AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 9th Lesson Floating Bodies

9th Class Physical Science 9th Lesson Floating Bodies Textbook Questions and Answers

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Question 1.
A solid sphere has a radius of 2 cm and a mass of 0.05 kg. What is the relative density of the sphere? (AS 1)
Answer:
Radius of the sphere = 2 cm
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 1

Question 2.
A small bottle weighs 20 g when empty and 22 g when filled with water. When it is filled with oil it weighs 21.76 g. What is the density of oil? (AS 1)
Answer:
Weight of water = 22 – 20 = 2 gm
Weight of oil = 21.76 – 20 = 1.76 gm
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 2

Question 3.
An ice cube floats on the surface of a glass of water (density of ice = 0.9 g/cm3). When the ice melts will the water level in the glass rise? (AS 1)
Answer:
Yes, the water level rises.

Reason :
The ice cube floats on water, because its density is less than the density of water. When ice cube melts, it becomes water, so that the water level rises.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

Question 4.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance sink or float when placed on the surface of water? What will be the mass of water displaced by the substance? (AS 1)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 9
∴ Weight of the water displaced by the substance = 20 g

Question 5.
Find the pressure at a depth of 10 m in water if the atmospheric pressure is 100 kPa. [1Pa = 1 N/m²] [100kPa = 105 Pa = 105 N/m² = 1 atm.] (AS 1)
Answer:
Depth ‘h’ = 10 m ; Atmospheric pressure P0 = 100 kPa
Density of water p = 1 gm/cm³ = 1 kg/m³
Pressure at a depth ‘h’ is P = P0 + hρg
= 100 + 10 x 1 x 9.8
= 100 + 98
= 198 kPa

Question 6.
Why do some objects float on the water? And some sink? (AS 1)
Answer:

  • Floating or sinking of objects on water depends on two factors.
    a) Relative density
    b) Weight of the water displace by the object
  • If the relative density of an object is greater than 1, the object sinks otherwise it floats.
  • Eventhough the relative density is greater than 1, if the weight of the water displaced by the object equal to the weight of the object itself, the object floats on the water.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

Question 7.
Explain density and relative density and write their formulae. (AS 1)
Answer:
Density :
Density is defined as mass per unit volume.
Density = \(\frac{\text { Mass }}{\text { Volume }}\)

Unit of density is gm/cm³ or kg/m³.

Relative density :
Relative density of an object is the ratio of density of the object to the density of water.
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 3

Question 8.
What is the value of density of water? (AS 1)
Answer:
Value of density of water = 1 gm/cm³ (or) 1 kg/cm³.

Question 9.
Find the relative density of wood. Explain the process. (Lab Activity 1) (AS 3)
Answer:
Aim :
To find the relative density of wood.

Materials required :
Overflow vessel, 50 ml measuring cylinder, spring balance, wooden block, water.
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 4

Procedure:

  1. Weigh the 50 ml measuring cylinder and note its weight.
  2. Weigh the wooden block and note its weight.
  3. Pour water in the overflow vessel until it starts dripping from its beak.
  4. When water stops dripping from the beak, place the 50 ml measuring cylinder under it.
  5. Slip the wooden block gently into the overflow vessel, ensuring that the water does not splash out.
  6. Once the wooden block is in the overflow vessel, water flows out of the beak and collects in the 50 ml cylinder.
  7. Wait till the flow of water from beak, stops.
  8. Weigh the cylinder with the water that overflowed and record the weight.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 5

Question 10.
Which is denser, water or milk? (AS 2)
Answer:
The density of water is 1 gm/cc and that of milk is 1.02 gm/cc. Hence milk is slightly denser than water.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

Question 11.
What is buoyancy? (AS 1)
Answer:
Buoyancy is the upward force that a fluid exerts on an object less denser than itself.
(or)
Buoyancy is the ability of an object to float in a liquid.

Question 12.
Classify the following things into substances having relative density > 1 and relative density < 1. Wood, iron, rubber, plastic, glass, stone, cork, air, coal, ice, wax, paper, milk, kerosene, groundnut oil, soap. (AS 1)
Answer:

Relative density > 1 Relative density < 1
Iron Wood
Glass Rubber
Stone Plastic
Milk Cork
Soap Air
Coal
Ice
Wax
Paper
Kerosene
Groundnut oil

Question 13.
How can you appreciate the technology of making ships float, using the material which sink in water? (AS 6)
Answer:

  • We know that a piece of iron sinks in water.
  • The relative density of iron is 8.5 which is many times more than water.
  • A ship that floats on water is made up of tonnes of iron. It is really wonderful.
  • According to the Archimedes principle of buoyancy, any object can float when its weight is equal to the weight of water displaced by it.
  • Hence ships are made with a larger surface area so that it displaces the water, whose weight is equal to the weight of loaded ship.
  • It involves very sharp scientific calculations and a large engineering technology.
  • Really such type of technologies are highly appreciable and the scientists who formulates all these are also really great.

Question 14.
Can you make iron float? How? (AS 3)
Answer:
Yes, we can make iron to float on water.
Procedure:

  1. Take a piece of iron and drop it in a vessel of water.
  2. We observe that the iron piece sinks in water.
  3. Take a thin foil of iron and fold it into four folds.
  4. Drop it in water. It sinks.
  5. Now unfold the foil and bend it in the form of a bowl. [You can use an iron tin also]
  6. Now drop the bowl in water.
  7. If floats on water.

Reason:
The weight of water displaced by iron bowl (iron tin) is less than the weight of the iron bowl (iron tin).

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

Question 15.
How can you find the relative density of a liquid? (Lab Activity – 2) (AS 3)
Answer:
Aim :
To find the relative density of a liquid.

Materials required :
Small bottle of 50 ml capacity (the bottle should weigh not less than 10 gm), spring balance, any liquid (milk or oil or kerosene) about 50 ml.

Procedure:

  1. Find the weight of empty 50 ml bottle.
  2. Fill the bottle with water and weigh it.
  3. Find the weight of 50 ml water.
  4. Remove water from the bottle and fill it with any liquid (say milk).
  5. Weigh the bottle with liquid.
  6. Weight of 50 ml liquid = Weight of bottle with liquid – Weight of empty bottle

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 6

Question 16.
Find the relative density of different fruits and vegetables and make a list. (AS 3)
Answer:

  • Follow the procedure in Q.No (11) of A.S – (3).
  • Use different fruits and vegetables in the place of wooden block.
  • Write the observed values in the following table.
Name of the fruit or vegetable Relative density
Cabbage 0.36
Cauliflower 0.26
Bottleguard 0.56
Potato 0.67
Onion 0.59
Chilli 0.29
Bitterguard 0.4
Apple 1.22
Grape 1.04
Orange 0.34

Question 17.
Make a lactometer with ball point refill. What would you do to make the refill stand vertically straight? (Activity – 2) (AS 5)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 7

  • Take an empty ball pen refill. It should have a metal point.
  • Take a boiling tube and fill it with water.
  • Put the refill in with metallic point inside the water.
  • Use a pen to mark the point on the refill to show the part which is above the water surface.
  • Pour out the water from the boiling tube and fill it with milk.
  • Float the refill in the milk.
  • Puf the record mark on the refill, at the point showing the part which is above the surface of the milk.
  • These two marks are not at the same place. This is the improvised lactometer.
  • We have to attach a small weight at the bottom of the refill to make it stand vertically straight.

Question 18.
Draw the diagram of a mercury barometer. (AS 5)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 8

Question 19.
Write a note on Pascal’s discovery in helping to make hydraulic jacks. (AS 6)
Answer:
Pascal’s principle:
External pressure applied to an enclosed body of fluid is transmitted equally in all directions throughout the fluid volume and the walls of the containing vessel.

Use :

  1. This principle is used in designing and working of hydraulic jacks.
  2. Hydraulic jacks are useful to lift heavy objects like cars and other vehicles in automobile work shops when the vehicles are to be required or repaired.
  3. Here very less force is used to lift such heavy vehicles.

Appreciation :

  1. Hydraulic jacks are used not only in automobile work shops, but also in so many industries where heavy weights are to be lifted with a little force.
  2. All the comforts which we are enjoying now are the efforts of scientists who discovered the laws and principles.
  3. Hence we have to appreciate the efforts of Pascal for his contribution in designing these principles.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

Question 20.
Write a note on Archimedes discovery of force of buoyancy. (AS 6)
Answer:
Archimedes principle:
When a body is immersed in a fluid, it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

Use :
The principle is used to determine the purity of metals.

Appreciation :

  1. This principle was discovered by Archimedes eventually when he was taking bath.
  2. With this principle he could solve the problem assigned by the king to him.
  3. So many problems in our life may be solved by so many scientific principles discovered by various scientists.
  4. You might have heard about falling of the statue of Budha in Hussain Sagar.
  5. That statue was lifted by using the principle of buoyancy.
  6. Archimedes is thought to be so important as a mathematician that scientists honoured him.
    a) A large hole or crator on the moon is named after Archimedes.
    b) Some mountains on the moon are called the monte – Archimedes.
  7. Hence the efforts of Archimedes in discovering such type of principle may be highly appreciated.

Question 21.
You found the relative densities of some solids and some liquids by doing some activities. List the solids and liquids in increasing order of relative density. (AS 4)
Answer:

Substance Relative density
Kerosene 0.81
Rubber 0.94
Milk 1.02
Glass 1.29
Iron 8.5

Question 22.
Iron sinks in water, wood floats in water. If we tie an iron piece to wood piece of the same volume, buritlle and drop it in water, would bundle sink or float? Make a guess and find out whether your guess is correct or wrong with an experiment. Give reasons. (AS 2, AS 3)
Answer:
The body sinks in water.

Reason :
The combined mass of the system increases, so the combined density also increases. Hence the body sinks in water.

Question 23.
Air brakes in automobiles work on Pascal’s principle. What about air brakes? Collect the information about the working process of air brakes. (AS 4)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 10

  • Air brakes works on the principle of conversion of energy. Generally in trains, while it is moving -fit .produces kinetic energy. This kinetic energy has to be reduced to make the train stop. Here air is used to reduce kinetic energy by converting it into heat energy.
  • The system of air brakes in trains has been shown in the figure.
  • The important parts are compressor, main reservoir, driver’s brake value, brake pipe, triple value, auxiliary reservoir, brake cylinders, and brake block.

Working:

  1. When the driver placed the value in application position, the air pressure in the brake pipe escapes.
  2. The loss of pressure is detected by the slide value in the triple value.
  3. Now a connection between the auxiliary reservoir and the brake cylinder has been opened and the air in the auxiliary reservoir feeds through into the brake cylinder.
  4. The air pressure forces the piston to move against the spring pressure and causes the brakes to be applied to the wheels.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies

Question 24.
Where do you observe Archimedes principle in daily life? Give two examples.
Answer:
Daily life application of Archimedes principle :

  1. Archimedes principle of buoyancy is applied in our daily life in many ways.
  2. Fish, human swimmers, ice bergs and ships float follow Archimedes principle of buoyancy.
  3. Rise of balloon in air also follows Archimedes principle.
  4. While dragging water from a well, the bucket filled with water seems to be weightless till it reaches the surface of the water in the well. This is also due to buoyancy.
  5. Swimming of duck in water is also an example of Archimedes principle.

Question 25.
Where do you observe Pascal’s principle in daily life? Give a few examples.
(OR)
Write any one application of Pascal’s principle in daily life.
Answer:
Daily life application of Pascal’s principle :
Pascal’s principle is applied in the working of

  1. Hydraulic jacks
  2. Hydraulic lifts
  3. Hydraulic pumps
  4. Hydraulic cranes
  5. Siphons
  6. Artesian wells
  7. Water towers and dams

9th Class Physical Science 9th Lesson Floating Bodies InText Questions and Answers

9th Class Physical Science Textbook Page No. 141

Question 1.
a) Did kerosene float above the water or did water float above the kerosene?
Answer:
Kerosene floats above the water.

b) Which objects float in kerosene?
Answer:
Plastic buttons, match stick, tiny paper balls, wax, etc. floats in kerosene.

c) Which objects sink in kerosene but float on water?
Answer:
Wax sink in kerosene but float on water.

d) Which objects sink in water?
Answer:
Pins, small pebbles, sand, etc. sink in water.

e) Draw a diagram of the tube, showing the results of your activity.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 18

f) Why did different objects behave differently?
Answer:
The density of objects is the main reason to behave differently.

9th Class Physical Science Textbook Page No. 144

Question 2.
1) What is the relative density of wood?
Answer:
0.8

2) What is the relative density of glass?
Answer:
1.2

3) Which is denser, rubber or plastic?
Answer:
Rubber is denser than plastic.

4) Which is denser, wood or cork?
Answer:
Wood is denser than cork.

5) Do objects that have a relative density less than 1 Sink in water or float on it?
Answer:
Float on water.

6) Do the objects that sink in water have a relative density less than 1 or more than 1?
Answer:
More than 1.

7) Classify the above materials as denser than stone and less denser than the stone.
Answer:
Denser than the stone : iron, nails.
Less denser than the stone :
eraser, wood, glass slides, plastic cube, Aluminium, glass marbles, cork.

8) What relationship do you find between the relative density of objects and floating sinking of the objects?
Answer:
If the relative density is less than 1 the object of less density will float above the other in the water.

9th Class Physical Science Textbook Page No. 145

Question 3.
1) Which liquid will float on top if groundnut oil is poured over water?
Answer:
Groundnut oil float on the water.

2) If we put a wooden block in kerosene, will it float or sink? Give reasons for your answer.
Answer:

  • Wooden block will sink in the kerosene.
  • Because, the density of the wooden block is more than the kerosene.

3) A piece of wax floats in water but the same piece of wax sinks in a liquid say liquid ‘X’. Will the relative density of liquid . ‘X’ be less than 1 or greater than 1? How can you say?
Answer:

  • The relative density of liquid ‘X’ is less than 1.
  • The density of wax is less than 1 because it floats on the water.
  • The density of liquid ‘X’ is less than the density of the wax this means less than 1. because wax sinks in that liquid ‘X’.

4) If we mix some water in milk, will the relative density of the mixture be less than or more than the relative density of milk?
Answer:

  • We know that relative density of milk is more than 1 and density of milk is more ‘ than water.
  • If we mix some water in milk, the density of the mixture will decreases and the relative density will also be less than the relative density of milk.

5) If we take two bottles of equal volume and pour pure milk in one and milk mixed with water in the other, which one will be heavier?
Answer:

  • Pure milk is heavier than water mix milk.
  • This is due to density of milk is more than water.

9th Class Physical Science Textbook Page No. 151

Question 4.
Why is the height of mercury column nearly 76 cm in the tube?
Answer:
Air pressure is the weight of air in the atmosphere above the reservoir (bowl of mercury). So, the level of mercury continues to change until the weight of mercury in the glass tube is exactly equal to the weight of the air above the reservoir, which is 76 cm.

9th Class Physical Science Textbook Page No. 154

Question 5.
What happens if we replace this cylindrical liquid column with another object which is made up of a material whose density is equal to the density of liquid?
Answer:
We know that the pressure difference in the liquid,
P2 – P1 = hρg
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 19

Since F = PA and W = mg
We get F = W (Values of displaced liquid)

  1. Here F’ is the force applied on the object and ‘W’ is the weight of the liquid.
  2. So, the”force applied on the object by the liquid is equal to the weight of the displaced liquid.

9th Class Physical Science Textbook Page No. 155

Question 6.
Why does the stone lose weight when it is immersed?
Answer:

  • Suspend a stone from a spring balance. Note the reading of the spring balance.
  • Take a beaker half-filled with water.
  • Now immerse the stone in the water, note the reading of the spring balance.
  • We notice that the stone, when immersed appears to lose some weight.
  • The immersed stone appears to lose weight because the force of buoyancy.
  • Thus the apparent loss of weight must be equal to the force of buoyancy acting on the immersed stone.

9th Class Physical Science Textbook Page No. 142

Question 7.
Let us suppose you have two blocks and you do not know what material they are made of. The volume of one block is 30 cm3 while the other is 60 cm3. The second block is heavier than the first. Based on this information, can you tell which of the two blocks is denser?
Answer:
No, we cannot say which of the two blocks is denser, because any one of the quantity
i. e., either volume or weight must be same.

9th Class Physical Science Textbook Page No. 153

Question 8.
a) What would happen if Toricelli’s experiment is done on moon?
b) A stopper is inserted in the small hole of the glass tube of the mercury barometer below the top level of the mercury in it. What happens when you pull out the . stopper from the glass tube?
c) Why don’t we use water instead of mercury in Toricelli experiment? If we are ready to do this experiment, what length of tube is needed?
d) Find the weight of the atmosphere around the earth (take the radius of earth as 6400km)
Answer:
a) If Toricelli’s experiment had been done on moon, the height of mercury column will be zero. Because there is no atmosphere on the moon.

b) The mercury level does not change, because there is vacuum above the mercury level. Hence no pressure is on the mercury. Also, the weight of mercury column must be equal to the air pressure above the reservoir. Hence there will be no change in the height of the mercury column.

c) We cannot use water instead of mercury in Toricelli experiment because, if we want to use water, we have to take the glass tube of length nearby above 10 m, which is inconvenient.
If water is taken,
P0 = ρhg
1 01 yin5
1.01 × 105 = 1 × 10³ × h × 9.8 ⇒ h = = 0.1030 × 10² = 10.3 m

d) Weight of atmosphere = Atmospheric pressure x Surface area of the earth
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 17

9th Class Physical Science Textbook Page No. 157

Question 9.
a) Why is it easier for you to float in saltwater than in freshwater?
b) Why is there no horizontal buoyant force on a submerged body?
c) Two solid blocks of identical size are submerged in water. One block is iron and the other is aluminium. Upon which is the buoyant force greater?
d) A piece of iron when placed on a block of wood, this makes the wood to float lower in the water. If the iron piece is suspended beneath the wood block, would it float at the same depth? Or lower or higher?
Answer:
a) Salt water is denser than freshwater.
b) Buoyant force is the upward force only. The body is submerged means, its weight is more than the buoyant force. Here there will be no horizontal buoyant force.
c) Buoyant force on iron block is more than that of aluminium block.
d) It floats on higher depth than in the first case.

9th Class Physical Science 9th Lesson Floating Bodies Activities

Activity – 1

Question 1.
Comparing density – relative density.
Take two test tubes of the same size and fill one to the brim with water and the other with oil.
a) Which will weigh more?
Answer:
The test tube with oil will weigh more.

b) Which liquid is denser?
Answer:
Oil is denser than water.
Take two equal sized blocks made of wood and rubber.

c) Which of these two blocks is heavier?
Answer:
Wooden block is heavier.

d) Which one is denser?
Answer:
Wooden block is denser than rubber.

Activity – 3

Question 2.
Do the objects denser than water float in it? Prove it with an activity.
Answer:

  • Collect some objects listed in the table below.
  • Place them one by one in a glass of water and observe whether they sink or float in water.
  • Record your observations in the table.

Observations:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 20

  • Find the relative densities of all the objects.
  • We observe that some objects floats and some objects like iron nail, glass marble, stone sinks in water.
  • Geometry box made of iron floats on water, though its relative density is greater than 1. This is due to its larger surface area.
  • Geometry box made of iron floats on water though it is made up of a substance denser than water.
  •  We can say that the floating or sinking of an object does not depend not only on its relative density, but also its surface area which displaces the water.

