plants.
Answer:
Arenchyma.

D) Food material in parenchyma.
Answer:
Storage tissue

E) Pores essential for gaseous exchange and transpiration.
Answer:
Stomata

Question 4.
Compare and contrast the following. Xylem and Phloem. (AS 1)
Answer:

Xylem	Phloem
1. Xylem transports water and minerals from roots to the apical parts of the plant.	It transports food material from the leaves to growing parts of the plant.
2. Xylem consists of trachieds, vessels, xylem fibres and xylem parenchyma.	Phloem consists of sieve tubes, sieve cells, companion cells, phloem fibres and parenchyma.
3. Only xylem parenchyma is living.	Sieve tubes, sieve cells, companion cells and phloem parenchyma are living.
4. Trachieds, vessels, xylem fibres are dead tissues.	Phloem fibres are dead tissue.
5. Xylem gives mechanical strength to the plant.	Phloem does not give mechanical strength to the plants.
6. Conduction of water by xylem is unidirectional i.e., from roots to apical parts of the plant.	Food material conduction is bidire­ctional i.e., from leaves to storage organs or growing points or from storage organs to growing parts of plants.

Meristematic tissue and Dermal tissue :
Answer:

Meristematic tissue	Dermal tissue
1. Cells are small having thin cell wall.	1. The walls of the cells are thicker.
2. They are capable of dividing.	2. They are not capable of dividing.
3. This is present at shoot tip, root tip and where branches arise.	3. It is present as epidermis, mesodermis and endodermis.
4. It helps in the growth of the plant.	4. It protects the plant from water loss due to transpiration.

Question 5.
Give reasons to the following. (AS 1)
A) Xylem is a conductive tissue.
Answer:

1. Xylem conducts water and mineral salts from the soil to the apical parts of the plants.
2. It transports materials away from the root.
3. Conduction of water by xylem is unidirectional i.e., from roots to apical parts of the plants.

B) Epidermis gives protection.
Answer:

1. Epidermis usually consists of a single layer of cells.
2. The walls of the cells of epidermis are thicker.
3. The epidermis protects the plants from loss of water, mechanical damage, and invasion by parasitic and disease causing organisms.

Question 6.
“Bark cells are impervious to gases and water”. What experiment will you perform to prove this? (AS 3)
Answer:

• In big trees dermal tissue forms several layers above the epidermis. It is called Bark.
• The several layers of bark does not allow any things like gases and water to pass through it.
• Hence bark cells are impervious to gases and water.

Question 7.
Though Chlorenchyma, Arenchyma, and storage tissues are parenchyma in nature, why do they have different (specific) names? (AS 1)
Answer:

• Chlorenchyma, Arenchyma and storage tissue are parenchymatous tissues,
• These three parenchymatous tissues are modified to perform various functions.
• The parenchyma which contains chloroplasts is called chlorenchyma. It performs photosynthesis.
• The parenchyma which contain large air cavities or spaces is called Arenchyma. It helps the plant to float.
• The parenchyma which stores water or food or waste products is called storage tissue.

Question 8.
Draw and label the diagram of the T.S of stem. (AS 5)
Answer:

Question 9.
Describe the functions of Meristem, Xylem and Phloem. (AS 1)
Answer:
1. Functions of Meristem:

1. It is a dividing meristematic tissue. It divides continuously. The cells formed from meristems later they differentiated as components of other tissues.
2. It brings about overall growth and repair.

2. Functions of xylem :

1. It conducts water and mineral salts from the root to apical parts of the plant like stems and leaves.
2. It gives mechanical support to the plant.

3. Functions of phloem :
Phloem conducts food material from the photosynthetic parts of the plants to other parts.

Question 10.
While observing internal parts of plants, how do you feel about its structure and functions? (AS 6)
Answer:

• While observing the internal parts of plants I felt that there are different types of tissues to perform various functions.
• For examples xylem and phloem of stem and stomata present on the outer layers of leaf are mend for performing different functions.
• Hence 1 felt that cells are organised from tissues and tissues are organised to perform various functions making the plant alive.

Question 11.
If you want to know more about tissues in plants, what questions you are going to ask? (AS 2)
Answer:
i) Which plant tissue provides both mechanical strength and flexibility?
ii) Which structure protects the plant body against the invasion of disease causing organisms?
iii) What will happen if apical meristem is destroyed or cut?
iv) What is the tissue present in the husk of coconut?
v) Why plants need different types of tissues?

Question 12.
Collect information about dermal tissues of plants in what way they help to them? Display it on wall magazine. (AS 4)
Answer:

• Dermal tissue usually consists of a single layer of tissues showing variations in the types of cells.
• On the basis of their location and function dermal tissues are studied as three different types epidermis (outer layer), mesodermis (middle layer) and endodermis (inner- most layer).
• Small pores are seen in the epidermis of the leaf called stomata.
• Cells of the roots have long hair like parts, called root hairs.
• Gum is secreted from the dermal layer of gum tree.
• The dermal layer protects the plants from loss of water, mechanical damage and invasion by parasitic and disease causing organisms.

9th Class Biology 2nd Lesson Plant Tissues Activities

Activity – 1

Question 1.
Parts of the plants and their functions.
Fill in the table.

Function	Name of the parts
Absorption of water from soil
Exchange of gases
Photosynthesis
Transpiration
Reproduction

Answer:

Function	Name of the parts
Absorption of water from soil	Roots
Exchange of gases	Stomata of leaf
Photosynthesis	Leaf
Transpiration	Stomata of leaf
Reproduction	Flower

Activity – 2

Question 2.
How do you observe the cells in onion peel under microscope? Draw and label the diagram. Write your observations.
Answer:
Observing cells in onion peel:
Take a piece of onion peel. Place it on the slide.

Put a drop of water and then a drop of glycerine on it. Gently cover it with a cover slip. Observe it under microscope.

Observations:

1. All the cells are similar in shape and structure.
2. Intercellular spaces are present.
3. Cells are arranged in rows.
4. Each cell has cell wall, nucleus and cytoplasm.

Activity – 3

Question 3.
Observe the Cells in a leaf peel.
Answer:

• Take a betel leaf or a Tradescantia leaf.
• Tear it with a single stroke. So that a thin edge be seen at the torn end.
• Observe the thin edge where the leaf has been torn under the microscope in the” same manner as you had observed the onion peel.

Draw a diagram what you have observed and compare it with figure.

Observations:
We may have observed that the cells are present in groups with certain arrangement. With the help of following activities, we shall try to find out whether these arrangements have special roles to play in the plant body.

a) Are all the cells similar?
Answer:
All the cells are not similar in shape and size.

b) Is there any difference in their arrangement?
Answer:
They are arranged compactly living the small gaps called stomata. Stomata are surrounded by bean shaped cells (Guard cells).

c) What can we infer from the above activities?
Answer:
From the above activities it infers that some of the cells may modify or arranged in a different way to perform specific functions.

d) Have you noticed that the cells are in groups in both the activities?
Answer:
The cells are arranged in groups.

e) Compare and write a note on the arrangements of the cells that you have observed in both of the activities.
Answer:
In the second activity cells are almost of same size and shape but in this activity cells shape and size varies from one another depending on their function.

Activity – 4

Question 4.
How do you observe root tips of onion? Draw the diagram you observe under microscope. Write your observations.
Answer:

1. Take a transparent bottle filled with water. Take the onion bulb slightly larger than the mouth of the bottle.
2. Put the onion bulb on the mouth of the bottle.
3. Observe the growth of roots for few days till they grow to nearly an inch.
4. Take the onion out and cut some of the root tips.
5. Take an onion root tip. Place it on the slide. Put a drop of water and then a drop of glycerin on it.
6. Cover it with a cover slip. Put the 2-3 layer of filter paper on the cover slip.
7. Tap the cover slip gently press with the blunt end of the needle or brush to spread the material.
8. Observe under the microscope.