Activity – 4

Question 3.
Show that for a floating object, the weight of the object is equal to the weight of water displaced by it.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 11

  • Take a beaker and weigh it. Note down its weight.
  • Fill water in an overflow jar, wait until the water stops dripping from the outlet of the overflow jar.
  • Now place the beaker below the outlet of the overflow jar.
  • Take a wooden block, moisten it with water and gently drop it into the overflow jar.
  • Water will flow out of the overflow jar and collects in the beaker kept under the overflow jar.
  • Measure the weight of beaker with water.
  • Subtract the weight of beaker from this. The value gives the weight of water displaced by wooden block. Note it.
  • Now remove wooden block from the overflow jar, make it cry and weight.
  • We can observe that the weight of wooden block is equal to weight of water displaced by it.
  • Do the same experiment with some other substances and record your observations in the table.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 12

Activity – 5

Question 4.
Explain how does metal aluminium floats on the water with an activity.
Answer:

  • Take a small sheet of aluminium foil.
  • Fold it four or five times, pressing the foil tight after each fold.
  • We can find that the aluminium foil will sink in the water.
  • Now unfold the aluminium foil and make it as a small bowl.
  • We can find that the bowl will float on the water.
  • The metal bowl displace larger amount of water than a metal piece.
  • Weight of the displaced water is more than the metal sheet.
  • So, bowl will float on the water.

Activity – 6

Question 5.
Prove that the water exerts upward force on objects.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 13

  1. Take an empty plastic bottle.
  2. Put the cap on it tightly.
  3. Place the bottle in a bucket of water.
  4. The bottle will float.
  5. Push the bottle into the water by your hand as shown in figure.
  6. We feel some upward force.
  7. Try to push it further down. We feel increase in the upward force.
  8. Now release the bottle.
  9. It bounce back to the surface of water.
  10. Here the upward force of water a real, observable force.
  11. This force acting on unit area of the surface of an object is called static pressure of the water.

Activity – 7

Question 6.
Describe an activity to observe the air pressure.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 14

  1. Take a glass tumbler.
  2. Stick some cotton at the bottom of it.
  3. Immerse it inversely in water upto the bottom of the container as shown in figure.
  4. Take out the tumbler from water.
  5. We observe that the cotton attached at the bottom of the tumbler is not wet.
  6. This is due to the force of air which is applied on water by the air present in the tumbler and stops water from entering the tumbler.
  7. This force on unit area of water is the pressure of air.

Activity – 8

Question 7.
How can you measure the force of buoyancy and how much?
(OR)
Why does and stone lose weight when it is immersed in water?
Answer:

  • Suspend a stone from a spring balance.
  • Note that the reading of the spring balance.
  • The reading gives the weight of the stone.
  • Take a beaker half filled with water.
  • Now immerse the stone in the water.
  • Note the reading of the spring balance.
  • The reading gives the weight of the immersed stone.
  • We may notice that the stone, when immersed, appears to lose some weight.
  • The immersed stone appears to lose weight because the force of BUOYANCY, exerted on the stone by the water in the upward direction.
  • Thus the apparent loss of weight must be equal to the force of buogancy acting on the immersed stone.
  • The lose of weight of stone is equal to weight of the water displaced by the stone, this is the force of buoyancy.

Activity – 9

Question 8.
‘State and prove Archimedes principle of buoyancy.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 15
Archimedes principle :
Archimedes principle states that when a body is immersed in a fluid, it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies 16
Proof:

  1. Suspend a stone from a spring balance.
  2. Note the reading on the spring balance. This reading is the weight of the stone.
  3. Take an overflow jar with water and place a graduated beaker below the beak as shown in the figure.
  4. Now immerse the stone in the water and note the reading on the spring balance.
  5. Measure the volume of the water that overflows into the graduated beaker.
  6. The reading of the spring balance gives the weight of the immersed stone.
  7. The beaker reading gives the volume of water displaced by the stone.
  8. The difference in the two readings of spring balance gives the apparent loss of weight of the stone.
  9. Now weigh the water in the graduated beaker.
  10. We observe that the apparent loss of weight of the stone is equal to the weight of the water displaced by the stone.
  11. Hence Archimedes principle is proved.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 11 Sound Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 11th Lesson Sound

9th Class Physical Science 11th Lesson Sound Textbook Questions and Answers

Improve Your Learning

Pick out the correct answer :

Question 1.
When we say sound travels in a medium (AS 1)
A) the medium travels
B) the particles of the medium travel
C) the source travels
D) the disturbance travels
Answer:
D) the disturbance travels

Question 2.
A sound wave consists of (AS 1)
A) number of compression pulses only
B) number of rarefaction pulses only
C) number of compression and rarefaction pulses one after the other
D) vacuum only
Answer:
C) number of compression and rarefaction pulses one after the other

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 3.
Hertz stands for oscillations per (AS 1)
A) second
B) minute
C) hour
D) milli second
Answer:
A)second

Question 4.
When we increase the loudness of sound of a TV, the property of sound that changes is (AS 1)
A) amplitude
B) frequency
C) wavelength
D) speed
Answer:
A) amplitude

Question 5.
The characteristic of the sound that describes how the brain interprets the frequency of sound is called (AS 1)
A) pitch
B) loudness
C) quality
D) sound
Answer:
A) pitch

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 6.
In a stethoscope, sound of heart beats travel through stethoscope tube (AS 1)
A) by bending along the tube
B) in a straight line
C) undergoing multiple reflections
D) all of the above
Answer:
C) undergoing multiple reflections

Question 7.
Explain the following terms : (AS 1)
a) amplitude
b) wavelength
c) frequency
Answer:
a) Amplitude :
The maximum variation in density or pressure from the mean value is called amplitude.
(or)
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 8
The maximum disturbance of particles of a medium from their mean position is called amplitude.

b) Wavelength :
The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength of a sound wave, denoted by W. Wavelength is measured in ‘meters’.

c) Frequency :

  1. The number of oscillations of the density of the medium at a place per unit time is called the frequency of the sound wave.
  2. Frequency is denoted by ‘o’.
  3. The S.I. unit of frequency is ‘Hertz’.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 8.
Deduce the relation between wavelength, frequency, and speed of sound. (AS 1)
Answer:
1) Speed of sound can be defined as the distance by which a point on the wave, such as a compression or rarefaction, travels in unit time.

2) Let the distance travelled by a wave in T seconds = X metres

3) The distance travelled by a wave in 1 second = \(\frac{\lambda}{\mathrm{T}}\) meters

4) Thus by definition of speed of wave, v = \(\frac{\lambda}{\mathrm{T}}\) ……….. (1)

5) We know that frequency and time period are related as υ = \(\frac{1}{T}\) ……….. (2)

6) From (1) and (2) we get v = λ .υ
∴ Speed of sound = Frequency × Wavelength

Question 9.
How are multiple reflections of sound helpful to doctors and engipeeps? (AS 7)
Answer:
1) Doctors use multiple reflections of sound to hear the sounds produced with in the body using stethoscope.
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 11
2) Doctors can see the images of patient’s organs like liver, gall bladder, uterus, etc. to know the abnormalities in their functioning, using ultrasounds.

3) Engineers use the reflections of sound in designing concert halls and cinema halls.

4) Generally the ceilings of concert halls, conference halls, cinema halls are designed such that sound after reflection reaches all corners of the hall as shown in the figure.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 10.
Name two quantities that vary periodically at a place in air as a sound wave travels through it. (AS 1)
Answer:
The two quantities that vary periodically at a place in air as a sound wave travels through it are density and pressure of particles.

Question 11.
Which has larger frequency – infrasonic sound or ultrasonic sound? (AS 7, AS 2)
Answer:

  • Infrasonics are the sounds of frequency less than 20 Hz.
  • Ultrasonics are the sounds of frequency greater than 20 kHz.
  • Hence the ultrasonics have larger frequency.

Question 12.
The grandparents and parents of two-year-old girl are playing with her in a room. A sound source produces a 28 kHz sound. Who in the room is most likely to hear the sound? (AS 2, AS 7)
Answer:

  • The two-year-old girl is able to hear the sound.
  • Children can hear sounds of somewhat higher frequencies up to 30 kHz.

Question 13.
Does the sound follow same laws of reflection as light does? (AS 1)
Answer:

  • Reflection of sound follows the same laws as the reflection of light when sound is reflected.
  • The directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 14.
Why is soft furnishing avoided in concert halls? (AS 7)
Answer:

  • Sound reflects like the reflection of light.
  • But unlike to light, sound reflects more on rough surfaces than soft surfaces.
  • In concert halls, sound must undergo multiple reflections, so as to reach all corners of the hall.
  • Hence for better reflection, soft furnishing is avoided in concert halls.

Question 15.
Two sources A and B vibrate with the same amplitude. They produce sounds of frequencies 1 kHz and 30 kHz respectively. Which of the two waves will have larger power? (AS 1)
Answer:
Frequency of source A = 1 kHz; Frequency of source B = 30 kHz

As the speed of wave increases with frequency and both the waves have same amplitude, the sound produced from source B has larger power.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 16.
What do you understand by a sound wave? (AS 1)
Answer:

  • Sound is produced from a vibrating body.
  • It travels through air in the form of a wave.
  • Sound waves are longitudinal.

Question 17.
Define the wavelength of a sound wave. How is it related to the frequency and the wave speed? (AS 1)
Answer:
Wavelength :
The distance between two consecutive compressions or rarefactions is called wavelength.

Relation between wavelength, frequency, and wave speed :

  1. Speed of sound can be defined as the distance by which a point on the wave, such as a compression or rarefaction, travels in unit time.
  2. Let the distance travelled by a wave in T seconds = λ metres
  3. The distance travelled by a wave in 1 second = \(\frac{\lambda}{\mathrm{T}}\) meters
  4. Thus by definition of speed of wave, v = \(\frac{\lambda}{\mathrm{T}}\) ………… (1)
  5. We know that frequency and time period are related as o = \(\frac{1}{T}\) ……….. (2)

AP Board 9th Class Physical Science Solutions Chapter 11 Sound 1

Question 18.
Explain how echoes are used by bats to judge the distance of an obstacle in front of them. (AS 1)
Answer:

  • Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves.
  • The high pitched ultrasonic squeaks of the bat are reflected from the obstacles or prey and returned to bat’s ear.
  • The nature of reflections tells the bat where the obstacle or prey is and what it is like.
  • The bats use ultrasound for navigation and location of the food in dark.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 19.
With the help of a diagram describe how compression and rarefaction pulses are produced in air near a source of sound. (AS 5)
Answer:

  • Consider a vibrating membrane of a musical instrument like a drum or tabla.
  • As it moves back and forth, it produces a sound.
  • The figure shows the membrane at different instants and the condition of the air near it at those instants.
  • As the membrane moves forward, it pushes the particles of air in the layer in front of it.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound 9

  • So, the particle of air in the layer get closer to each other, hence the density increases.
  • This layer of air pushes and compresses the layer next to it and so on.
  • We call this disturbance as compression pulse.
  • When the membrane moves backward, it drags back the layer of air near it. Hence the density decreases.
  • The particles of air in the next layer on the right move into fill this less dense area.
  • This is a rarefaction pulse moves to right.
  • As the membrane moves back and forth repeatedly, compression and rarefaction pulses are produced, one after the other.
  • These two pulses travel one behind the other, carrying the disturbance with it.

Question 20.
How do echoes in a normal room affect the quality of the sounds that we hear? (AS 7)
Answer:

  • Echo is a reflected sound, arriving at the position of listener more than 0.1s after the direct sound.
  • Quality is the characteristic of a sound which enables us to distinguish between musical notes emitted by different musical instruments.
  • In a normal room, if echo is formed, we can hear multiple sounds, at same time.
  • Our ear cannot perceive and judge the sound from where it is coming.
  • So, quality of sound does not work here.

Question 21.
Explain the working and applications of SONAR. (AS 1)
Answer:

  • SONAR stands for Sonographic Navigation And Ranging.
  • This is a method for detecting and finding the distance of objects under water by means of reflected ultrasonic waves.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound 2
Working of SONAR:

  1. SONAR system consists of a transmitter and a detector in the “Observation Centre” on board of a ship.
  2. From the observation centre, ultrasonic waves of high frequency are sent in all directions under the water through transmitter.
  3. These waves travel in straight lines till they hit an object such as a submarine, a sunken ship, etc.
  4. The waves are then reflected and are received back by the receiver at the observation centre.
  5. The study of these reflected waves gives information about the direction of the object located.
  6. The time between sending ultrasonic wave and receiving its echo, the distance of the object is calculated.
  7. Reflections from various angles can be utilized to determine the shape and size of the object.

Mathematical expression :

  1. Let’d’ be the distance between SONAR and an underwater object.
  2. ‘t’ be the time between sending an ultrasonic wave and receiving its echo.
  3. ‘u’ be the speed of sound in water.
  4. The total distance covered by the wave from the SONAR to the object and back is 2d.
  5. From the equation s = ut ⇒ 2d = ut ⇒ d = \(\frac{ut}{2}\).

Application:
Marine geologists use this method to determine the depth of the sea and to locate underwater hills and valleys.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 22.
Find the time period of a source of a sound wave whose frequency is 400 Hz. (AS 1)
Answer:
Frequency υ = 400 Hz
Time period T =?
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 3

Question 23.
A sound wave travels at a speed of 340 m/s. If its wavelength is 2 cm, what is the frequency of the wave? Will it be in the audible range? (AS 1)
Answer:
Speed of sound v = 340 m/s.; Wavelength λ = 2 cm = 0.02 m.
Frequency υ =?
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 4
The audible range of sound wave is 20 Hz to 20 kHz.
Hence this is in the audible range.

Question 24.
Given that sound travels in air at 340 m/s, find the wavelength of the waves in air produced by a 20 kHz sound source. If the same source is put in a water tank, what would be the wavelength of the sound waves in water? (AS 7)
Speed of sound in water = 1,480 m/s.
Answer:
In air:
Speed of sound wave (v) = 340 m/s ; Frequency of source of sound (o) = 20 kHz
Wavelength of the sound wave λ =?
v = υλ
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 5
∴ Wavelength of the sound wave in air = 17 m

Same source is kept in water :
∴ Speed of sound in water (v) = 1480 m/s
Frequency of sound wave (p) = 20 kHz
Wavelength of sound wave λ =?
v = υλ
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 6
∴ Wavelength of the sound wave in water = 74 m

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 25.
A man is lying on the floor of a large, empty hemispherical hall, in such a way that his head is at the centre of the hall. He shouts “Hello!” and hears the echo of his voice after 0.2 s. What is the radius of the hall ? (Speed of sound in air 340 m/s) (AS 7)
Answer:
Let the distance travelled by the sound wave = 2d m
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 12

As the head of the man is at the centre of hemispherical room, then ‘d’ is the radius of the hall.
∴ Radius of the hall = 34 m

Question 26.
“We know that sound is a form of energy. So, the large amount of energy produced due the sound pollution in cosmopolitan cities can be used to our day-to-day needs of energy. It also helps us to protect biodiversity in urban areas”. Do you agree with this statement? Explain.
Answer:

  1. Sound is a form of mechanical energy.
  2. So, the mechanical energy can be converted into electrical energy.
  3. Experiments are going on this concept.
  4. If this is successful, we have the following benefits.
    a) Sound pollution can be controlled.
    b) Conventional methods of producing electrical energy from coal or water will cause in loss of biodiversity. This can be avoided.
    c) Natural resources like water can be protected.
    d) Increasing needs of energy can be overcome by this method.

Question 27.
How do you appreciate efforts of a musician to produce melodious sound using a musical instrument by simultaneously controlling frequency and amplitude of the sounds produced by it.
Answer:

  • The sounds which produce pleasing effect on the ear are called musical sounds.
  • Any instrument which produces musical sound is called musical instrument.
  • The person who plays a musical instrument to produce melodious sound is called a musician.
  • The musician must have control on breathing, concentration on the output of the sound, which is a very hard task.
  • For this the musician needs a lot of practise.
  • With the musician’s practise and knowledge over musical notes only we can hear melodious sound otherwise it could only be a noise.
  • Hence the efforts of a musician are highly appreciable.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 28.
You might have observed that sometimes your pet dog starts barking though no one is seen near in its surroundings or no disturbance heard nearby. Does this observation raise any doubts in your mind about the peculiar behaviour of dog after your understanding about ‘range of hearing the sound’. If yes, write them.
Answer:

  • Dogs can hear sounds of frequencies up to about 50 kHz, which is ultrasonic.
  • After hearing this ultrasonics, a dog will bark panicly, though no one is seen near.
  • I understood this after studying about ‘range of hearing the sound’.
  • Before the knowledge of ‘range of hearing the sound’, I felt that the dogs are barking by seeing some devils, which is a misconception.
  • Scientific knowledge helps us to know reasons for many misconcepts.

Question 29.
Find out the names of animals (and their photographs from internet) which communicate using infra-sonic or ultra-sonic sound and prepare a scrap book.
Answer:
Scrap book
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 10

 

9th Class Physical Science 11th Lesson Sound InText Questions and Answers

9th Class Physical Science Textbook Page No. 184

Question 1.
How does sound reach our ears from the source of its production?
Answer:
Soufid travels in the form of waves. It reaches our ears from the source of its production, in the form of waves.

Question 2.
Does it travel by itself or is there any force bringing it to our ears?
Answer:
Sound does not travel by itself. When a sound is produced, the kinetic energy of the source, vibrates the nearest particles in the medium. These particles transfers energy and finally, it reaches our ears.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 3.
What is sound? Is it a force or an energy?
Answer:
Sound is a form of energy.

Question 4.
Why don’t we hear sounds when our ears are closed?
Answer:
When our ears are closed, the energy in the form of waves reaches our ear, but it cannot make the eardrum to vibrate. Hence we cannot hear the sound.

Question 5.
Why is the light ray dancing, after sound is made in the tin?
Answer:
The dancing of light ray, after the sound is made in the tin shows that the sound travels in the form of waves from the source of its production.

Question 6.
What do you infer from this?
Answer:
Sound is a form of energy which can travel in the form of waves through the medium.

Question 7.
Can we say that sound is a form of mechanical energy?
Answer:
Yes, sound is a form of mechanical energy.

9th Class Physical Science Textbook Page No. 185

Question 8.
Do you hear any sound?
Answer:
We cannot hear any sound.

Question 9.
Do you see any vibrations in the tuning fork?
Answer:
Yes, we can sense the vibrations in the tuning fork.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 10.
What do you conclude from the above activity?
Answer:
We observe that vibrating tuning fork produces sound.

Question 11.
Can you produce sound without vibration in the body?
Answer:
We cannot produce sound without vibration in the body.

Question 12.
Give some examples of vibrating bodies which produce sound.
Answer:
Drums, tabla, calling bell, school bell, etc.

Question 13.
What part of our body vibrates when we speak?
Answer:
When we speak, vocal cord vibrates in our body.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 14.
Do all vibrating bodies necessarily produce sound?
Answer:
All vibrating bodies produce sound, but we cannot hear some of them, due to the limit of audible frequency.

9th Class Physical Science Textbook Page No. 186

Question 15.
If sound travels in the form of a wave then what is the pattern?
Answer:
Sound travels in the form of longitudinal or transverse waves in the air or in the other material.

9th Class Physical Science Textbook Page No. 188

Question 16.
What do you say about sound waves in air by the above activity?
Answer:
From the above activity, we can say that there involves change in the density of medium while sound waves are travelling in air.

Question 17.
Are they longitudinal or transverse?
Answer:
Sound waves in air are longitudinal.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 18.
Does sound get reflected at the surface of a solid?
Answer:
Yes, sound gets reflected at the surface of a solid as in the case of reflection of light.

Question 19.
What happens if you lift your tube slightly above the table?
Answer:
If we lift the tube slightly we cannot hear the sound clearly.

Question 20.
Are able to listen to the sound? If not why?
Answer:
We are unable to hear the sound. If we lift one of the pipes then the pipe carrying incident sound, the pipe carrying reflected sound will not be in the same plane. Hence we cannot hear the sound.

9th Class Physical Science Textbook Page No. 195

Question 21.
Do hard surfaces reflect sound better than soft ones?
Answer:
Generally, hard surfaces reflect sound better than soft surfaces. But sound reflects quite well from rough surfaces than polished surfaces.

9th Class Physical Science Textbook Page No. 187

Question 22.
Do compressions and rarefactions in sound wave travel in same directions or in opposite directions? Explain.
Answer:

  • Compressions and rarefactions in a sound wave will be in opposite direction.
  • In a compression, all the particles come close, so the density and pressure increases.
  • In a rarefaction, all particles drag back, so the density and pressure decreases.
  • In a microscopic view of particle, the compression and rarefaction travel in opposite directions.

9th Class Physical Science Textbook Page No. 191

Question 23.
Does the frequency of sound waves depend on the medium in which it frawels? How?
Answer:
Yes.
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 7
2) As speed of wave differs from medium to fnedium, the frequency also changes, keeping the wavelengths constant.

Question 24.
The frequency of source of sound is 10 Hz. How many times does it vibrate in one minute?
Answer:
Number of vibrations per second = 10
Number of vibrations in one minute = 10 × 60 = 600

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 25.
Gently strike a hanging bell (temple bell) and try to listen to the sound produced by it with a stethoscope keeping it both at bottom portion and top portion of the bell. Is the pitch and loudness of the sound same at the two portions? Why?
Answer:
No. The bob of the bell strikes at the bottom portion of the bell. Hence the pitch and loudness are high.

Question 26.
During a thunderstorm if you note a 3 second delay between the flash of lightning and sound of thunder. What is the approximate distance of thunderstorm from you ?
Answer:
Time taken to reach the sound = 3 s ; Speed of sound in air = 343.2 m/sec.
Distance of thunderstorm = 343.2 x 3 = 1029.6 m

9th Class Physical Science Textbook Page No. 194

Question 27.
Two girls are playing on identical stringed instruments. The strings of the both instruments are adjusted to give notes of same pitch. Will the quality of two notes be same? Justify your answer.
Answer:
If the two girls are playing with same instruments, then the quality is same. If the girls are playing with different stringed instruments the quality will be different.

Quality is the characteristic which enables us to distinguish between musical notes emitted by different musical instruments.

Question 28.
What change, would you expect in the characteristic of a musical sound when we increase its frequency one instance and amplitude in another instance?
Answer:
When frequency is increased, the pitch of sound increases. When amplitude is increased, its loudness increases.

9th Class Physical Science Textbook Page No. 195

Question 29.
What could be the reason for better reflection of sound by rough surfaces than polished surfaces?
Answer:

  • Sound reflects better on rough surfaces than polished surfaces.
  • The rough surface reflects sound in all directions, so it can reach in many directions.

9th Class Physical Science Textbook Page No. 196

Question 30.
Why is an echo weaker than the original sound?
Answer:

  • Echo is the reflected sound.
  • While it travels back, it gradually loses its energy.
  • Hence the echo is weaker than original sound.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound

Question 31.
In a closed box if you say hello, the sound heard will be Hellooooo ………. What does it mean?
Answer:
In a closed box, the multiple reflections of sound has no way to go out side. Hence we will hear the sound as hellooooooo ………

9th Class Physical Science Textbook Page No. 197

Question 32.
What is the advantage of having conical openings in horns, megaphones, etc?
Answer:
The conical openings in horns, megaphones, etc. will guide the reflected sound waves in forward direction and spreads towards the audience.

Question 33.
Why do we put cushions on the chairs, carpet on the floor, straw materials on the walls in cinema halls?
Answer:

  • These materials absorb unnecessary reflections of sound, so that we can hear more clearly.
  • As these material absorb reflected sound waves, the people outside the cinema halls, cannot hear the sound produced in the cinema hall.

9th Class Physical Science Textbook Page No. 199

Question 34.
What is the benefit of using ultrasound over light waves in the above applications?
Answer:

  • Light waves cannot penetrate in the internal organs like liver, kidney, etc.
  • Light wave after reflection do not form image.
  • Ultrasound waves can penetrate through internal organs like liver, kidney, etc.
  • After reflection, ultrasound waves produce image. So, in ultrasonography and surgeries, ultrasound waves are better than light waves.

9th Class Physical Science 11th Lesson Sound Activities

Activity 1

Question 1.
How can .you say that the sound is a form of energy?
Answer:

  1. Take a tin can and remove both ends to make a hollow cylinder.
  2. Take a balloon and stretch it over the can.
  3. Wrap a rubber band around the balloon.
  4. Take a small mirror and stick it on the balloon.
  5. Take a laser light and let it fall on the mirror.
  6. After reflection the light spot is seen on the wall.
  7. Now shout directly into the open end of the can and observe the dancing light.

AP Board 9th Class Physical Science Solutions Chapter 11 Sound 13
Observations:

  1. When sound is made, the energy produced from the sound vibrates the membrane of the balloon, resulting in the dancing of light ray.
  2. This shows that sound is a form of energy which travels in air.

Activity – 2

Question 2.
Prove that the sound is produced from a vibrating source.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 14

  1. Attach a small piece of steel wire to one of the prongs of a tuning fork as shown in the figure.
  2. Strike the tuning fork with a rubber hammer.
  3. While it is vibrating, try to draw a straight line on a piece of smoked glass as quick as possible with, it.
  4. Keep the end of the wire in such a way just it touches the glass.
  5. A line is formed in the form of a wave.
  6. Repeat the experiment when the tuning fork is not vibrating and observe the difference in the line.

Observations:

  1. We have produced vibrations in the tuning fork by striking it with a hammer.
  2. Thus the vibrating tuning fork produces sound.
  3. Thus the sound is produced by vibrating bodies.

Activity – 3

Question 3.
а) How do you demonstrate the formation of compressions and rarefactions in a slinky?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 15

 

  1. Take a slinky.
  2. It is a spring-shaped toy which can be extended or compressed very easily.
  3. Lay it down on a table or the floor as shown in figure.
  4. Ask a friend to hold one end.
  5. Pull the other end to stretch the slinky and then move it to and fro along its length.
  6. We will see alternate compressions and rarefactions of the coil.
  7. This is similar to the pattern of varying density produced in a medium when sound passes through it.

b) Mow do you demonstrate the formation of crests and troughs in a slinky?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 16

  1. Hang a slinky from a fixed support.
  2. Hold it gently at the lower end and quickly move your hand sideways and back.
  3. This will cause a hump on the slinky near the lower end.
  4. The hump travels upwards on the slinky as shown in the figure.
  5. The humps formed alternately are known as crests and troughs.

Activity – 4

Question 4.
Describe an activity to listen the reflected sound.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 11 Sound 17

  1. Take two long, identical tubes and place them on
  2. Ask your friend to speak softly into one tube while you use the other tube to listen.
  3. Adjust the tube until you hear the best sound.
  4. You will find that you hear your friend’s voice best when the tube makes equal angles with a normal to the wall.
  5. This shows that reflection of sound follows the laws of reflection of light.
  6. Lift your tube slightly above the table.
  7. You will not be able to listen the voice clearly, because the plane carrying the incident wave and reflected wave are changed.
  8. Repeat the experiment by placing flat objects of different materials (steel and plastic trays, a card board, a tray wrapped with cloth, etc.) against the wall and observe the changes in the sound.

Observations:

  1. Reflection of sound follows the laws of reflection of light.
  2. When the plane carrying incident wave and reflecting wave changes, the reflected sound cannot be heard clearly.
  3. Hard or rough surfaces reflect the sound better than soft surfaces.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 8th Lesson Gravitation

9th Class Physical Science 8th Lesson Gravitation Textbook Questions and Answers

Improve Your Learning

Question 1.
What path will the moon take when the gravitational interaction between the moon and earth disappears? (AS 2)
Answer:
The force of attraction between moon and earth is given by F = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
M = mass of the earth ; m = mass of the moon ; R = radius of the earth
Here the gravitational interaction between moon and earth disappears.
∴ G = 0 ⇒ F = 0

  • Therefore the moon neither revolves around the earth nor fall into the earth.
  • It takes a straight path away from the earth.

Question 2.
A Car moves with constant speed of 10 m/s in a circular path of radius 10 m. The mass of the car is 1000kg. Who or what is providing the required centripetal force for the car? How much is it? (AS 1)
Answer:
Speed of the car (v) = 10 m/s ; Radius of the path (r) = 10 m
Weight of the car (m) = 1000kg
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 1
The required centripetal force is provided by the friction between the tyres of the car and the road.

Question 3.
A small metal washer is placed on the top of a hemisphere of radius R. What minimum horizontal velocity should be imparted to the washer to detach it from the hemisphere at the initial point of motion? (AS 1, AS 7)
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 11
Answer:
Radius of hemisphere = R; Mass of hemisphere = M
Let the radius of washer = r and mass = m
Distance between hemisphere and washer = R + r
The centripetal force required to rotate the washer = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
The gravitational force of washer due to hemisphere is = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
But the necessary centripetal force must be equal to the gravitational force
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 12

Question 4.
Explain why a long pole is more beneficial to the tight rope walker if the pole has slight bending. (AS 1, AS 7)
Answer:

  • You must have noticed the circus artists doing rope walking.
  • During this act, they carry a long bamboo pole in their hands.
  • The reason for this is that the line joining the centre of gravity and centre of equilibrium must fall within the rope for achieving the stable equilibrium.
  • Thus when an artist finds that he is falling towards left, he shifts bamboo pole towards right, so that his centre of gravity stay undisturbed.
  • Thus he can balance himself on the rope.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 5.
Why is it easier to carry the same amount of water in two buckets, one in each hand rather than in a single bucket? (AS 7)
Answer:
It is because in the later case centre of gravity of our body shifts towards the bucket and there is a tendency that the line joining the centre of gravity and centre of equilibrium may fall outside our feet.

However in the former case the centre of gravity not only gets lowered, but also it is at such a point that line joining the C.G. and C.E. falls within our feet.
Hence one is a stable equilibrium.

Question 6.
What is the speed of an apple dropped from a tree after 1.5 second? What distance will it cover during this time? Take g = 10 m/s². (AS 1)
Answer:
An apple is dropped from a tree.
∴ Initial velocity u = 0 ; Time t = 1.5 s
a = g = 10 m/s2 ; Final velocity v = ?
v = u + at = 0 + 10 x 1.5 = 0 + 15 = 15 m/s
Distance covered (s) = ut + \(\frac{1}{2}\) at² = 0 × 1.5 + \(\frac{1}{2}\) × 10 × 1.5 × 1.5 = 0 + 5 × 2.25 = 11.25 m

Question 7.
A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entrie time of motion? What is the velocity at 5 seconds after the projection? Take g = 10 m/s². (AS 1)
Answer:
Initial speed u = 40 m/s ; g = 10 m/s²
Maximum height reached (h) = \(\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{40 \times 40}{2 \times 10}\) = 80 m
Entire time of motion (T) = \(\frac{2 \mathrm{u}}{\mathrm{g}}=\frac{2 \times 40}{10}\) = 8 s
Entire time of motion is 8 seconds.
∴ It starts to fall down after 4 seconds.
At 5 seconds the body is in downward direction.
u = 0 m/s, a = g = 10 m/s², t = 5 – 4 = 1 sec.
v = u + at = 0 + 10 × 1 = 10 m/s
∴ The velocity at 5 seconds is 10 m/s downward.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 8.
A boy is throwing balls into the air one by one in such a way that when the first ball thrown reaches maximum height he starts to throw the second ball. He repeats this activity. To what height do the balls rise if he throws twice in a second? (AS 1, AS 7)
Answer:
The boy throws the second ball when the first ball reaches its maximum height. He throws twice in a second.
Time of ascent of first ball is 1/2 sec.
After 1/2 sec, the first ball starts to fall down and the second ball starts from ground.
Let the distance travelled = s meters.; Initial velocity = u m/s
Time of ascent (t1) = 1/2 sec.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 13
They reach a height of 15/4 meters.

Question 9.
A man is standing against a wall such that his right shoulder and right leg are in contact with the surface of the wall along his height. Can he raise his left leg at this position without moving his body away from the wall? Why? Explain. (AS 7)
Answer:

  • When the right leg and right shoulder are in contact with the surface of the wall along the height of a man, his weight is towards the wall.
  • The centre of gravity will be away from the foot. Just like in the case of carrying a bucket full of water with one hand.
  • The line joins the centre of gravity and centre of equilibrium is not perpendicular to the horizontal.
  • Hence he cannot lift his left leg without moving his body along the wall.

Question 10.
A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the height from which it is dropped. Take g = 10 m/ s². (AS 1)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 6
For last 6 m, distance travelled
s = 6 m ; u = ? ; t = 0.2 sec ; a = g = 10 m/s²
s = ut + \(\frac{1}{2}\) at²
6 = u (0.2) + \(\frac{1}{2}\) × 10 × (0.2)²
6 = (0.2) u + (5 × 0.04)
6 – 0.2 = 0.2 u
5.8 = 0.2 u ⇒ u = \(\frac{5.8}{0.2}\) = 29
∴ u = 29 m/s
This u will be the final velocity while travelling the distance x.
∴ s = x, v = 29 m/s, a = g = 10 m/s², u = 0 m/s
v² – u² = 2as ⇒ 29² – 0 = 2 × 10 × x ⇒ x = \(\frac{841}{20}\) = 42.05 m
∴ Total distance = x + 6 = 42.05 + 6 = 48.05 m
∴ The ball is dropped from a height of 48.05 m.

Question 11.
A ball is dropped from a balloon going up at a speed of 5 m/s. If the balloon was at a height 60 m. At the time of dropping the ball, how long will the ball take to reach the ground? (AS 1, AS 7)
Answer:
At t = 0, the stone was going up with a velocity of 5 m/s. After that it moves as a freely falling body, with downward acceleration ‘g’.
If it reaches the ground at time t1,
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 14

Question 12.
A ball is projected vertically up with a speed of 50 m/s. Find the maximum height, the time to reach the maximum height, and the speed at of the maximum height. (g = 10 m/ s²) (AS 1)
Answer:
Initial speed u = 50 m/s ; g = 10 m/s²
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 2
After reaching maximum height, the velocity becomes ‘zero’.

Question 13.
Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circle in equal time. What is the ratio of their speeds and centripetal accelerations? (AS 1)
Answer:
Masses of cars : m1 and m2; Radius of circles : r1 and r2
Given that their time period is equal.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 3

Question 14.
Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart. Find the gravitational force of attraction between them. (AS 1)
Answer:
Masses of balls M1 and M2 = 10 kg each.; Distance d = 10 cm = 0.1 m
Gravitational force of attraction between them is
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 4

Question 15.
Find the free-fall acceleration of an object on the surface of the moon, if the radius of the moon and its mass are 1740 km and 7.4 × 1022 kg respectively. Compare this value with free fall acceleration of a body on the surface of the earth. (AS 1)
Answer:
Radius of the moon = 1740 km = 1740 × 10³ m
Mass of the moon = 7.4 × 1022 kg ; G = 6.67 × 10-11 Nm² kg-2
Free fall acceleration of a body on the surface of the moon
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 7

Question 16.
Can you think of two particles which do not exert gravitational force on each other? (AS 2)
Answer:
Two particles which do not exert gravitational force on each other will be mass less particles. But every particle has even a little mass. Hence we cannot find two particles which do not exert gravitational force on each other.

Question 17.
An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration? (AS 7)
Answer:

  • According to Newton’s third law, when an apple is freely falling, the force on the apple due to earth is equal to the force on the earth due to apple.
  • The force is due to gravity, which causes acceleration in the body.
  • As the insect is inside the freely falling apple, it feels that the earth is falling towards it with an acceleration ’g’.
  • According to the insect, the acceleration is due to the force on the apple due to the earth.
  • Actually the earth is not falling towards the apple.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 18.
A scooter weighing 150 kg together with its rider moving at 36 kin/hr is to take a turn of radius 30 in. What force on the scooter towards the center is needed to make the turn possible ? Who or what provides this? (AS 1)
Answer:
Weight of the scooter with rider = 150 kg
Speed = 36 km/hr = 36 × \(\frac{5}{18}\) = 10 m/sec.
Radius of the turning = 30 m
The force needed on the scooter towards the centre is centripetal force.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 8
∴ A centripetal force of 500 N is required on the car.
This force is provided by the friction between the tyres of the car and road.

Question 19.
The bob of simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this moment. (Take g = 9.8 m/sec²) (AS 1)
Answer:
Mass of the bob = 100 g = 0.1 kg ; Length of the string = 1 m
Speed of the bob v = 1.4 m/s ; Let the tension in the string be T.
The forces acting on the bob are
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 9
a) Weight of the bob mg, downwards
b) Tension in the string ‘T’ upward
Weight of bob = \(\frac{\mathrm{mv}^{2}}{\mathrm{l}}\)
Tension in the string T = g cos θ
∴ According to Newton’s third law
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 10

Question 20.
How can you find the centre of gravity of India map made with steel? Explain. (AS 3)
(OR)
Describe an activity to know the center of gravity of a India map.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 5

  1. Take an India map made of steel.
  2. Make holes at four different corners i.e., top, bottom, left, and right, and name them as A, B, C, and D as shown in figure.
  3. Suspend the map through a hole A, by means of a string to a nail ‘P’ as shown in figure.
  4. Suspend a plumb line from the nail P.
  5. Draw the line AX along the plumb line.
  6. Similarly suspend the map through other holes B, C and D and draw lines BY, CZ, DW along the plumb line.
  7. These lines (more than two) intersect at a point.
  8. This point (G) is the centre of gravity of the map.

Question 21.
Explain some situations where the center of gravity of man lies out side of the body. (AS 1)
Answer:
Centre of gravity of a human being is located interior to the second sacral vertebra.
Centre of gravity of man lies out side the body in the following situation :
a) While doing sit-ups.
b) While carrying a load like bucket full of water with one hand.
c) While walking on a narrow base like walking on a rope or pole or narrow wall, etc.
d) While walking with one leg.
e) As age increases its position changes.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 22.
Where does the centre of gravity of atmosphere of the earth lie? (AS 2)
Answer:
The earth’s atmosphere is about 10,000 km thick, but most of its bulk is contained in the first 11 km above the earth’s surface. Since the earth and its atmosphere are roughly spherical, the centre of the earth is also the centre of gravity of the earth’s atmosphere.

Question 23.
Where does the center of gravity lie, when a boy is doing sit-ups? Explain. (AS 7)
Answer:

  • The centre of gravity of a boy when he start erect, falls in the foot.
  • When he is doing sit-ups, the centre of gravity shifts from foot to his base.
  • The weight vector also move from the base.
  • Hence the boy stretches his hands or bends slightly towards the earth. While doing sit-ups, in order to make the weight vector pass through base, so that he acquire stability.

9th Class Physical Science 8th Lesson Gravitation InText Questions and Answers

9th Class Physical Science Textbook Page No. 126

Question 1.
What is that force?
Answer:
The force acting on these objects to make them move around another object, instead of moving in straight line in the gravitational force.

Question 2.
Is the motion of the earth around the sun uniform motion?
Answer:
The earth takes 365.25 days to complete one rotation around the sun. Hence it is uniform motion.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 3.
Is the motion of the moon around the earth uniform motion?
Answer:
Moon takes 27.3 days to complete one rotation around the earth, which does not change. Hence it is uniform motion.