Observations:

1. All the cells are not similar in shape and structure.
2. Cells are arranged in rows.
3. Meristems are present below the root cap.

Activity – 5

Question 5.
Observe the roots of onion which had been cut off? Write your findings.
Answer:

• Take the onion and cut the end of the roots. Mark the cut end of roots with a permanent marker.
• Put the onion bulb on the mouth of the bottle.
• Leave the set up aside at least four to five days.
• Take care that there is enough water in the glass so that the roots are submerged.

Observations :

1. By removing the tip of the onion root, having a particular arrangement of cells, the growth of the root in length is stopped.
2. Cells are present in groups.

Activity – 6

Question 6.
Write down the arrangement of cells in the given table :
Answer:

Arrangement of the cells (Tissues)	Shoot tip	Root tip
1. At the tip	Apical meristems are present.	Meristems below the root cap are present.
2. At the lateral side	Lateral meristems are present.	Lateral meristems are present.
3. At the point of branching	Intercalary meristems are present.	Meristems are absent.

Activity – 7

Question 7.
Observe temporary mount of T.S of Dicot stem under microscope and draw, label the diagram. Write your findings.
Answer:
Observing mount of T.S of Dicot Stem : Prepare a temporary mount of the T.S of dicot stem observe it under microscope.

Findings :

1. In the T.S of dicot stem meristematic tissue, vascular tissue, dermal tissue and ground tissue are present.
2. All the cells are not similar in shape and structure.

Activity – 8

Question 8.
Observe Rheo leaf peel under microscope. Draw and label the parts. Write your findings.
Answer:

1. Take a fresh leaf of Rheo leaf.
2. Tear it with a single stroke, so that a thin whitish edge can be seen at torn end.
3. Slowly remove it and observe the peel under microscope.

Findings:

1. Cells are structurally similar. They are compactly arranged without intercellular.
2. It is the dermal tissue of the plant.
3. It has an stomatal opening.

Activity – 9

Question 9.
Observe some tissues.
Answer:
Take permanent slides of Chlorenchyma, Arenchyma, Storage Tissue from your laboratory and observe them under the microscope. Find out the characteristics and differences and record them in your notebook.

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

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

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

Review of Your Previous Knowledge

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

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

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

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

Improve Your Learning

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

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

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

1. Gold
2. Silver
3. Platinum.

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

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

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

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

2. Washing:

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

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

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

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

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

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

1. Haematite (Fe₂O3)
2. Magnetite (Fe₃O₄).

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

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

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

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

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

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

Froth floatation process for the concentration of sulphide ores

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

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

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

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

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

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

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

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

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

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

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

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

1. Electrolytic Reduction
2. Smelting.

(OR)

1. Distillation
2. Poling.

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

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

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

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

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

Procedure :

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

Observation :

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

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

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

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

Extraction of Gold :

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

Extraction of Platinum:

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

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

ii)

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

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

Uses of activity series the extraction of metals :

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

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

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

Applications in daily life :

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

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

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

Daily life examples for washing :

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

Fill In The Blanks

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

1. Froth flotation
2. activity series
3. distillation
4. air, water
5. calcination

Multiple Choice Questions

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

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

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

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

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

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

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

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

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

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

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

10th Class Chemistry Textbook Page No. 238

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

10th Class Chemistry Textbook Page No. 240

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

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

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

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

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

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

1. Concentration or Dressing
2. Extraction of crude metal
3. Refining or purification of the metal.

10th Class Chemistry Textbook Page No. 248

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

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

10th Class Chemistry Textbook Page No. 251

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

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

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

10th Class Chemistry Textbook Page No. 239

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

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

10th Class Chemistry 10th Lesson Principles of Metallurgy Activities

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

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

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

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

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

Review of Your Previous Knowledge

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

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

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

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

Improve Your Learning

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

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

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

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

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

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

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

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

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

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

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

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

TV screen Activity :