9th Class Physical Science Textbook Page No. 127

Question 4.
Does the velocity of the body change in uniform circular motion? Why?
Answer:
The velocity of the body in uniform circular motion is constant.

If the velocity is not constant, the time period changes from time to time and it cannot be treated as uniform circular motion.

Question 5.
Does the body in uniform circular motion have an acceleration? What is the direction of acceleration?
Answer:
The body in uniform circular motion have an acceleration, which is directed towards the centre of the circle, known as centripetal acceleration.

9th Class Physical Science Textbook Page No. 129

Question 6.
Do you know what questions arose in his mind from this observation?
Answer:
When apple fell to the ground, Newton might have thought like this:
a) Why did the apple fall on the earth?
b) Why doesn’t the apple go up in the sky?
c) Whether all the objects fall to ground ?
d) What makes them to fall on the ground?

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 7.
Why did the apple fall to the ground?
Answer:
The apple fell to the ground due to the gravitational attraction of the earth.

Question 8.
Why does the moon not fall to the ground?
Answer:
The gravitational force in the moon due to earth is equal to the gravitational force on the earth due to moon. Hence the moon does not fall to the ground.

Question 9.
What makes the moon to move in a circular orbit around the earth?
Answer:
The gravitational force between moon and earth act as centripetal force and makes the moon to revolve around the earth in uniform circular motion.

9th Class Physical Science Textbook Page No. 135

Question 10.
How do you feel during the free-fall of your body from the height?
Answer:

  • I feel weightlessness during the free-fall of my body from the height.
  • This is due to the gravitational force of earth on the body.

9th Class Physical Science Textbook Page No. 136

Question 11.
Try to balance ladder on your shoulder. When does it happen?
Answer:
When the centre of gravity of the ladder and the centre of gravity of our body lies in the same line then we can balance the ladder on our shoulders.

9th Class Physical Science Textbook Page No. 128

Question 12.
Can an object moVe along a curved path if no force acts on it?
Answer:

  • If a body has to move in a curved path, its direction of velocity must change continuously.
  • This is done by the centripetal force.
  • Hence an object cannot move along a curved path if no force acts on it.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 13.
As a car speeds up when rounding a curve, does its centripetal acceleration increase? Use an equation to defend your answer.
Answer:
Centripetal acceleration \(a_{c}=\frac{v^{2}}{r}\)
As v increases, its centripetal acceleration also increases.

Question 14.
Calculate the tension in a string that whirls a 2 kg – toy in a horizontal circle of radius 2.5 iv when it moves at 3 m/s.
Answer:
Mass of the toy m = 2 kg ; Radius of the circle = 2.5 m ; Speed of the toy = 3 m/s
As the toy is moving in a horizontal circle, the necessary centripetal force is provided due to the tension in the string.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 31

9th Class Physical Science Textbook Page No. 131

Question 15.
In figure, we see that the moon falls around earth rather than straight into it. If the magnitude of velocity were zero, how would it move?
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 27
Answer:
If the magnitude of velocity were zero, the moon would move towards earth due to acceleration due to gravity.

Question 16.
According to the equation for gravitational force, what happens to the force between two bodies if the mass of one of the bodies doubled?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 28
∴ The gravitational force will be doubled.

Question 17.
If there is an attractive force between all objects, why we do not feel ourselves gravitating toward massive buildings in our vicinity?
Answer:

  • Earth is massive than the building.
  • Hence the gravitational force between ourselves and earth is more than that of between ourselves and building.
  • Hence we do not feel gravitating towards massive building in our vicinity.

Question 187.
Is the force of gravity stronger on a piece of iron than on a piece of wood if both have the same mass?
Answer:
Yes. The force of gravity is stronger on a piece of iron then on a piece of wood if both have the same mass.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 19.
An apple falls because of the gravitational attraction of the earth. What is the gravitational attraction of apple on the earth?
Answer:

  • We know that acceleration due to gravity, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
    Here M is the mass of the earth.
  • In this mass of the object which is freely falling has no effect on ‘g’.
  • Hence the gravitational attraction of apple on the earth is negligible.

9th Class Physical Science Textbook Page No. 133

Question 20.
Give an example for the motion of an object of zero speed and with non-zero acceleration?
Answer:
Protons and neutrons inside the nucleus.

Question 21.
Two stones are thrown into air with speeds 20 m/s, 40 m/s respectively. What accelerations are possessed by the objects?
Answer:
Stone -1 :
Initial velocity u = 20 m/s
After a time t, it reaches to ground then final velocity v = 0 m/s
Accelerahon = \(\frac{0-20}{t}=\frac{-20}{t} \mathrm{~m} / \mathrm{s}^{2}\)

Stone – 2 :
Initial velocity u = 40 m/s
Final velocity v = 0 m/s
Accelerahon = \(\frac{0-40}{t}=\frac{-40}{t} \mathrm{~m} / \mathrm{s}^{2}\)

9th Class Physical Science Textbook Page No. 135

Question 22.
When is your weight equal to mg?
Answer:
When we are on the surface of the earth, our weight is equal to ‘mg’ on earth.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 23.
Give example of when your weight is zero?
Answer:

  1. When we are at the centre of the earth, our weight is zero.
  2. When we are freely falling, we feel weightlessness.

9th Class Physical Science Textbook Page No. 138

Question 24.
Where does the centre of gravity of a sphere and triangular lamina lie?
Answer:
1) Centre of gravity of a sphere is a point where the whole mass of the sphere is assumed to be concentrated which is called centre.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 29

2) Centre of gravity of a triangular lamina lies where the whole mass of the triangle is assumed to be concentrated which is centroid.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 30

Question 25.
Can an object have more than one centre of gravity?
Answer:
Basing on the position or distribution of mass, an object can have more than one centre of gravity.

Question 26.
Why doesn’t the leaning tower of Pisa topple over?
Answer:

  • The centre of gravity of the leaning tower of Pisa is very close to earth.
  • Even though it is leaning, the line of action of total weight passes through the base. The base area of leaning tower of Pisa is very large. So it doesn’t topple over.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Question 27.
Why must you bend forward when carrying a heavy load on your back?
Answer:
If we bend forward, then the line of action of total weight passes through the body. So we are stable. Hence we must bend forward when carrying a heavy load on our back.

9th Class Physical Science 8th Lesson Gravitation Activities

Activity – 1 Uniform circular motion

Question 1.
Describe an activity to observe uniform circular motion.
Answer:
Uniform Circular Motion :
Uniform circular motion is a motion of the body with a constant speed in circular path.

Material required :
Electric motor, old C.D., small wooden block, battery, connecting chords, stop clock.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 15

Procedure:

  1. Take an electric motor and fix a disc to the shaft of the electric motor.
  2. Place a small wooden block on the disc, as shown in figure.
  3. Switch on the motor.
  4. Find the time required to complete ten revolutions by the block.
  5. Repeat the same two to three times.
  6. We observe that the wooden block moves in a circular path with a constant speed.
  7. So, this motion of wooden block is called uniform circular motion.

Activity – 2

Question 2.
Define centripetal acceleration and derive an expression for centripetal acceleration.
Answer:
Centripetal acceleration :
The acceleration which can change only the direction of velocity of a body is called centripetal acceleration.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 16
Derivation :

  1. Let a body is in uniform circular motion.
  2. Draw velocity vectors at successive time intervals as shown in the figure.
  3. Transfer tails of each velocity vector to coincide at a single point without changing their direction as shown in the figure.
  4. In the above figure, the directed line joining two vectors represents change in velocity (Av).
  5. Let us consider the change in velocity during the course of a complete revolution of a body i.e., the sum of the magnitudes of the changes in velocity during a complete revolution will be equal to the sum of the sides of the depicted polygon.
  6. The smaller the sides of our polygon, the closer they cling to the circle of radius v, consequently the magnitude of change in velocity of the body during the course of revolution, will be equal to the circumference ’2πv’ of the circle.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 17

9) As the vertex angle of isosceles triangle decreases, the angle between the change in velocity approaches to 90°.

10) Therefore, the acceleration of a body in uniform circular motion is directed perpendicular to its velocity.

Activity – 3 Free fall

Question 3.
Accelerahon is independent of masses.
Place a small paper on a book. Release the book with the paper from certain height from the ground.
a) What is your observation?
Answer:
The book and paper will fall on the ground at same time, if the paper is kept in the book. If the paper is kept on the book and released, then the paper will be separated from book and falls on the ground little bit later than the book.

b) Drop the book and paper separately, what happens?
Answer:
Book reaches the ground first, then the paper reaches.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Activity – 4

Question 4.
What is the direction of ‘g’?
Throw a stone vertically up. Measure the time required for it to come back to earth’s surface with stop clock.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 18
a) What happens to its speed when it moves up and down?
Answer:

  • When the stone moves up, its velocity decreases gradually and finally it becomes zero, because it is moving against the direction of gravitation.
  • When its velocity becomes zero, it starts to fall down due to gravitational force of earth, and its velocity increases.

b) What is the direction of acceleration?
Answer:

  • While the stone is moving up, the direction of acceleration is in upward direction, which is against to the direction of gravitational force.
  • While the stone is falling down, the direction of acceleration is in downward direction, which is in the direction of gravitational force.

Activity – 5

Question 5.
Describe the method of measuring the weight of a free fail body.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 19

  • Take a spring balance and suspend it to the ceiling and put some weight to it.
  • Note the reading of the spring balance.
  • Now drop the spring balance with load from certain height to fall freely.
  • Carefully observe the change in the position of indicator on the spring balance scale while it is in free-fall.
  • We observe that the indicator of spring balance shows zero while it is in free-fall.
  • This is due to the weightlessness of a free-fall body.

Activity – 6

Question 6.
How do you demonstrate the changes during the free-fall of a body?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 20

  • Take a transparent tray and make holes on opposite sides.
  • Take two or three rubber bands and tie them tightly, close to each other between the holes.
  • Now place a stone on the bands as shown in the figure.
  • We observe that the bands bend due to the weight of the stone.
  • Now drop the tray with stone.
  • We observe that the bands do not bend. They straight.
  • When in equilibrium on a firm surface, weight is evidenced by a support force. Then weight equals to mg.
  • When the body falls freely then it experiences weightlessness.
  • Even in this weightless condition, there is still a gravitational force acting on the body causing downward acceleration.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Activity – 7

Question 7.
Give one activity that showing center of gravity and balancing.
(OR)
How do you balance spoon and fork?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 21

  • As shown in the figure arrange a fork, spoon and wooden match stick together.
  • The combination will balance nicely on the edge of the glass.
    The weights of the spoon and fork will pass through the centre of gravity and the base (glass).

Activity – 8

Question 8.
Can you get up without bending?
Sit in a chair comfortably as shown in figure. Try to get up from the chair without bending your body or legs.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 22
a) Are we able to do so? If not why?
Answer:

  1. We cannot get up from the chair without bending our body or legs.
  2. When we sit in a chair, we are in weightless condition.
  3. We need a support force to get up from the chair.
  4. This support force can be obtained by bending our body or legs.

Activity – 9

Question 9.
How do you balance a ladder on your shoulder?
Answer:

  • Take a ladder.
  • Find the mid stick by counting the sticks of the ladder.
  • Put that mid stick on the shoulder and balance the ladder.
  • That is the center of gravity of a ladder.

Activity – 10

Question 10.
How can you locate the centre of gravity of a uniform object?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 23

  1. Take a meter scale.
  2. Suspend it from various points.
  3. We observe that it bends to one side.
  4. Suspend it from the mid point.
  5. We observe that the scale will be in horizontal position without bending to any side.
  6. At this point, the scale behaves as if its entire weight is concentrated at this point.
  7. The support given at this single point gives support to the entire scale.
  8. This point is its centre of gravity.

Activity – 11

Question 11.
Identifying the centre of gravity of a ring.
The center of gravity of any freely suspended object lies directly beneath the point of suspension.

If a vertical line is drawn through the point of suspension, the center of gravity lies somewhere along that line. To determine exactiy where it lies along the line, we have only to suspend the object from the some other point and draw a second vertical line through that point of suspension. The center of gravity lies where the two lines intersect.
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 24

a) Where does the center of gravity of a ring lie?
Answer:
The centre of gravity of a ring lies at its geometric centre.

b) Does the center of gravity of a body exist outside the body?
Answer:
For the bodies which are in right angled triangle shape, the centre of gravity falls out side the body. Hence we can say the CG of a body may be outside or inside the body.

c) Does center of gravity of an object exist at a point where there is no mass of the object?
Answer:
The centre of gravity exist at a point where there is no mass of the object, for the objects like annual rings.

AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation

Activity – 12

Question 12.
Shift of the centre of gravity and its effects.
Try to touch your toes as shown in figure (a). Try this again when standing against a wall as shown in figure (b).
AP Board 9th Class Physical Science Solutions Chapter 8 Gravitation 25
a) Are you able to touch your toes in second case as shown in figure? If not why?
Answer:

  • We cannot touch our toes in the second case as shown in figure.
  • When we stand erect with the support of wall, the centre of gravity shifts its place from second sacral vertebra to the place near to abdomen.

b) What difference do you notice in the center of gravity of your body in above two positions ?
Answer:

  • In the first position, the centre of gravity is at a place near to the centre of back-bone.
  • In the second position, the centre of gravity shifts from the place near to the centre of backbone to the place near to abdomen.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 10th Lesson Work and Energy

9th Class Physical Science 10th Lesson Work and Energy Textbook Questions and Answers

Improve Your Learning

Question 1.
Define work and write its units. (AS 1)
Answer:
Work : Work can be defined as the product of the force (F) and the distance (s) moved along the direction of the force.
Work done = Force × displacement = F × s
[This formula is used in only translatory motion of the object]

  1. Work is a scalar quantity.
  2. Unit of work is ‘N – m’ or ‘Joule’ (J).

Question 2.
Give few examples where displacement of an object is in the direction opposite to the force acting on the object. (AS 1)
Answer:

  • When a ball is thrown up, the motion is in upward direction, whereas the force due to earth’s gravity is in downward direction.
  • If a ball is moving on plain ground, will get stopped after sometime, due to frictional force acting on it in opposite direction.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 3.
Identify the wrong statement among the following. Rewrite them by making necessary corrections. (AS 1)
a) Work and energy have different units.
Answer:
Correction :
Work and energy have same units.

b) When an aeroplane takes off, the work done by its weight is positive.
Answr:
Correction : When an aeroplane takes off, the work done by its weight is negative.

c) The potential energy of spring increases when it is extended and decreases when it is compressed.
Answer:
Correct.

d) If the work done by external forces on a system is negative then the energy of the system decreases.
Answer:
Correct.

e) When a body is falling freely from a height, its kinetic energy remains constant.
Answer:
Correction : When a body is falling freely from a height, its K.E. increases and P.E. decreases, but the total energy of the body remains constant.

f) The unit of power is watt.
Answer:
Correct.

Question 4.
What is mechanical energy? (AS 1)
Answer:
The sum of the kinetic energy and the potential energy of an object is called its mechanical energy.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 5.
State the principle of conservation of energy. (AS 1)
Answer:
Energy can neither be created nor destroyed. It can only be changed from one form to another.

Question 6.
When you lift a box from the floor and put it on an almirah the potential energy of the box increases but there is no change in its kinetic energy. Is it violation of conservation of energy? Explain. (AS 7)
Answer:

  • When the box is on the floor, it has no K.E., but has some P.E.
  • Its P.E. increases as it is lifted and put on an almirah.
  • In this event, the energy produced by the person converts into K.E. and the total M.E. remains same.
  • Hence it is not the violation of conservation of energy.

Question 7.
One person says that potential energy of a particular book kept in an almirah is 20 J and other says it is 30 J. Is one of them necessarily wrong? Give reasons. (AS 2, AS 1)
Answer:
Two cases will arise in this situation.
Case – I :
One of them will be wrong, if both of them kept the book in the same shelf of the almirah.

Case – II :
Both of them will be correct, if the first person keeps the book in lower shelf and the second one keeps the book in upper shelf of the almirah.

Question 8.
In which of the following cases is the work done positive or zero or negative? (AS 1)
a) Work done by the porter on a suitcase in lifting it from the platform on to his head.
Answer:
Positive

b) Work done by the force of gravity on suitcase as the suitcase falls from porter’s head.
Answer:
Positive

c) Work done by the porter standing on platform with suitcase on his head.
Answer:
Zero

d) Work done by force of gravity on a ball thrown up vertically up into the sky.
Answer:
Negative

e) Work done by force applied by hands of a man swimming in a pond.
Answer:
Negative

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 9.
What is potential energy? Derive an equation for gravitational potential energy of a body of mass ‘m’ at a height ‘h’. (AS 1)
Answer:
Potential energy:
The energy possessed by an object because of its position or shape is called its potential energy.
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 1

Potential energy of a body at height (Or) Gravitational potential energy :

  1. The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.
  2. Consider an object of mass ‘m’ raised to height ’h’ from the ground.
  3. A force is required to do this.
  4. The minimum force required to raise the object is equal to the weight of the object (mg).
  5. The object gains energy equal to the work done on it.
  6. Let the work done on the object against gravity be ‘W’.
    ∴ W = Force × displacement
    = mg × h
    W = mgh
  7. The object gains the energy equal to the mgh’ units.
  8. This energy is the potential energy of the object at a height ‘h’.
    ∴ P.E. = mgh

Question 10.
When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground? (AS 7)
Answer:

  • Let us imagine an apple of mass ‘m’ falls from a height h.
  • Before starting to fall down, its velocity is zero. Hence its energy is purely potential.
  • When it falls down, its gravitational potential energy gradually converts into kinetic energy, but the entire energy in the system remains same.
  • When it strikes the ground, its energy is fully kinetic, till its velocity becomes zero.

Question 11.
Let us assume that you have lifted a suitcase and kept it on a table. On which of the following does the work done by you depend or not depend? Why?
a) The path taken by the suitcase
b) The time taken by you in doing so
c) The weight of the suitcase
d) Your weight
Answer:
We know that P.E. = mgh
a) The work done by you does not depend on the path taken by the suitcase, because the height from the ground to the top of the table is fixed in this case, (h constant)

b) The work done by you does not depend the time taken, in doing so, because the work done is same. If you take less time your power will be more otherwise the power is less.

c) The work done by you depends on the weight of the suitcase. As the weight of the suitcase increases, the work done will also increase (See the formula P.E. = mgh).

d) The work done by you depends on your weight, we know that a person with more weight can Mft the suitcase easily.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 12.
When you push your bicycle up an incline, the potential energy of the bicycle and yourself increase. Where does this energy come from? (AS 7)
Answer:
The energy inside our body causes increase in the P.E. of the bicycle and ourself.

As we are pushing the bicycle upon incline, the work done by the gravitational force is negative. Hence the energy loses.

Question 13.
Why does a person standing for a long time get tired when he does not appear to be doing any work? (AS 7)
Answer:
1) A person standing for a long time, does not appear to be doing any work, because there is no displacement.
2) Eventhough work has not been done, the person gets tired.
3) This is due to the utilization of energy produced in his body.

Question 14.
What is kinetic energy? Derive an expression for the kinetic energy of a body of mass ‘m’ moving at a speed ‘v’. (AS 1)
Answer:
Kinetic energy :
The energy possessed by an object due to its motion is called kinetic energy.

Numerical expression for K.E.:
1) Let us assume that an object of mass (m) is at rest on a smooth horizontal plane as shown in figure.
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 2
2) Let be displaced through a distence ‘s’ from the point A to B by a force (F) acting upon it in the direction of the displacement.
3) In the horizontal direction the net force ‘Fnet‘ is equal to the force applied ‘F’.
∴ W = Fnet.s = F.s …………. (1)
4) Let the work done on the object cause a change in its velocity from ‘u’ to ‘v’ anr the ‘acceleration produced be ‘a’.
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 4

8) As we Have assumed that object is at rest, then the initial velocity u = 0.
∴ W = \(\frac{1}{2}\) mv²

9) We know that K.E. of a body moving with certain velocity is equal to work done on the object to acquire that velocity from rest.

10) Thus the K.E. of an object of mass’m’ and moving with velocity V is equal to
K.E = \(\frac{1}{2}\) mv²

Question 15.
A free-fall object eventually stops on reaching the ground. What happens to its kinetic energy? (AS 1)
Answer:

  • Its kinetic energy becomes ‘zero’.
  • As the ball stops on reaching the ground, its final velocity ‘v’ becomes zero.
  • Hence the K.E. = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\).m.0 = 0

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 16.
A man carrying a bag of total mass 25 kg climbs up to a height of 10 m in 50 seconds. Calculate the power delivered by him on the bag. (AS 1)
Answer:
Mass of the bag = 25 kg ; Height ‘h’ = 10 m
Acceleration due to gravity g = 9.8 m/s²
The P.E. of the man = mgh = 25 × 9.8 × 10 = 2450 J
The work done by the man is equal to the P.E. possessed by the man.
∴ The work done by the man (W) = 2450 J ; Time (t) = 50 sec.
∴ Power delivered by the man on the bag = \(\frac{\mathrm{W}}{\mathrm{t}}=\frac{2450 \mathrm{~J}}{50 \mathrm{sec}}\) = 49 watts

Question 17.
A 10 kg ball is dropped from a height of 10 m. Find (a) the initial potential energy of the ball, (b) the kinetic energy just before it reaches the ground and (c) the speed just before it reaches the ground.
Answer:
Mass of the ball = 10 kg ; It is dropped from a height h = 10 m
Acceleration due to gravity g = 9.8 m/sec²
a) The initial P.E. of the ball = mgh = 10 × 9.8 × 10 = 980 J
b) We know that the energy is conserved in a freely falling object at anywhere in its journey.
∴ The K.E. just before it reaches the ground = 980 J

c) K.E. just before it reaches the ground = 980 J ;
Speed = v m/s
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 6

Question 18.
Calculate the work done by a person in lifting a load of 20 kg from the ground and placing it 1 m high on a table. (AS 1)
Answer:
Mass of a load = 20 kg ; Height h = 1 m
Acceleration due to gravity = 9.8 m/s²
The work done by a person in lifting the load is equal to the P.E.
P.E. = mgh = 20 × 9.8 × 1 = 196 J

Question 19.
Find the mass of a body which has 5 J of kinetic energy while moving at a speed of 2 m/s. (AS 1)
Answer:
KE = 5 J
Speed v = 2 m/sec
Mass of the body = m kg.
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 5

Question 20.
A cycle together with its rider weighs 100 kg. How much work is needed to set it moving at 3 m/s?
Answer:
Mass of the bicycle with its rider = 100 kg ; Speed = 3 m/s Amount of work needed to set it moving is equal to its K.E.
K.E. = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 100 × 3 × 3 = 450 J
∴ 450 J work is needed.

Question 21.
When the speed of a ball is doubled its kinetic energy (AS 1)
A) remains same
B) gets doubled
C) becomes half
D) becomes 4 times
Answer:
D) becomes 4 time

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 22.
Two bodies of unequal masses are dropped from the top of a building. Which of the following is equal for both bodies at any instant?
A) Speed
B) Force of gravity
C) Potential energy
D) Kinetic energy
Answer:
B) Force of gravity

Question 23.
A man with a box on his head is climbing up a ladder. The work done by the man on the box is ……….
A) Positive
B) Negative
C) Zero
D) Undefined
Answer:
A) Positive

Question 24.
A porter with a suitcase on his head is climbing up steps with uniform speed. The work done by the “weight of the suitcase” on the suitcase is ……
A) Positive
B) Negative
C) Zero
D) Undefined
Answer:
B) Negative

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 25.
How will the increasing energy needs and conservation of energy influence international peace, cooperation and security? Discuss.
Answer:
1) The increasing energy needs and conservation of energy influence international peace, co-operation and security.

2) For example, many great nations depend on the gulf countries and many countries for fuel (Petrol, diesel, etc.). The need of fuel brings co-operation among the countries. Also the need of energy (electricity, fuel, etc.) link all nations to a small circle so that there comes the peace, co-operation and security.

3) Atomic energy, thermal energy, chemical energy, etc. bring unity among all nations.

Question 26.
How would you assess the role of energy conversion occurring naturally in maintaining ecological balance of nature?
Answer:

  • The chemical changes in the sun is providing us the heat energy and light energy to the earth.
  • The heat energy of the sun evaporates the water into water vapour. This helps us to get heavy rainfall.
  • The solar energy is taken by the plant to change CO2, water and chloroplast into starch which is the food of the plant. This process is called photosvnthesis.
  • In the photosynthesis process the plant releases O2 (oxygen).
  • The leguminous (Pea plants) plants help the bacteria present at their roots. In return they help the plants to fix nitrogen in them The nitrogen cycle is maintained.
  • From the above examples we can understand that the energy conversion is helpful in maintaining ecological balance of nature.

Question 27.
Collect pictures showing various situation where potential energy possessed by an object depends on its shape and position. Prepare a scrap book.
Answer:
Note : Students can collect pictures of different objects at different heights, with different shapes.
e.g.: 1) A stone on a hill.
2) A book on a table and a pen on a table, etc.
Paste these pictures in your scrap book.

Question 28.
Draw a diagram to show conservation of mechanical energy in case of free falling body.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 3

9th Class Physical Science 10th Lesson Work and Energy InText Questions and Answers

9th Class Physical Science Textbook Page No. 161

Question 1.
How are these works being done?
Answer:
All the works mentioned above are being done by utilizing human muscular energy or electrical energy.

Question 2.
What do you need to do these works?
Answer:
Both human beings or machines need energy to do work.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 3.
Where does the energy spent go ultimately?
Answer:
The energy spent will be converted into another form to complete the task.

Question 4.
Is there any transfer of energy while work is being done?
Answer:
Yes. We can observe the transfer of energy while work is being done. For example, in fan, electrical energy is converted into mechanical energy. In heater, electrical energy is converted into heat energy, etc.

Question 5.
Can we do any work without transfer of energy?
Answer:
No, we cannot do any work without transfer of energy.

9th Class Physical Science Textbook Page No. 162

Question 6.
What is work?
Answer:
When a force is applied on an object and a displacement is observed, the work is said to be done.

Question 7.
Why is there difference between general view of work and scientific view of work?
Answer:
In general view, work is said to be done while reading, writing, eating or even standing for some time. We cannot observe any displacement in these cases. But in scientific view, work is said to be done when there is a displacement in the direction of force applied.

9th Class Physical Science Textbook Page No. 163

Question 8.
Observe the following examples.
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 14 AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 15 AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 16 AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 17
a) Are all the people mentioned in the above examples doing work ?
Answer:
No.
In example (1) and (2) there is displacement. Hence they are doing work.
In example (3) and (4) there is no displacement. Hence they are not doing any work.

b) How do you define work?
Answer:
Work is defined as the product of force and displacement.
W = F × s

9th Class Physical Science Textbook Page No. 165

Question 9.
What would be the work done when the force on the object is zero?
Answer:
We know work done W = F × s
Here F = 0
⇒ W = 0 × s = 0
∴ The work done is zero.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 10.
What would be the work done when the displacement of the object is zero?
Answer:
Work done W = F × s
Here s = 0
⇒ W = F × 0 = 0
∴ The work done is zero.

Question 11.
Can you give some examples, where the displacement of the object is zero?
Answer:

  • When a boy pushes a wall, the displacement in the wall is zero.
  • When a boy stands for some time at one place, then there is no displacement in the boy.
  • Wind causes the swing in trees, but the tree does not have any displacement.

9th Class Physical Science Textbook Page No. 166

Question 12.
What happens to the speed of the ball while it moves up?
Answer:
The speed of the ball while it moves up decreases gradually, due to negative gravitational force acting on it.

Question 13.
What is the speed at its maximum height?
Answer:
The speed of the ball at its maximum height is zero.

Question 14.
What happens to the speed of the ball during its downward motion?
Answer:
During its downward motion, the ball starts from zero velocity*and its velocity gradually increases due to the positive gravitational force acting on it.

9th Class Physical Science Textbook Page No. 167

Question 15.
What is energy?
Answer:
Energy is the capability to do work.

Question 16.
How can we decide that an object possess energy or not?
Answer:
The capacity of doing work by an object on another object depends on position and state of the object which is doing work.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 17.
What will happen to the plate? Why?
Answer:
The plate breaks into pieces, because of the potential energy of the metal ball.

Question 18.
What changes do you notice? Why?
Answer:
The toy car is at rest before winding the key but the same toy gets energy to move when the key attached to it is wound up.

9th Class Physical Science Textbook Page No. 168

Question 19.
What could be the reason?
Answer:

  • The reason for breaking up of the plate is the potential energy of the metal ball.
  • The reason for moving the toy car is the potential energy of the spring in the toy.

Question 20.
Where does this energy go?
Answer:
The energy is transferred from one object to another.

Question 21.
Is there any energy transfer between the object doing the work and the object on which work has been done?
Answer:
Yes. In case of the toy car, the potential energy of the wounded spring is transferred to car and made the toy car to move.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 22.
Can any force do work without energy transfer?
Answer:
No.

9th Class Physical Science Textbook Page No. 169

Question 23.
Where do we get energy from?
Answer:
The Sun is the biggest natural and primary source of energy for us. Many other secondary sources are derived from the sun. We can also get energy from the interior of the earth and from tides of the sea.

Question 24.
Can you think of other sources of energy?
Answer:
The other natural sources of energy are wind energy, tidal energy, gravitational energy, etc. The artificial sources of energy are electrical energy, heat energy, muscular energy, etc.

Question 25.
Are there sources of energy which are not dependent on the Sun?
Answer:
Energy can exists in several forms like mechanical energy, light energy, thermal energy, sound energy, electrical energy, magnetic energy, etc. These energies, though they are artificial, they ultimately depend on the Sun.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 26.
Do you know why a person gets tired standing at a place for long time?
Answer:
Though the person standing is not doing any work externally, a lot of work is being done inside the body.

The muscles of the body become stretched when he stands for long time and heart has to pump more blood to muscles. This leads to loss of energy inside the body and hence he gets tired.

9th Class Physical Science Textbook Page No. 176

Question 27.
How do green plants produce food?
Answer:
Green plants produce food from sunlight, by means of photosynthesis.

Question 28.
How are fuels like coal and petroleum formed?
Answer:
Dead plants hurried deep below the earth’s surface for millions of years get converted to fuels like petroleum and coal which have chemical energy stored in them.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 29.
What kind of energy conversions sustain the water cycle in nature?
Answer:
Water cycle and conversation of energies :
1) The Sun heats water in oceans and seas. Then water evaporates as water vapour into the air.
Here potential energy converts into kinetic energy.

2) Rising air currents take the vapour up into the atmosphere. There cooler temperatures cause it to condense into clouds.
Here kinetic energy converts into potential energy.

3) Air currents move water vapour around the globe, cloud particles collide, grow and fall out as rain.
Here potential energy converts into kinetic energy.

4) Run off and ground water are stored as fresh water in lakes. Some precipitation falls as snow or hail.
Here kinetic energy converts into potential energy.

9th Class Physical Science Textbook Page No. 178

Question 30.
Do all of us do the work at the same rate?
Answer:
The rate of work done by all of us will not same, because the capacity to do work differs from person to person depending on the weight, nature of work, etc.

Question 31.
Is the energy spent by the force doing work the same every time?
Answer:
No. The energy spent by the force doing work depends on the capacity of the machine or the capacity of a person.

Question 32.
Do the machines consume or transfer energy at same rate every time while doing a particular work?
Answer:
No. For example, consider two grinders of different wattages. The grinder with high wattage can perform fast. In this the energy transfer is faster. The grinder with less wattage can perform slowly. In this the energy transfer is slow.

9th Class Physical Science Textbook Page No. 179

Question 33.
Whose argument is correct?
Answr:
Raheem’s argument is correct, in view of work done.
Second labourer’s argument is correct in view of working hours.

Question 34.
Is work done in two cases same?
Answer:
The work done in two cases is same.

AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy

Question 35.
Why is there a change in rate of doing work?
Answer:
The rate of doing work is different, as the two labourers completed in different times.

9th Class Physical Science Textbook Page No. 165

Question 37.
A wooden chair is dragged on the level floor and brought to the same place. Let the distance covered be ‘s’ and frictional force acted on the chair by the floor be T. What is the work done by the frictional force?
Answer:
The displacement of the chair is zero. Hence the work done is zero.

9th Class Physical Science Textbook Page No. 166

Question 38.
Lift an object up from the ground. Work done by the force exerted by you on the object moves it in upward direction. Thus the force applied is in the direction of displacement. However there exists a force of gravitation on the object at the same time
• Which one of these forces is doing positive work?
• Which one is doing negative work?
• Give reasons.
Answer:

  • Force applied by us on the object, because the object moves in the direction of force applied.
  • Force of gravitation on the object is doing negative work. Here the object is moving against to gravitational force.

9th Class Physical Science Textbook Page No. 169

Question 39.
What would happen if nature does not allow the transfer of energy? Discuss with few examples.
Answer:
If nature could not allow the transfer of energy, the normal life is not possible, e.g.:
1) If the solar energy is not transferred into chemical energy, in which plants prepare their food, we cannot find any plant on the earth. Life without plants is impossible. We can not get food, cloth, shelter and fresh air for our survival.

2) Water cycle is also an example of transfer of energy in nature. If this does not take place, ground water cannot be recharged, we cannot find any water further.

9th Class Physical Science Textbook Page No. 172

Question 40.
a) Why is it easier to stop a lightly loaded truck than heavier one that has equal speed?
Answer:
Due to its light weight, the truck with lighter load is easy to stop.

b) Does the kinetic energy of a car change more when it goes from 10 m/s to 20 m/s or when it goes from 20 m/s to 30 m/s?
Answer:
Case – I :
K.E.(I) = \(\frac{1}{2}\) × m × 10 × 10 = 50m J
K.E.(F) = \(\frac{1}{2}\) × m × 20 × 20 = 200m J
Difference in K.E. = 200m – 50m = 150m J

Case – II :
K.E.(I) = \(\frac{1}{2}\) × m × 20 × 20 = 200m J
K.E.(F) = \(\frac{1}{2}\) × m × 30 × 30 = 450m J
Difference in K.E. = 450m – 200m = 250m J

c) A person starts from rest and begins to run. The runner puts a certain momentum into himself. What is the momentum of ground? And the runner puts a certain amount of kinetic energy into himself. What is the kinetic energy of the ground?
Answer:
Here two cases arise.
Case – I :
In view of the observer, ground has no velocity. Hence the momentum and kinetic energy of the ground will be zero.

Case – II :
In view of the runner, ground has a velocity equal to the velocity of the runner.

  1. The momentum of ground is equal to the momentum of the runner in magnitude but opposite in direction.
  2. The K.E. of ground is also equal to the K.E of the runner in magnitude but opposite in direction.

9th Class Physical Science Textbook Page No. 175

Question 41.
Does the international space station have gravitational potential energy?
Answer:
International space station is situated in space where the gravitational force is zero. Hence the gravitational potential energy is zero.

9th Class Physical Science Textbook Page No. 178

Question 42.
a) Someone wanting to sell you a super ball claims’ that it will bounce to a height greater than the height from which it is dropped. Would you buy this ball? If yes explain, if not explain.
Answer:
Yes. I will buy this ball. When the applied force is more, the ball bounce to a height greater than the hefght from which it is dropped.

b) A ball, initially at the top of the inclined hill, is allowed to roll down: At the bottom its speed is 4 m/s. Next, the ball is again rolled down the hill, but this time it does not start from rest. It has an initial speed of 3 m/s. How fast is it going when it gets to the bottom?
Answer:
A ball start from the top of an inclined plane u} = 0 m/s; v, = 4 m/s.
Let the acceleration a’ and distance ’s’.
v1² – u1² = 2as …………. (1)
Now the ball is rolled down from the top of an inclined plane with a velocity 3 m/s.
∴ u2 = 3 m/s; v2 =?
Here the acceleration of the inclined plane is ‘a’ and the distance is ‘s’.
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 13

9th Class Physical Science Textbook Page No. 179

Question 43.
The work done by a force F1 is larger than the work done by another force F2. Is it necessary that power delivered by F1 is also larger than that of F2? Why?
Answer:
Power delivered by F1 is may or may not be larger than F2.
Two cases arise.
Case – I :
If the displacement is same, as F1 > F2, the power delivered by F1 is larger than F2.

Case – II :
If the displacement is different, power delivered by F1 may not be larger than F2.

9th Class Physical Science 10th Lesson Work and Energy Activities

Activity – 1

Question 1.
Read Ex : 1, 2, 3, and 4 from page 162 and 163. Now fill the following table.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 7

Activity – 2

Question 2.
Explain the relation between work done on an object and energy of the object with an activity.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 8

  1. Take a hard spring and keep it on the table as shown in the figure.
  2. Now compress the spring with your palm and release it after few seconds.
  3. We will notice that when a spring is being compressed there is a change in its size.
  4. When it is released it gains some energy and may even jump from the table.
  5. The work done by your palm on the spring increases its energy and makes it to jump.
  6. Thus we can conclude that the object which does work loses energy and the object on which work has been done gains energy.

Activity – 3

Question 3.
List the energy sources.
Answer:
The main source of energy is the sun. Many other secondary sources of energy are electrical energy, magnetic energy, chemical energy, muscular energy, geothermal energy, energy from fossil fuels, etc.

Activity – 4

Question 4.
Explain the energy of moving objects with an activity.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 9

  1. Consider a metal ball and a hollow plastic block which are kept on a table side by side.
  2. Now, suppose that the ball is separated from the block and brought to one end of the table and pushed to roll on the table with speed ‘v’.
  3. We may notice that, the ball hits the plastic block and displaces it from point A to B.
  4. Thus a moving ball is more energetic than the ball at rest.
  5. We can conclude that a body possesses more energy when it is moving than when it is at rest.

Activity – 5

Question 5.
Explain potential energy with the help of a bow.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 10

  1. Take a bamboo stick and make a bow as shown in figure.
  2. Place an arrow made of a light stick on it with one end supported by the string of the bow and stretch the string gently and release the arrow.
  3. We notice that the arrow gets separated from the bow and falls down on the ground.
  4. Now place the arrow on the bow with one end supported by the string, applying more force and release the arrow.
  5. We observe that the arrow flies with great speed into the air.
  6. From this we can conclude that the bow in normal shape is not able to push the arrow, but when we stretch the string, it acquires energy to throw the arrow into air with great speed.
  7. The energy acquired by the bow due to change in its shape is known as its potential energy.

Activity – 6

Question 6.
Write an activity to show the energy in stretched rubber band.
Answer:
Take a rubber band hold it at one end and pull it from the other end. Then release the rubber band at one of the ends.

When we release the rubber band, it strikes our hand with great force. This force is due to the energy acquired by the rubber band by stretching it.