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

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

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

2) Direction of the magnetic force (F) :

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

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

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

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

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

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

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

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

Working and Theory :

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

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

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

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

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

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

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

Activity – 2

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

Procedure :

1. Connect the terminal of a coil to a sensitive galvanometer as shown in the figure.
2. Normally, we would not expect any deflections of needle in the galvanometer because there is to be no electromotive force in this circuit.
3. Now if we push a bar magnet towards the coil, with its north pole facing the coil, we observe the needle in the galvanometer deflects, showing that a current is set up in the coil.
4. The galvanometer does not deflect if the magnet is at rest.
5. 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.
6. If we use the end of south pole of a magnet instead of north pole in the above activity, the deflections are exactly reversed.
7. This experiment proves “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Precautions :

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

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

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

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

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

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

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

Procedure:

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,
1.5 v 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. Scrape the insulation com¬pletely on one end of the coil.
4. Scrape the insulation on top, left and right of the other end. The bottom should be insulated.
5. Now complete the electric mo¬tor as shown in the figure. “5

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

1. Connect the terminals of a coil to a sensitive galvanometer as shown in the figure.
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 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.

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

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

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

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

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

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

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

1. Generation of electricity
2. Transmission of electricity
3. Metal detectors in security checking
4. The tape recorder
5. Use of ATM cards
6. Induction stoves
7. Transformers
8. Induction coils (spark plugs in automobiles)
9. Break system in railway wheels
10. AC and DC generators
11. Windmills, etc.

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

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

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

Fill in The Blanks

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

1. weber/m² (or) Tesla
2. area
3. zero
4. ILB
5. Law of conservation of energy

Multiple Choice Questions

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

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

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

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

10th Class Physics Textbook Page No. 221

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

10th Class Physics Textbook Page No. 213

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

10th Class Physics Textbook Page No. 214

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

10th Class Physics Textbook Page No. 215

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

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

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

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

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

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

10th Class Physics Textbook Page No. 218

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

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

10th Class Physics Textbook Page No. 219

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

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

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

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

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

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

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

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

10th Class Physics Textbook Page No. 221

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

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

10th Class Physics Textbook Page No. 222

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

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

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

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

10th Class Physics Textbook Page No. 223

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

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

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

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

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

10th Class Physics Textbook Page No. 224

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

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

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

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

Question 32.
Why does the coil rotate?
Answer:

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

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

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

10th Class Physics Textbook Page No. 225

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

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

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

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

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

10th Class Physics Textbook Page No. 226

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

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

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

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

10th Class Physics Textbook Page No. 227

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

10th Class Physics Textbook Page No. 228

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

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

10th Class Physics Textbook Page No. 229

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

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

10th Class Physics Textbook Page No. 230

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

10th Class Physics Textbook Page No. 232

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

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

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

10th Class Physics Textbook Page No. 233

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

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

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

10th Class Physics Textbook Page No. 234

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

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

10th Class Physics 12th Lesson Electromagnetism Activities

Activity – 2

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

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

Activity – 3

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

Procedure:

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

Activity – 4

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

Procedure:

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

Activity 5

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

Procedure :

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

Activity – 6

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

Procedure :

1. Take a wooden plank covered with white paper.
2. Make holes on its surface.
3. Pass copper wire through the holes.
4. Join the ends of the coil to a battery through a switch.
5. Current passes through the coil, when we switch on the circuit.
6. Now sprinkle iron filings on the surface of the plank, around the coil. Then orderly pattern of iron filings is seen on the paper.
7. The iron filings arrange themselves in orderly way 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 end of the solenoid behaves like a north pole and the other behaves like south pole.
10. Outside the solenoid the direction of the field lines of force is from north to south while inside the direction is from south to north. Thus the magnetic field lines are closed loops.
11. Hence electric charges in motion produce magnetic field.