Activity – 7

Question 7.
Show that the object at some height possesses energy.
Answer:

  1. Take a heavy ball.
  2. Drop it on a thick bed of wet sand from different heights from 25 cm to 1.5 m.
  3. Observe the depression created by the ball on die bed of sand. Compare the depths of these depressions.
  4. When the ball is dropped from different heights, the depressions created by the ball on the bed of sand will be different. ‘
  5. As the height from which the ball was dropped increases, the depth of depression also increases.
  6. Thus we can conclude that a body at some height possesses energy.

Activity – 8

Question 8.
List out the energy conversions in nature and in day to day life.
Answer:
Discuss various ways of energy conversions in nature as well as in our day to day activities and make a separate list of situations for natural conversions of energy and energy conversions in day to day life and write them in the given tables.
Table

Sl.No. Situation of energy conversion in nature
1 Heat energy from the sun used for preparing food by plants gets con­verted into chemical energy.
2 Heat energy of the earth’s crust is useful in the formation of fossil fuels like petroleum, coal, etc.
3 Heat energy from the sun is converted into wind energy and wave energy.
4 Food we take will convert into energy.
5 Water cycle
Heat energy → Water vapour → water (rain)

Table

Situation of energy conversion Gadgets / appliances used for energy conversion
1. Conversion of electrical energy into mechanical energy Electric fan
2. Conversion of mechanical energy into electrical energy Dynamo
3. Conversion of chemical energy into electrical energy Cell
4. Conversion of chemical energy into mechanical energy Motor vehicles
5. Electrical energy into heat energy Electric stove
6. Electrical energy into light energy Bulb
7. Solar energy into electrical energy Solar lamp
8. Wind energy to electrical energy Wind mills
9. Gravitational potential energy to electrical energy Hydro electric power

Activity – 9

Question 9.
Write an activity which shows the conservation of mechanical energy.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 11

  1. Take a long thread say 50-60 cm long.
  2. Attach a small heavy object like a metal ball at one end.
  3. Tie other end to a nail fixed to the wall as shown in figure.
  4. Now pull the bob of the pendulum to one side to the position A1 and release it.
  5. The bob swings towards opposite side and reaches the point A2. It repeats the motion over and over again.
  6. The potential energy of the bob is minimum at A and reaches maximum at A1 because the height of the bob is maximum at that position.
  7. When the bob is released from this point (A1), its P.E. decreases and K.E. starts increasing slowly.
  8. When the bob reaches the position A, its K.E. reaches maximum, and P.E. becomes minimum.
  9. As the bob proceeds from A to A2, its P.E. increases slowly and becomes maximum at A2.
  10. The total P.E. and K.E. at any point on the path of motion during the oscillation of the pendulum.
  11. Thus the total mechanical energy in the system of pendulum remains constant.

Activity – 10

Question 10.
Calculate the total energy of free-fall at different heights.
An object of mass 20 kg is dropped from a height of 4 m. Compute the potential and kinetic energy in each case and write the values in the table. (Take g = 10 ms-2)
AP Board 9th Class Physical Science Solutions Chapter 10 Work and Energy 12
a) What do you say about total energy of system of freely falling body?
Answer:
In a freely falling body, the total energy of system (i.e. the sum of potential energy and kinetic energy) is same at any instance of its travel.

b) Is the mechanical energy conserved in the system?
Answer:
The sum of potential energy and kinetic energy at any instance of its travel is same. Hence we can say that the mechanical energy is conserved in the system.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 1 Motion Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 1st Lesson Motion

9th Class Physical Science 1st Lesson Motion Textbook Questions and Answers

Improve Your Learning

Question 1.
As shown in following figure, a point traverses the curved path.
Draw the displacement vector from given points A to B.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 1
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 2
As the point traverses from A to B, the displacement is the shortest distance between A and B. Hence the displacement vector will be as follows.

Question 2.
“She moves at a constant speed in a constant direction.” Rephrase the same sentence in fewer words using concepts related to motion. (AS 1)
Answer:
“She moves with constant velocity”.

Reason :
Constant speed in a constant direction is nothing but ‘constant velocity’.

Question 3.
What is the average speed of a Cheetah that sprints 100 m in 4 sec? What if it sprints 50 m in 2 sec? (AS 1, AS7)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 9

Question 4.
Correct your friend who says, “The car rounded the curve at a constant velocity of 70 km/h”. (AS 1)
Answer:
“The car rounded the curve at a constant speed of 70 km/h”.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Reason :
In a circular motion, speed remains constant but velocity changes.

Question 5.
Suppose that the three balls shown in figure below start simultaneously from the top of the hills. Which one reaches the bottom first? Explain.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 14
Answer:
Ball from first hill reaches the bottom first.
Reason :

  1. In the first hill, the ball has uniform rectilinear motion.
  2. So, the speed and velocity have same magnitude and direction.
  3. In the second and third hills, the ball takes curved path.
  4. So, the direction of velocity changes.

Question 6.
In the figure given below distance vs time graphs showing motion of two cars A and B are given. Which car moves fast? (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 3
Answer:
Car A moves fast.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 4
Reason :

  1. If we draw perpendiculars to X and Y axes from A and B respectively, we can observe that A covers large distance (S1) within a short time (t1).
  2. Find the slopes of the lines OA and OB at any instant. Slope of OA is high. Hence car A moves faster.

Question 7.
Draw the distance vs time graph when the speed of a body increases uniformly. (AS 5)
Answer:
Let us consider a car moves as shown in the table.

Time (t) sec Distance in meters
0 sec 0 meters
1 sec 3 meters
2 sec 6 meters
3 sec 9 meters
4 sec 12 meters
5 sec 15 meters

Now draw a s-t graph.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 16

Question 8.
Draw the distance-time graph when its speed decreases uniformly. (AS 5)
Answer:
Let us consider the movement of a car after applying brakes.

Time (t) sec Distance in meters
0 sec 20 meters
1 sec 18 meters
2 sec 16 meters
3 sec 14 meters
4 sec 12 meters
5 sec 10 meters

Now draw distance-time graph.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 16

Question 9.
A car travels at a speed of 80 km/h during the first half of its running time and at 40 km/h during the other half. Find the average speed of the car. (AS 1, AS 7)
Answer:
Let the total running time = x hrs
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 17

Question 10.
A car covers half the distance at a speed of 50 km/h and the other half at 40 km/h. Find the average speed of the car. (AS 1, AS 7)
Answer:
Let the total distance = x km.
First half is covered with a speed of 50 km/h.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 18

Question 11.
Derive the equation for uniform accelerated motion for the displacement covered in its nth second of its motion. (sn = u + a ( n – \(\frac{1}{2}\)) (AS 1)
Answer:
We know that distance travelled by an object in t seconds Is s = ut + \(\frac{1}{2}\) at²
∴ Distance travelled in ‘n’ seconds, s(n sec) = un + \(\frac{1}{2}\)an² ………. (1)
Distance travelled in (n – 1) seconds, s(n – 1) = u(n – 1) + \(\frac{1}{2}\) a(n – 1)² …….. (2)
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 10

Question 12.
A particle covers 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration. Find initial speed, acceleration and distance covered in next 2 sec. (AS 1, AS 7)
Answer:
Distance covered in first 5 sec = 10
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 19
To find the distance covered in next 2 sec, we have to find the initial speed after 8 sec
i.e., the final velocity after 8 sec.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 20

Question 13.
A car starts from rest and travels with uniform acceleration ‘α’, for some time and then with uniform retardation ‘β’ and comes to rest. The time of motion is “t”. Find the maximum velocity attained by it. (αβt/(α+β)) (AS 1, AS 7)
Answer:
Acceleration a = a m/sec²
Initial speed u = 0 m/sec
Let the time be t1 sec.
From equation v = u + at
⇒ v = 0 + αt1
\(\therefore \mathrm{t}_{1}=\frac{\mathrm{v}}{\alpha} \mathrm{sec}\)
Retardation a = – β m/sec²
Initial speed ‘u’ is equal to the final
velocity with acceleration ‘α’
= u = αt1 m/sec
Final velocity v = 0 m/sec
Let the time be t2 sec
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 21

Question 14.
A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/s², at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus? (AS 1, AS 7)
Answer:
Bus is at rest.
∴ u = 0; a = 1 m/sec²
Let the bus cover the distance ‘s’ in ‘n’ seconds.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 22
A man running with uniform velocity, v = 10 m/sec.
Distance covered by man in n seconds = 10 nm.
But after ‘n’ seconds the man catches the bus.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 23

∴ The minimum time in which the man catches the bus is 8 sec.

Question 15.
A body leaving a certain point “O” moves with a constant acceleration. At the end of the fifth second, its velocity is 1.5 m/s. At the end of the sixth second, the body stops and then begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point “0”. (AS 1)
Answer:
Velocity in 5th sec = 1.5 m/sec ; The body comes to rest in 6th sec.
∴ Final velocity in 6th sec, v = 0
∴ Acceleration in 6th sec is v = u + at ⇒ 0 = 1.5 + a. 1 ⇒ a = -1.5 m/sec²
[The velocity in 5th sec becomes the initial velocity for 6th sec]
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 11
After 6 sec, the body comes to rest.
∴ v = 0, a = -1.5 m/sec², u = ?, t = 6 sec.
v = u + at ⇒ 0 = u – 1.5 × 6 ⇒ u = 9 m/sec.
∴ Distance traversed by the body in 6 sec. i.e., before it stops.
s = ut + \(\frac{1}{2}\) at² = 9 × 6 + \(\frac{1}{2}\) × – 1.5 × 6² = 54 – 27 = 27m.
For backward journey,
u = 0 m/sec, t = 6 sec, a = -1.5 m/sec²
v = u + at ⇒ v = 0 – 1.5 × 6 ⇒ v = – 9
∴ Velocity for backward journey is – 9 m/sec.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 16.
Distinguish between speed and velocity.(AS 1)
Answer:

Speed Velocity
1. The distance covered in unit time is called average speed. 1. The displacement of an object per unit time is called average velocity.
2. Speed = \(\frac{\text { Distance }}{\text { Time }}\) 2. Velocity =\(\frac{\text { Displacement }}{\text { Time }}\)
3. Speed is scalar. 3. Velocity is vector.
4. Speed gives the idea of how fast the body moves. 4. Velocity gives the idea of how fast the body moves in specified direction.

Question 17.
What do you mean by constant acceleration? ((AS 1)
Answer:

  • Acceleration is the rate of change of velocity.
    2
  • It gives an idea how quickly velocity of a body is changing.
  • Acceleration is uniform, when in equal intervals of time, equal changes of velocity occurs.
  • For example, while driving a car, if we steadily increase the velocity from 30 km/h to 35 km/h in 1 sec and 35 km/h to 40 km/h in the next second and so on. In this case the acceleration is 5km/h, is said to be constant acceleration.
    AP Board 9th Class Physical Science Solutions Chapter 1 Motion 5

Question 18.
When the velocity is constant, can the average velocity over any time interval differ from instantaneous velocity at any instant ? If so, give an example; if not, explain why. (AS 2, AS 1)
Answer:
No. Here velocity is constant.
∴ Average velocity over any time interval is same and the instantaneous velocity at any instant is same.

Ex : Let us consider a car moves on a straight road with constant velocity say 10 m/s.
1) Now let the distance covered (AB) by the car in 1 s = 10 m.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 6

2) Distance covered in 2s (AC) = 20 m.
Average velocity from A to C is \(\frac{20 m}{2 s}\) = 10 m/s.

3) Instantaneous velocity at A or B or C at any point = 10 m/s.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 19.
Can the direction of velocity of an object reverse when its acceleration is constant? If so give an example; if not, explain why? (AS 2, AS 1)
Answer:
Yes. In case of a vertically projected body, while the body is moving up, the direction of velocity is upward, whereas while it is falling down, the direction of velocity is downward. Acceleration in both the cases is constant (numerically).

Question 20.
A point mass starts moving in a straight line with constant acceleration V’. At a time t after the beginning of motion, the acceleration changes sign, without change in magnitude. Determine the time t0 from the beginning of the motion in which the point mass returns to the initial position. (AS 1)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 45 AP Board 9th Class Physical Science Solutions Chapter 1 Motion 12

Question 21.
Consider a train which can accelerate with an acceleration of 20 cm/s² and slow down with deceleration of 100 cm/s². Find the minimum time for the train to travel between the stations 2.7 km apart. (AS 1)
Answer:
Let the Acceleration of the train a = 20 cm/s²
Deceleration of the train β = 100 cm/s²
Distance between the two stations s = 2.7 km = 27 × 104 cm
Let the minimum time for the train to travel between the two stations is t sec.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 13

Question 22.
You may have heard the story of the race between the rabbit and tortoise. They started from same point simultaneously with constant speeds. During the journey, rabbit took rest somewhere along the way for a while. But the tortoise moved steadily with lesser speed and reached the finishing point before rabbit. Rabbit woke up and ran, but rabbit realized that the tortoise had won the race. Draw distance vs time graph for this story. (AS 5)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 8

  1. OX – movement of tortoise.
  2. OABC – movement of rabbit
  3. Rabbit and tortoise start at O’.
  4. After time tj rabbit is at A and tortoise is at P.
  5. Rabbit takes rest up to time t2.
  6. After time t2, tortoise is at Q, but rabbit has no displacement.
  7. After time t3, the tortoise reaches the destination ‘X’.
  8. But rabbit reaches the destination after time t4.

Question 23.
A train of length 50 m is moving with a constant speed of 10 m/s. Calculate the time taken by the train to cross an electric pole and a bridge of length 250 m. (AS 1)
Answer:
Length of the train 50 m.; Speed of the train v = 10 m/s.
Distance travelled while crossing an electric pole = Length of the train = s = 50 m.
∴ Time taken to cross the electric pole ‘t’ = \(\frac{s}{v} \Rightarrow t=\frac{50}{10}\) = 5 s.
Length of the bridge = 250 m.
Distance travelled while crossing the bridge = Length of train + Length of bridge
= 50 + 250 = 300 m.
∴ Time taken to cross the bridge = \(\frac{300 \mathrm{~m}}{10 \mathrm{~m} / \mathrm{s}}\) = 30 sec.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 24.
Two trains each of having a speed of 30 km/h are headed at each other in opposite direction on the same track. A bird flies off one train to another with a constant speed of 60 km/h when they are 60 km apart till before they crash. Find the distance covered by the bird and how many trips the bird can make from one train to the other before they crash. (AS 1)
Answer:
Speed of each train = 30 km/hr
Speed of the bird = 60 km/hr
Distance between the two trains = 60 km
These two trains crash in one hour.
The bird flies a distance of 60 km till before the two trains crash.
The bird can make number of trips (infinity) before they crash.

Question 25.
A Stone dropped from top of a well reaches the surface of water in 2 seconds, find the velocity of stone while it touches the surface of water and what is the depth of the water surface from top of well (g=10m/s²) (Using V = U + at, S = Ut + 1/2 at²)
Answer:
Given that
t = 2s
u = 0 m/s [∵ free fall body]
v = ?
Depth s = ?
a = g = 10 m/s²

i) v = u + at
v = 0 + 10 × 2 = 20 m/s

ii) s = ut + – \(\frac{1}{2}\)at²
= 0 + \(\frac{1}{2}\) × 10 × 2²
= \(\frac{1}{2}\) × 10 × 4
= 20 m
Hence, velocity of stone while it touches the surface of water = 20 m/s
Depth of the water surface from the top of well = 20 m.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 26.
An object moving with 6m per second execute an acceleration 2 m/s² in next 3 seconds. How much distance it covered? (s = ut + 1/2 at²)
Answer:
u = 6 m/s; t = 3 sec; a = 2 m/s²
s = ut + \(\frac{1}{2}\) at²
= 6 × 3 + \(\frac{1}{2}\) × 2 × 3² = 18 + 9 = 27 m
The object covers 27 m in 3 sec.

Question 27.
A car stopped after travelling distance 8 m due to applying brakes at the speed of 40 m/s. Find acceleration and retardation of car in that period, (v² – u² = 2as)
Answer:
Here u = 40 m/s; v = 0 (vehical stopped); s = 8 m; a =?
v² – u² = 2as 0 – 40² = 2 × a × 8
a = \(\frac{-(40)(40)}{2 \times 8}\) =-100m/s
Acceleration = 100 m/s² with retordation on (-sign).

9th Class Physical Science 1st Lesson Motion InText Questions and Answers

9th Class Physical Science Textbook Page No. 1

Question 1.
If earth is in motion, why don’t we directly perceive the motion of the earth?
Answer:
Earth is in motion. We, the people on the earth also move with a speed equal to that of the earth. We cannot directly perceive the motion of the earth, because of this.

Question 2.
Are the walls of your classroom at rest or in motion? Why?
Answer:
The walls are at rest in view of our observation. When we discuss this in view of the motion of the earth, the walls are also in motion.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 3.
Have you ever experienced that the train in which you sit appears to move when it is at rest? Why?
Answer:
This happens when we sit in a stationary train and, the train on another track starts moving.

9th Class Physical Science Textbook Page No. 2

Question 4.
Why do we observe these changes?
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 39
Answer:
These changes are due to the point of observation. We know that earth is a sphere, the upward direction of the vertical position on its surface decisively depends upon the place on the earth’s surface, where the vertical is drawn.

Question 5.
Are the terms relative or not?
Answer:
The terms “longer”, “shorter”; “up” and “down”, etc. are relative to each other.

9th Class Physical Science Textbook Page No. 4

Question 6.
What answer may the passenger give to the driver?
Answer:
The car is in motion with respect to the observer on the road, but at rest with respect to the passenger. Because motion is a combined property of the observer and the body which is being observed.

Question 7.
How do we understand motion?
Answer:
A body is said to be in motion when its position is changing continuously with time relative to an observer.

9th Class Physical Science Textbook Page No. 6

Question 8.
Can you measure the average speed and average velocity?
Answer:
Yes, we can measure the average speed and average velocity.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 9.
How can you differentiate speed and average velocity?
Answer:

  1. Speed gives the idea of how fast the body moves.
  2. Velocity is the speed of an object in a specified direction.

9th Class Physical Science Textbook Page No. 7

Question 10.
Can you find the speed of the car at a particular instant of time?
Answer:
Yes, we can find the speed of the car at any instant of time by looking at its speedometer

Question 11.
What is the speed of the car at the instant of time ‘t3‘ for given motion?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 40
The instantaneous speed is represented by the slope of the curve at a given instant of time. We can find the slope of the curve at any point on it by drawing a tangent to the curve at that point. The slope of the curve gives speed of the car at that instant.

9th Class Physical Science Textbook Page No. 8

Question 12.
In what direction does an object move? Distance vs time graph
Answer:
The object moves in the direction tangential to the direction of the motion of the string.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 13.
Which motion is called uniform? Why?
Answer:
The motion of the body is said to be in uniform when its velocity is constant.

9th Class Physical Science Textbook Page No. 9

Question 14.
What is the shape of the graph?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 41
The shape of the graph for a body which is in uniform motion is a straight line as shown in the figure.

9th Class Physical Science Textbook Page No. 10

Question 15.
a) What is the shape of the graph?
Answer:
It is a curve.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 42

b) Is it a straight line or not? Why?
Answer:
The graph is not a straight line because the speed is changing irregularly.

Question 16.
Draw velocity vectors in the given figure at times t = 0, 1s, 2s, 3s.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 28
From the graph, we conclude that when the ball moves down the inclined plane its speed increases gradually but its direction remains constant.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 43

9th Class Physical Science Textbook Page No. 11

Question 17.
Draw velocity vectors at times t = 1s, 2s, 3s in the given figure.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 29
From the graph, we conclude that when the ball moves up the inclined plane its speed decreases gradually, but the direction of motion remains constant.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 44

Question 18.
Can you give few examples for motion of an object where its speed remains constant but velocity changes?
Answer:
For the bodies which are in uniform in circular motion the speed remain constant but velocity change. Ex : Rotation of earth, revolution of moon around the earth, etc.

Question 19.
Is the direction of motion constant? How?
Answer:
No, the direction of motion also changes continuously.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 20.
Can you give some more examples where speed and direction simultaneously change?
Answer:
Motion of a rocket, horizontally projected body, kicked football, a cricket ball bowled by a bowler, etc.

9th Class Physical Science Textbook Page No. 12

Question 21.
What is acceleration? How can we know that a body is accelerating?
Answer:

  1. Acceleration gives an idea how quickly velocity of a body changing.
  2. It is equal to the rate of change in velocity.
  3. While travelling in a bus or car, when the driver presses the accelerator, the passen¬gers sitting in the bus experience acceleration. Their bodies press against the seats due to acceleration.

Question 22.
At which point is the speed maximum?
Answer:
At B, the speed will be maximum.

Question 23.
Does the object in motion possess acceleration or not?
Answer:
Any object which is in motion possesses acceleration.

9th Class Physical Science Textbook Page No. 5

Question 24.
What is the displacement of the body if it returns to the same point from where it started? Give one example from daily life.
Answer:
When a body returns to the same point where it is started, then the displacement is zero.
Ex: A man starts from his home, goes to a market and returns home. Then his displacement is zero.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 25.
When do the distance and magnitude of displacement become equal?
Answer:
The distance and the magnitude of displacement become equal when the body moves along a straight line in one direction.

9th Class Physical Science Textbook Page No. 6

Question 26.
What is the average speed of the car if it covers 200 km in 5 h?
Answer:
Distance = 200 km ; Time = 5 h
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 34

Question 27.
When does the average velocity become zero?
Answer:
The average velocity of a body becomes zero when its displacement is zero.

Question 28.
A man used his car. The initial and final odometer readings are 4849 and 5549 respectively. The journey time is 25h. What is his average speed during the journey?
Answer:
Distance covered = 5549 – 4849 = 700 km.
Time = 25h.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 35

9th Class Physical Science Textbook Page No. 9

Question 29.
Very often you must have seen traffic police stopping motorists or scooter drivers who drive fast and fine them. Does fine for speeding depend on average speed or instantaneous speed? Explain.
Answer:
Instantaneous speed.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 30.
One airplane travels due north at 300 km/h and another airplane travels due south at 300 km/h. Are their speeds the same? Are their velocities the same? Explain.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 36

  1. Speed is same.
  2. Velocities are same in magnitude but differs in direction in the observer’s point of view.

Question 31.
The speedometer of the car indicates a constant reading. Is the car in uniform motion? Explain.
Answer:
Yes.

  1. The indicator in speedometer changes its position even for a small change in speed.
  2. As it indicates a constant reading, the car moves equal distances at equal intervals of time.
  3. Hence the motion is uniform.

9th Class Physical Science Textbook Page No. 11

Question 32.
An ant is moving on the surface of a ball. Does it’s velocity change or not? Explain.
Answer:
Velocity changes.
As the ant is moving on the surface of a ball, it has to go in circular motion.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion

Question 33.
Give an example of motion where there is a change only in speed but no change in direction of motion.
Answer:
Motion of a bus on the road.

9th Class Physical Science Textbook Page No. 13

Question 34.
What is the acceleration of a race car that moves at constant velocity of 300 km/h?
Answer:
Velocity = 300 km/h = \(300 \times \frac{5}{18}=\frac{500}{6}\) = 83.33 m/sec
As the velocity is constant, the acceleration is also constant.
∴ Acceleration = 83.33 m/sec².

Question 35.
Which has the greater acceleration, an airplane, that goes from 1000 km/h to 1005 km/h in 10s or a skateboard that goes from zero to 5km/h in 1 second?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 37

Question 36.
What is the deceleration of a vehicle moving in a straight line that changes its velocity from 100 km/h to a dead stop in 10 sec?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 38

Question 37.
Correct your Mend who says “Acceleration gives an idea of how fast the position changes.”
Answer:
“Acceleration gives the idea of how fast the position changes in a given direction.”

9th Class Physical Science 1st Lesson Heat Activities

Activity 1 Distance and Displacement

Question 1.
Draw a graph showing the difference between distance and displacement.
Answer:

  • Take a ball and throw it into the air with some angle to the horizontal.
  • Observe its path and draw it on paper.
  • The figure shows the path taken by the ball.
  • The distance ASB gives the distance travelled by the ball.
  • The length of \(\overrightarrow{\mathrm{AB}}\) gives the displacement of the ball.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion 24

Displacement:
Displacement is the shortest distance between initial and final points in a specified direction represented by a vector.

Distance :
Distance is the length of the path traversed by an object in a given time interval.

Activitie – 2 Drawing displacement vectors

Question 2.
Draw displacement vectors from A to B in the following situations.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 25
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 26

Activitie – 3 Measuring the average speed

Question 3.
How can the average speed be measured?
Answer:

  1. Select two positions (say A and B) 50 meters apart in the ground.
  2. Ask some students to stand at point A.
  3. Ask another group of students with stop watches to stand at B.
  4. When you clap your hand, the students at A start running towards the point B in any direction or path.
  5. At the same time the students at B start their stop watches.
  6. Observe that for each runner there is a student at B to measure the time taken for completing the race.
  7. Note the time taken by each student to cover the distance between the points A and B in the table given below.
    Student Time taken to reach B (Sec.) Average speed (50 ft) m/s
    A1 t1 ………
    A2 t2 ………
    A3 t3 ………
  8. The student who took the least time to reach B (from A) is said to be the fastest runner.
  9. The student who is fastest runner has the greatest average speed.
  10. Thus we measure the average speed of any moving body.

Activity – 4 Observing the direction of motion of a body

Question 4.
Show that the direction of velocity is tangent to the path at a point of interest when a body is in uniform circular motion.
Answer:

  • Carefully whirl a small object on the end of the string in the horizontal plane.
  • Release the object while it is whirling on the string.
  • We observe that the body along the tangent at the point where we released the body.
  • Try to release the object at different points on the circle and observe the direction of motion of object after it has been released form the string.
  • We will notice that the object moves on a straight line along the tangent to the circle at the point where we released it.

Activity – 5 Understanding uniform motion

Question 5.
Describe uniform motion.
Answer:
1) Consider a cyclist moving on a straight road.
2) The distance covered by him with respect to time is given in the following table.

Time (t in seconds) Distance (s in meters)
0 0
1 4
2 8
3 12
4 16

3) Draw distance vs time graph for the given values in the table.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 27
4) The graph will be as shown in the figure.
5) The straight line graph shows that the cyclist covers equal distances in equal inter¬vals of time.
6) If the direction of motion of the cyclist is assumed as constant, then we conclude that velocity is constant.
7) The motion of the body is said to be uniform when its velocity is constant.

Activity – 6 Observing the motion of a ball on an inclined plane

Question 6.
Describe an activity to explain the situation that “the speed changes but the direction of motion remains constant”.
Answer:
1) Set up an inclined plane as shown in the figure.
2) Take a ball and release it from the top of the inclined plane.
3) The positions of the ball at various times are shown in the figure given below.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 28
4) On close observation we find that when the ball moves down on the inclined plane, its speed increases gradually, and the direction of motion remains constant on inclined plane.
5) Now push the ball till it acquires certain speed and release it with that speed from the bottom of the inclined plane.
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 29
6) We observe that the ball moves upward to a certain distance and comes back to the bottom.
7) From this we conclude that the speed changes but the direction of motion remains constant.

Activity – 7 Observing uniform circular motion

Question 7.
Explain with an example where “speed remains constant, but its velocity changes”.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 30

  • Whirl continuously a stone which is tied to the end of the string.
  • Draw its path of motion and velocity vectors at different positions as shown in figure.
  • Assume that the speed of stone is constant.
  • We observe that the path of the stone is a circle, and the direction of velocity changes at every instant of time, but the speed is constant.
  • In this activity, we observe that though speed remains constant, its velocity changes.

Activity – 8 Observing the motion of an object thrown into air

Question 8.
Explain an activity to observe where speed and direction of motion change continuously.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 1 Motion 31

  • Throw a stone into the air by making some angle with the horizontal.
  • The path of the stone and velocity vectors are as shown in the figure.
  • Here we observe that the speed of stone is not uniform as it traverses different distances at different intervals of time and finally comes to rest.
  • The direction of motion is also not constant, as shown by the velocity vectors.
  • In this activity, we noticed that the speed and direc¬tion of motion both change continuously.

Lab Activity

Question 9.
Describe an activity to find the acceleration and velocity of an object moving on inclined track.
Answer:
Aim : To find the acceleration and velocity of an object moving on an inclined track. Materials required: Glass marbles, book, digital clock, long plastic tubes and steel plate.
Procedure:

  • Take a long plastic U type flat electrical wire cassing channel of length nearly 200 cm. Use this channel as track.
  • Mark the readings in cm, along the track.
  • Place one end of the track on a book and the other end on the floor.
  • Keep a steel plate on the floor at the bottom of the track.
  • Consider the reading at the bottom of the track as zero.
  • Take a marble having enough size to travel in the track freely.
  • Now release the marble freely from a certain distance say 40 cm.
  • Start the digital clock when the marble is released.
  • It moves down on the track and strikes the steel plate.
  • Stop the digital clock when a sound is produced.
  • Repeat the same experiment for the same distance 2 to 3 times and note the values of times in the table.

AP Board 9th Class Physical Science Solutions Chapter 1 Motion 32

  • Repeat the same experiment or various distances.
  • Draw s -1 graph for above values.
  • Do the above experiment by various slopes of the track and find acceleration in each case.

Conclusions :

  1. As the slope increases, acceleration increases.
  2. When iron block is used, we obtain the same conclusion as above. (The numerical values are less than the numerical values when marble is used)

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

These TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions will help the students to improve their time and approach.

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

Question 1.
What is venation ? What is its use?
Answer:
The arrangement of veins in the lamina of leaf is called venation. Venation acts as a skeleton of the leaf and gives it a shape and support.

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

Question 2.
What is transpiration?
Answer:
Plants release excess water in their body through stomata and some other parts as well. This water is released in the forms of vapour and this process is called transpiration.

Question 3.
What are the functions of root?
Answer:
Roots absorb water and minerals from the soil and also help in anchoring the plant body to the soil.

Question 4.
Name the types of venations we see in the plants.
Answer:
There are two types of venations in different plants.

  • Reticulate venation
  • Parallel venation

Question 5.
How many types of roots are present in plants and name them.
Answer:
There are two types of root systems.

  • Tap root system
  • Fibrous root system

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

Question 6.
What happens when stomata are not present in leaf?
Answer:
If stomata are absent in leaf plants cannot take carbon dioxide which is useful for preparation of food.
Excess of water taken by plants cannot he expelled out. Exchange of gases will be stopped. As a result plant cannot perform its functions.

Question 7.
What would happen if there are no veins in the leaf?
Answer:
If veins are absent the leaves would lose their specific shape and support from the stem. The food that is prepared in the leaf cannot be supplied to the other parts of the plant. The water and minerals coming from the roots (via) stem cannot reach the leaves due to absence of veins.

Question 8.
Prove the experiment. “The excess water is evaporated from plants by the process by transpiration”. With the help of the procedure write the result.
(Or)
Write the procedure you followed for conducting an experiment on transpiration. Write the required material for experiment. What did you observe at last?
Answer:
Material required for transpiration experiment: Well watered potted plant, 2 polythene bags, twine thread.

Procedure: A well watered potted plant is picked up from the school garden. One of the leafy branches of the plant is enclosed in a polythene bag and tied up at its mouth. Another polythene bag is tied up at its mouth without keeping plant. The preparations are kept in the sun for few hours.

Observation : It is observed that tiny droplets of water appear on the inner side of leafy branch enclosed in polythene bag, whereas the other bag has no such droplets.

Question 9.
Collect any 5 plants and observe their paris. Write your observations in the table as it is given below.
Answer:

S.No. Name of the plant Root
Yes/No
Stem
Yes/No
Leaves
Yes/No
Flowers
Yes/No

1. We have to observe the following.

  • Presence or absence of flowers, roots, leaf, skin.
  • Common parts in the collected plants.

I collected the following plants and tabulated as the following:
Answer:
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 1

Question 10.
Collect the following plants, observe the roots. Classify the type of root system in them. (Tap root system and fibrous root system)
Tabulate the information as given below. Name of the plants :
a. Grass
b. Tulasi
c. Ummetta
d. Maize (Jonnalu)
e. Beans.

S.No. Plant name Root system
Tap root system / fibrous

Answer:
I collected the given plants from the surroundings. I observed the root systems of the plants. They are tabulated as given below.

S.No. Plant name Root system
Tap root system/fibrous root system
1. Grass Fibrous root system
2. Tulasi Tap root system
3. Datura (Ummetta) Tap root system
4. Maize (Jonnalu) Fibrous root system
5. Beans Tap root system

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

Question 11.
Teacher asked you to collect any 5 branches of 5 different plants. Observe the leaf modifications (Leaf base, Petiole, Lamina, shape and Edges of the leaves). Tabulate the information as shown. I was asked to find out the similarities among those collected leaves.

S.No. Plant
name
Leaf base Yes/No Petiole
Yes/No
Lamina
Yes/No
Shape of leaf Yes/No Edge of leaf Yes/No

Answer:
Aim : To find out similarities among the collected plant leaves.
Procedure : I visited my surroundings. I collected the branches of 5 different plants. I observed their leaf modifications. All the information is tabulated as given below.
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 2

Question 12.
VI class boys brought stems of 5 differents plants. They are
1. Beans
2. Guava
3. Brinjal
4. Neem
5. Cucumber.
They were asked to find out their position of growth. They tabulated the information. How can you get the information with the following table ?

SNo. Plant name Stem grows Vertically / Horizontally Branches are Present/ Absent

Answer:

S.No. Plant name Stem grows
Vertically / Horizontally
Branches are Present/Absent
1. Beans Vertically or horizontally Creepers are seen
2. Guava Vertically Branches present
3. Brinjal Vertically Branches present
4. Neem Vertically Branches present
5. Cucumber Horizontally Creepers are seen

Question 13.
Read the following information and answer the given questions.
The leaf lamina usually consists of a midrib/ veins and veinlets arranged in the form of a network. The long vein present in the middle of the lamina is called midrib. The branches arising from the midrib are called veins and the even finer divisions are veinlets. The arrangement of veins in the lamina is called venation. Venation acts as a skeleton of the leaf and give it a shape and support.

1. Which structures form a network in the leaf ?
Answer: Midrib, veins and veinlets.

2. What is the use of venation to the leaf ?
Answer: Venation acts as a skeleton of the leaf.

3. What are the veins ?
Answer: The branches arising from the midrib of leaf are called veins.

4. Which is the long vein in lamina ?
Answer: Midrib.

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

Question 14.
Draw the diagrams of tap root system and fibrous root system of plants.
Answer:
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 3
Question 15.
Draw any three storage roots (tubers). Write the definition of tubers.
Answer:
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 4
Definition of tubers : The roots which store the food material in storage tissue are called tubers.

Question 16.
Draw a neat labelled diagram of stomata in leaf. How is it useful to plant?
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 5
Use of stomata in plants : Stomata are useful in the exchange of gases between the plant and atmosphere.

Question 17.
Mention the indicated parts of the following and write any function of the leaf.
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 6

Answer:

  • Mid rib of the leaf
  • Leaf lamina

Function of leaf: Leaf prepares food by the process of photosynthesis.

Question 18.
a. Draw the following plant and identify the parts A, B, C.
Answer:
A – Roots B – Stem C – Leaf

TS 6th Class Science Important Questions 9th Lesson Plants: Parts and Functions

b. How do roots help the plant?
Answer:
Roots help in the absorption of water and minerals from the soil.
TS 6th Class Science Important Questions 9th Lesson Plants Parts and Functions 7

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom? Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 5th Lesson What is inside the Atom?

9th Class Physical Science 5th Lesson What is inside the Atom? Textbook Questions and Answers

Improve Your Learning

Question 1.
What are the three subatomic particles? (AS 1)
Answer:
The three sub-atomic particles are electrons, protons, and neutrons.

Question 2.
Compare the subatomic particles electron, proton, and neutron. (AS 1)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom 1

Question 3.
What are the limitations of J.J. Thomson’s model of the atom? (AS 1)
Answer:
The main limitation of J.J. Thomson’s model of atom was that he is unable to explain, how the positively charged particles are shielded from negatively charged particles without getting neutralized.

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 4.
What were the three major observations Rutherford made in the gold foil experiment? (AS 1)
Answer:
The three major observations Rutherford made in the gold foil experiment were

  1. Most of the space inside the atom is empty.
  2. All the positive charge must be concentrated in a very small space within the atom called nucleus.
  3. The size of the nucleus is very small as compared to the size of the atom.

Question 5.
Sketch Rutherford’s atomic model. Why is Rutherford’s model of the atom called the planetary model? (AS 5)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom 5
Rutherford’s model is called planetary model because the motion of the electrons around the nucleus resembles the motion of the planets around the sun.

Question 6.
Put tick (✓) against correct choice and cross (✗) against wrong choice: (AS 1)
i) In Rutherford’s gold foil experiment, majority of alpha particles passed directly through the gold foil. This observation leads to which conclusions?
a) The positively charged region of the atom is very small. (✗)
b) The atom must consist of empty space. (✓)
c) The alpha particles makes a direct hit on the positively charged region of the atom. (✗)
d) The positively charged region of the atom is very dense. (✗)

ii) In Rutherford’s gold foil experiment, occasionally the alpha particles veered from a straight line path. This observation leads to which conclusion?
a) The positively charged region of the atom is very small. (✗)
b) The majority of the space in the atom is empty. (✗)
c) The alpha particle makes a direct hit on the positively charged region of the atom. (✗)
d) The positively charged region of the atom is very dense. (✓)

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 7.
Which one of the following is the correct electronic configuration of sodium? (AS 1)
a) 2, 8
b) 8, 2, 1
c) 2, 1, 8
d) 2, 8, 1
Answer:
d) 2, 8, 1

Question 8.
Give the main postulates of Bohr’s model of an atom. (AS 1)
Answer:
The main postulates of Bohr’s model of atom are
AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom 2

  1. Only certain special, discrete orbits of electrons are allowed inside the atom. These orbits or shells are called energy levels.
  2. While revolving in these discrete orbits the electrons do not radiate energy and this helps that the electrons do not crash into the nucleus.
  3. These orbits or shells are represented by the letters K, L, M, N, ……… or the numbers n = 1, 2, 3, …………

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 9.
Compare all the proposed models of an atom given in this chapter. (AS 1)
Answer:
In this chapter, four atomic models were discussed. The main postulates of those models are

1. Dalton’s proposal :
a) Atoms are indivisible.
b) Atoms of an element are all identical to each other and different from the atoms of other elements.

2. Thomson’s proposal:
a) An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
b) The total mass of the atom is considered to be uniformly distributed throughout the atom.

3. Rutherford’s proposal:
a) All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom.
b) Negatively charged electrons revolve around the nucleus in well defined orbits.
c) Size of the nucleus is very small as compared to the size of the atom.