Activity – 8

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

Procedure :

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

Activity – 9

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

Procedure :

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

 

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

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

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

Review of Your Previous Knowledge

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

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

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

Improve Your Learning

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

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

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

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

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

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

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

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

Materials required :

1. A bulb
2. A battery
3. Key
4. Insulated wire
5. Multimeter

Procedure :

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

Observation :

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

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

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

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

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

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

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

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

i) Junction law :

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

ii) Loop law:

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

This law is based on the conservation of energy.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Procedure :

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

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

How to Make Rheostat:

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

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

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

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

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

We conclude that the resistance is constant.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Fill in The Blanks

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

1. electrical energy
2. less
3. 6 Ω
4. volt
5. Ampere
6. 12 Ω
7. \(\frac{11}{12} \Omega\)
8. 1 ampere

Multiple Choice Questions

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

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

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

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

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

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

10th Class Physics Textbook Page No. 179

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

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

10th Class Physics Textbook Page No. 180

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

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

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

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

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

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

10th Class Physics Textbook Page No. 181

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

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

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

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

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

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

10th Class Physics Textbook Page No. 183

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

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

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

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

10th Class Physics Textbook Page No. 184

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

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

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

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

10th Class Physics Textbook Page No. 185

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

10th Class Physics Textbook Page No. 186

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

10th Class Physics Textbook Page No. 187

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

10th Class Physics Textbook Page No. 189

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

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

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

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

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

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

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

10th Class Physics Textbook Page No. 190

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

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

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

10th Class Physics Textbook Page No. 191

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

10th Class Physics Textbook Page No. 192

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

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

10th Class Physics Textbook Page No. 193

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

10th Class Physics Textbook Page No. 195

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

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

10th Class Physics Textbook Page No. 196

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

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

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

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

10th Class Physics Textbook Page No. 197

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

10th Class Physics Textbook Page No. 199

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

10th Class Physics Textbook Page No. 201

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

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

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

10th Class Physics Textbook Page No. 202

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

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

10th Class Physics Textbook Page No. 203

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

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

10th Class Physics Textbook Page No. 203

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

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

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

10th Class Physics 11th Lesson Electric Current Activities

Activity – 1

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

Materials required:

1. A bulb
2. a battery
3. a switch
4. few insulated copper wire

Procedure (1) :

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

Procedure (2) :

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

Observation (2): The bulb does not glow.

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

Observation (3) : The bulb does not glow.

Result:
The battery contains charges which glow the bulb.

Activity – 3

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

• Copper rod
• Nichrome rod
• Battery
• Ammeter
• Key
• Manganin Wire

Procedure :

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

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

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

Activity – 5

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

Materials required :

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

Procedure:

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

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

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

Activity – 6

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

Materials required :

1. Bulbs
2. Voltmeters
3. Insulated wires
4. Ammeter
5. Key

Procedure :

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

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

Activity – 7

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

1. Bulbs
2. Ammeters
3. Buttery
4. Key
5. Wires

Procedure :

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

Observation :

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

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

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

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

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

Improve Your Learning

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

is called esterification.

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

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

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

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

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

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

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

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

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

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

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

1. Allotropy is a property shown by the elements.

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

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

4. Example for allotropes : Allotropes of carbon.

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

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

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

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

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

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

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

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

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

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

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

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

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

Justification :
In the following, C₃H₆ undergoes addition reaction.

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

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

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

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

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

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

Here CH₃COOH is used as preservative for pickles.

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

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

2) Ethane (C2H₆):

3) Ethene (C₂H₄):

4) Ethyne (C₂H₂)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Fill In The Blanks

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

1. unsaturated compounds
2. ethanol
3. vinegar
4. ester
5. H₂
6. – OH (Alcohol)
7. propene
8. ethyne (C₂H₂)
9. ethene (C₂H₄)
10. 3
11. substitutional

Multiple Choice Questions

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

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

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

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

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

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

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

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

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

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

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

10th Class Chemistry Textbook Page No. 254

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

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

10th Class Chemistry Textbook Page No. 255

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

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

Electronic configuration of carbon (excited state):

10th Class Chemistry Textbook Page No. 256

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

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

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

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

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

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

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

10th Class Chemistry Textbook Page No. 257

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

10th Class Chemistry Textbook Page No. 258

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

10th Class Chemistry Textbook Page No. 259

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

10th Class Chemistry Textbook Page No. 260

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

10th Class Chemistry Textbook Page No. 262

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

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

10th Class Chemistry Textbook Page No. 265

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

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

10th Class Chemistry Textbook Page No. 266

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

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

10th Class Chemistry Textbook Page No. 269

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

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

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

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

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

10th Class Chemistry Textbook Page No. 272

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

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

10th Class Chemistry Textbook Page No. 273

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

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

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

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

10th Class Chemistry Textbook Page No. 274

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

10th Class Chemistry Textbook Page No. 278

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

10th Class Chemistry Textbook Page No. 279

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

10th Class Chemistry Textbook Page No. 280

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

10th Class Chemistry Textbook Page No. 281

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

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

10th Class Chemistry Textbook Page No. 283

Question 25.
What are esters?
Answer:

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

10th Class Chemistry Textbook Page No. 284

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

10th Class Chemistry Textbook Page No. 286

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

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

10th Class Chemistry Textbook Page No. 280

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

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

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

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

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

Activity – 1

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

Sol. 2-Methyl butane

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

Sol. 3-Methyl, but-l-ene

5)

Sol. Prop-l-yne

Activity-2

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

2. But-l-yne
Sol. CH₃ – CH₂ – C = CH

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

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

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

Activity – 3

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

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

Activity – 4

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

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

 

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

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

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

Improve Your Learning

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

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

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

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

Covalency of an element:

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

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

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

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

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

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

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

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

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

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

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

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

5.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Uses of covalent compounds :

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

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

b) Water (H₂O):

The formation of water molecule can be shown like this also

c) Chlorine (Cl₂):

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

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

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

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

b) Calcium chloride (CaCl₂)

c) Carbon dioxide (CO₂) :

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

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

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

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

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

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

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

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

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

Role of octet in chemical properties of elements :

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

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

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

Formation of O₂ molecule :

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

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

a) Formation of BeCl₂ (Beryllium chloride) molecule :

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

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

f) Both the bonds are of same strength.

b) Formation of BF₃ molecule :

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

Fill in the Blanks

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

1. valence electrons
2. Helium
3. covalent bonds
4. Linus Pauling
5. covalent

Multiple Choice Questions

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

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

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

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

10th Class Chemistry Textbook Page No. 153

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

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

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

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

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

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

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

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

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

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

10th Class Chemistry Textbook Page No. 155

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

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

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

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

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

10th Class Chemistry Textbook Page No. 157

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

10th Class Chemistry Textbook Page No. 158

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

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

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

10th Class Chemistry Textbook Page No. 159

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

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

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

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

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

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

2) Lewis electron dot symbol for MgCl₂:
MgCl₂

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

Anion formation :

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

3) Lewis electron dot symbol for (Na₂O) :

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

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

4) Lewis electron dot symbol for (AlCl₃):

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

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

10th Class Chemistry Textbook Page No. 163

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

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

10th Class Chemistry Textbook Page No. 164

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

10th Class Chemistry Textbook Page No. 165

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

10th Class Chemistry Textbook Page No. 168

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

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

10th Class Chemistry Textbook Page No. 170

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

10th Class Chemistry Textbook Page No. 172

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

10th Class Chemistry 10th Lesson Chemical Bonding Activities

Activity – 1

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