4. Neils Bohr’s proposal :
a) Electrons are revolving around the nucleus in special, discrete orbits called en¬ergy levels or shells.
b) While revolving in these discrete orbits the electron do not radiate energy and this helps that the electrons do not crash into the nucleus.
c) These orbits or shells are represented by the letters K, L, M, N,…. or the numbers n = 1, 2, 3,

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 1o.
Define valency by taking examples of nitrogen and boron. (AS 1)
Answer:
Valency: The number of electrons present in outer most orbit of an atom is called its valency. Valency of Nitrogen :
a) Atomic number of nitrogen is 7.
b) The distribution of electrons is 2, 5.
c) The outer most orbit has 5 electrons.
d) Hence its valency should be 5. But it is easier to nitrogen to gain 3 electrons than to loose 5 electrons for becoming octet.
e) Hence the valency of nitrogen is ‘3’.

Valency of Boron :
a) Atomic number of boron is 5.
b) The distribution of electrons is 2, 3.
c) The outer most orbit has 3 electrons.
d) Hence the valency of boron is 3.

Question 11.
State the valencies of the following elements : magnesium and sodium. (AS 1)
Answer:
Magnesium :
a) Atomic number of magnesium is 12.
b) Distribution of electrons is 2, 8, 2.
c) Hence the valency is 2.

Sodium :
a) Atomic number of sodium is 11.
b) Distribution of electrons is 2, 8, 1.
c) Hence the valency is 1.

Question 12.
If Z = 5, what would be the valency of the element? (AS 2)
Answer:
1) If Z = 5, the distribution of electrons is 2, 3.
2) Hence the valency is ‘3’.

Question 13.
Write the atomic number and the symbol of an element which has mass number 32 and the number of neutrons 16 in the nucleus. (AS 1)
Answer:
Mass number (A) = 32 ; Number of neutrons (N) = 16
Number of protons (Z) = A- N = 32-16 = 16
∴ Atomic number =16
The element is sulphur.
The symbol of sulphur is S’.

Question 14.
Cl- has completely filled K and L shells. Explain. (AS 1)
Answer:
Atomic number of Cl is 17, but Cl has one more electron when compared with Cl atom. Distribution of electrons in Cl is

K L M
2 8 8

K shell can accommodate 2 electrons and L shell cab accommodate 8 electrons accord¬ing to the formula 2n2.
Hence the K and L shells are completely filled.

Question 15.
What is the main difference among the isotopes of the same element? (AS 1)
Answer:
The main difference between isotopes of the same element is
a) The number of neutrons is different.
b) Their physical properties are different but the chemical properties are similar.

Question 16.
For the following statements, write T for True and F for False. (AS 1)
a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
c) The mass of an electron is 1836 times that of proton.
Answer:
a) False
b) False
c) True

Question 17.
Fill in the missing information in the following table. (AS 4)
AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom 3
Answer:
AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom 4

Question 18.
How do you appreciate the efforts made by scientists to explain the structure of atom by developing various atomic models? (AS 6)
Answer:

  • Structure of atom, till today it is mysterious and challenging the scientists.
  • We have to appreciate the scientists right from Lavoisier, who proposed law of conservation of mass, Proust who proposed law of constant proportions, John Dalton for his first model of atom, Rutherford for giving planetary model of atom and Neils Bohr for his model of atom.
  • Till today scientists are trying to know the existence of more and sub-atomic particles besides electrons, protons and neutrons.
  • Hence the efforts of scientists are highly appreciable, for making our lives comfortable and leaving many challenges before us to unveil or discover them.

Question 19.
Geeta got a doubt, “Why does atomic nucleus contain proton and neutrons? Why can’t electrons and neutrons be in it”. Can you help to clarify her doubt? Explain. (AS 1)
Answer:
Nucleus contains protons and neutrons inside it but not electrons and neutrons. If it would have happen, then

  1. the alpha particles in the Rutherford’s alpha particle scattering experiment would have not been deflected or scattered.
  2. the idea of nucleus would have not been evolved because the mass of electron is negligible, it is most unstable.

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 20.
Collect information about various experiments conducted and theories proposed by scientists starting from John Dalton to Neils Bohr. Prepare a story with a title “History of atom”. (AS 4)
Answer:
History of atom
John Dalton proposed atomic theory based on law of conservation of mass and law of constant proportion as :

  1. Atoms were indivisible.
  2. Atoms of an element are all identical to each other and different from the atoms of other elements.

Later on various experiments conducted by Thomson, Goldstein, etc. proved that atom is divisible and consists sub-atomic particles like electrons, protons and neutrons. Based on this J.J. Thomson proposed a model of atom in 1898.

According to Thomson,

  1. An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
  2. The total mass of the atom is considered to be uniformly distributed throughout the atom.
  3. The negative and the positive charges are supposed to be balance out and the atom as a whole is electrically neutral.

This model is also called as plum pudding model or watermelon model.

Thomson’s student Ernest Rutherford conducted alpha particle scattering experiment and got the results which were not in favour of Thomson’s model. Based on his experiment, Rutherford proposed a model of atom. According to him,

  1. All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom. The electrons were not a part of nucleus.
  2. Negatively charged electrons revolve around the nucleus in well – defined orbits like planets revolve around the sun.
  3. The size of nucleus is very small as compared to the size of the atom.

This model could not account for stability of atom, as revolving electron must lose energy and eventually crash into the nucleus, as a result matter would not exist in the form that we see it now.

In 1913, Neils Bohr proposed another model to overcome Rutherford’s defect. According to Bohr,

  1. Only certain special, discrete orbits of electrons are allowed in side the atom. These orbits or shells are called energy levels.
  2. While revolving in these discrete orbits the electrons do not radiate energy and this helps that the electrons do not crash into the nucleus.
  3. These orbits or shells are represented by K, L, M, N ………… or the numbers 1, 2, 3,

This model could not predict the spectra of atoms.

Hence this journey continues

9th Class Physical Science 5th Lesson What is inside the Atom? InText Questions and Answers

9th Class Physical Science Textbook Page No. 75

Question 1.
Why are the atoms of different elements different?
Answer:
Nature and properties of elements depends on the arrangement of atoms. We know that different elements behave differently. This is due to the difference in their atoms.

Question 2.
Is there anything inside atom that make them to be same or different?
Answer:
The arrangement of sub atomic particles inside the atom is responsible to make them to be same or different.

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 3.
Are atoms indivisible?
Answer:
No, atom is divisible. There are many sub-atomic particles inside the atom according to the recent experiments.

9th Class Physical Science Textbook Page No. 77

Question 4.
If an atom consists of sub-atomic particles like protons, neutrons and electrons, how are they arranged in the atom ?
Answer:
The arrangement of sub-atomic particles like protons, neutrons and electrons has been explained by many scientists like Rutherford, Neils Bohr, etc. According to them, atom consists a central mass called nucleus. Nucleus consists protons and neutrons. Electrons revolve around the nucleus in fixed shells.

9th Class Physical Science Textbook Page No. 80

Question 5.
Why is atom stable?
Answer:
In an atom, the number of protons in the nucleus is equal to the number of electrons out side the nucleus. Hence the positive and negative charges in an atom are equal. So, atom is electrically neutral. So, atom is stable. But the stability of atom was explained by Neils Bohr in a different way.

Question 6.
Can you suggest any other arrangement of subatomic particles in the atom which prevents the revolving electron to fall into the nucleus?
Answer:
Electrons have to revolve around the nucleus in definite orbits such that the centripetal and centrifugal forces acting on the electron must be equal in magnitude and opposite direction. Then the revolving electron do not fall into the nucleus.

9th Class Physical Science Textbook Page No. 82

Question 7.
How many electrons can be accommodated in each shell of an atom?
Answer:
The number of electrons that can be accommodated in each shell of an atom depends on the shell number. First shell (K) consists 2 electrons, second (L) shell consists 8 electrons, third (M) shell consists 18 electrons, fourth shell (N) consists 32 electrons, and so on.

Question 8.
Can a particular shell have just one electron?
Answer:
No, shell has just one electron.

AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom?

Question 9.
What is the criteria to decide number of electrons in a shell?
Answer:
The number of electrons in a shell can be decided by using a formula 2n². (Where n is shell number).

9th Class Physical Science Textbook Page No. 83

Question 10.
What is the valency of oxygen that you can calculate by the method discussed above?
Answer:

  • Oxygen has 8 electrons in its atom. The distribution of electrons is 2, 6.
  • The outer most shell consists 6 electrons, this number is hear to 8.
  • Hence the valency of oxygen is 8 – 6 = 2.

9th Class Physical Science Textbook Page No. 85

Question 11.
Should we consider the number of neutrons as a characteristic of an atom?
Answer:
The mass of an atom which is a characteristic of an atom depends on the number of neutrons and protons that its nucleus contains. Hence the number of neutrons can be considered as a characteristic of an atom.

9th Class Physical Science Textbook Page No. 76

Question 12.
An atom is electrically neutral. But the electrons present in it are negatively charged particles. If only negative charges were present, the atom would not be neutral. Then, why are atoms considered to be neutral?
Answer:

  • This is the idea before Rutherford’s model.
  • According to Rutherford’s model, number of protons inside the nucleus and number of electrons outside the nucleus are equal.
  • Hence the net negative charge is equal to net positive charge. So, the atom is electrically neutral.

9th Class Physical Science Textbook Page No. 80

Question 13.
Compare Rutherford and Thomson’s models of the atom on the following basis :
1) Where is the positive charge placed?
2) How are the electrons placed?
3) Are they stationary inside the atom or moving?
Answer:

  1. According to Thomson, the positive charge is uniformly distributed throughout the atom. Whereas according to Rutherford, the positively charged protons are inside the nucleus.
  2. According to Thomson, electrons are embedded in the positively charged atom, but according to Rutherford, electrons are revolving around the nucleus in welldefined orbits.
  3. According to Thomson, electrons are stable inside the atom but according to Rutherford, electrons are moving inside the atom.

9th Class Physical Science Textbook Page No. 83

Question 14.
Phosphorus and sulphur show multiple valency. See table 2. Why do some elements show multiple valency? Discuss with your Mends and teachers.
Answer:

  • For sulphur, the number of electrons in outer most orbit is 6.
  • Hence the valency should be (8 – 6 =) 2.
  • But sulphur exists in so many forms.
  • In the excited state, these 6 electrons also tend to participate in the bond formation.
  • Hence sometimes it shows the valency 6. Ex : SO2, SO3, etc.
  • Same situation happens for phosphorus. Ex . PCl3, PCl5, etc.

9th Class Physical Science 5th Lesson What is inside the Atom? Activities

Activity – 1

Question 1.
Sketch the structure of atom as you imagine.
AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom 6
We learnt about electron, proton and neutron.
a) Suppose you had to arrange them in an atom, how do you do it?
Answer:
Many arrangements are possible. Think that atom looks like a room, we can arrange the particles in alternating rows.

b) In how many ways can you arrange these sub-atomic particles in a spherical shape?
Answer:
Protons are positively charged, electrons are negatively charged and neutrons are
neutral. Hence neutrons and protons can be kept nearer and electrons can be kept farther or near the edge of the sphere. This is only an assumption. We can arrange in so many ways like this.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 4th Lesson Atoms and Molecules

9th Class Physical Science 4th Lesson Atoms and Molecules Textbook Questions and Answers

Improve Your Learning

Question 1.
Draw the diagram to show the experimental setup for the law of conservation of mass. (AS 5)
(OR)
Draw the experimental arrangement used in verifying law of conservation of mass. Write the law of conservation of mass.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 1 AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 2 AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 3

Question 2.
Explain the process and precautions in verifying law of conservation of mass. (AS 5)
(OR)
Explain the procedure to prove in a chemical reaction the mass neither destroyed.
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 1 AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 2 AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 3
Answer:
Aim :
To verify law of conservation of mass.

Material required :
Sodium sulphate,Barium chloride, distilled water, conical flask, spring balance, small test tube, rubber cork, thread, retort stand.

Procedure:

  1. Prepare a solution of sodium sulphate by dissolving approximately 2 gm of sodium sulphate in 100 ml distilled water in a 250 ml conical flask.
  2. Prepare a Barium chloride solution by dissolving approximately 2 gm of potassium iodide in 100 ml water in another conical flask.
  3. Take 100 ml solution of sodium sulphate in 250 ml conical flask.
  4. Also take 4 ml solution of Barium chloride in test tube.
  5. Hang the test tube in the flask carefully without mixing the solutions. Put a cork on the flask.
  6. Weigh the flask with its contents carefully by spring balance.
  7. Now tilt and swirl the flask, so that the two solutions mix.
  8. Weigh the flask again by the spring balance.

Observations:

  1. Weight of flask and contents before mixing = m1 g
  2. Weight of flask and contents after mixing = m2 g

Conclusion :

  1. We have observed that the two weights i.e., mj and m2 are equal.
  2. This proves the law of conservation of mass.

Precautions:

  1. Care should be taken while handling chemicals.
  2. Glass apparatus may slip and break down. Hence make sure that they should not slip from your hands.
  3. Contents of the conical, ffhsk should not mix before weighing first time.
  4. Tie a thick thread to the conical flask, so that it will not slip while weighing.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 3.
15.9g of copper sulphate and 10.6g of sodium carbonate react together to give 14.2g of sodium sulphate and 12.3 g of copper carbonate. Which law of chemical combination is obeyed? How? (AS 1, AS 2)
Answer:
Reactants:
Mass of copper sulphate = 15.9 g ; Mass of sodium carbonate = 10.6 g
Total mass of reactants = 15.9 + 10.6 = 26.5 g

Products:
Mass of sodium sulphate = 14.2 g ; Mass of copper carbonate = 12.3 g
Total mass of products = 14.2 + 12.3 = 26.5 g
∴ Total mass of reactants is equal to total mass of products. This is the “Law of conservation of Mass”.

Question 4.
Carbon dioxide is added to 112 g of calcium oxide. The product formed is 200 g of calcium carbonate. Calculate the mass of carbon dioxide used. Which law of chemical combination will govern your answer? (AS 1, AS 2)
Answer:

  1. Let x g of carbon dioxide is added to 112 g of calcium oxide.
  2. The product is 200 g of calcium carbonate.
  3. According to law of conservation of mass, ,
    Total mass of reactants = Total mass of products
    x+ 112 = 200 g
    x = 88 g
    ∴ 88 g of carbon dioxide is used.

Question 5.
0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.144 g of oxygen and 0.096 g of boron. Calculate the percentage composition of the compound by weight. (AS 1)
Mass of compound of oxygen and boron = 0.24 g
On analysis,
Mass of oxygen in the compound = 1.44 g
Mass of boron in the compound = 0.096 g
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 4

Question 6.
In a class, a teacher asked students to write the molecular formula of oxygen. Shamita wrote the formula as 02 and Priyanka as O. Which one is correct? State the reason. (AS 1, AS 2)
Answer:
Shamitha’s answer is correct.
Reason :

  1. Oxygen is diatomic.
  2. Two atoms of oxygen combine to form oxygen molecule.
  3. Hence the formula of oxygen molecule will be ‘O2‘.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 7.
Imagine what would happen if we do not have standard symbols for elements. (AS 2)
(OR)
Is it necessary to use symbols for elements? Write your opinion.
Answer:

  • Chemistry involves a lot of reactions.
  • If we do not have symbols, we have to write their names to represent the reactions.
  • This is very tedious work.
  • To avoid this, we need standard symbols to elements, which are universally accepted.
  • In advanced studies, balancing of equations, atoms present in a compound, etc. will not be understood without symbols.
  • Simply chemistry will not be developed unless symbols, formulae, etc. are not known.

Question 8.
Mohith said “H2 differs from 2H.” Justify. (AS 1)
Answer:
H2 is the hydrogen molecule in which two hydrogen atoms are combined to form one hydrogen molecule.

2H is the hydrogen atom. Here 2 hydrogen atoms are ready to participate in chemical reaction.

Question 9.
Lakshmi gives a statement “CO and Co both represent element”. Is it correct? State reason. (AS 1, AS 2)
Answer:
Lakshmi’s statement is incorrect.

Reason :

  1. CO stands for carbon monoxide, a compound, which consists carbon and oxygen atoms.
  2. This can be identified with the help of both C and O are capital (upper case) letters.
  3. Co stands for cobalt, an element.
  4. This can be identified with the help of ‘C’ is capital (upper case) letter and ‘o’ small (lower case) letter.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 10.
The formula of water molecule is H2O. What information do you get from this formula? (AS 1)
Answer:

  • Water is a combination of hydrogen and oxygen.
  • Two hydrogen atoms and one oxygen atom combine to form one water molecule.
  • Molecular weight of water molecule is 18. [Hydrogen 1, Oxygen 16. H2O ⇒ 2 × 1 + 16=18]
  • 18 g of water molecule contains 6.022 × 1023 particles in it.
  • Valency of hydrogen is 1 and oxygen is 2.

Question 11.
How would you write 2 molecules of Oxygen and 5 molecules of Nitrogen? (AS 1)
Answer:
2 molecules of oxygen → 2O2
Reason :

  1. Oxygen is diatomic element.
  2. Two oxygen atoms combine to form one oxygen molecule.
  3. The formula of oxygen molecule is O2.

5 molecules of nitrogen → 5N2
Reason :

  1. Nitrogen is also diatomic element.
  2. Two nitrogen atoms combine to form one nitrogen molecule.
  3. Molecular formula of nitrogen is N2.

Question 12.
The formula of a metal oxide is MO. Then write the formula of its chloride. (AS 1)
Answer:

  • The valency of oxide is 2 i.e., O-2.
  • The formula of a metal oxide is given as MO.
  • Hence the valency of the given metal must be 2 i.e., M+2.
  • Valency of chloride is 1 i.e., C.
  • Therefore according to criss-cross method, the formula of given metal chloride will be MCl2.

Question 13.
Formula of calcium hydroxide is Ca(OH)2 and zinc phosphate is Zn3(PO4)2. Then write the formula to calcium phosphate. (AS 1)
(OR)
Formula of calcium hydroxide is Ca(OH)2 and zinc phosphate is Zn3(PO4)2. Then write the valencies of calcium and phosphate and then write the formula of calcium phosphate.
Answer:

  • Formula of calcium hydroxide is Ca(OH)2.
  • From criss-cross method we know that the valency of calcium is 2 i.e., Ca+2 and hydroxide is 1 i.e., OH.
  • Formula of zinc phosphate is Zn3(PO4)2.
  • Valency of Zn is 2 i.e., Zn+2, and valency of phosphate is 3 i.e., PO4-3.
  • Now the formula of calcium phosphate according to criss-cross method is Ca3(PO4)2.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 14.
Find out the chemical names and formulae for the following common household substances. (AS 1)
a) Common salt
b) Baking soda
c) Washing soda
d) Vinegar
Answer:

Common household substance Chemical name Formula
a) Common salt Sodium chloride NaCl
b) Baking soda Sodium bicarbonate NaHCO3
c) Washing soda Sodium carbonate Na2CO3
d) Vinegar Impure dilute acetic acid CH3COOH

Question 15.
Calculate the mass of the following. (AS 1)
a) 0.5 mole of N2 gas
b) 0.5 mole of N atoms
c) 3.011 × 1023 number of N atoms
d) 6.022 × 1023 number of N2 molecules
Answer:
a) 0.5 mole of N2 gas :
Mass of one mole of N2 gas = 28 g. (∵ Molecular wt. of N2 = 28)
Mass of 0.5 mole of N2 gas = 28 × 0.5 = 14 g

b) 0.5 mole of N atoms :
Mass of one mole of N atoms = 14 g (∵ Atomic wt. of N = 14)
Mass of 0.5 mole of N atoms = 14 × 0.5 = 7 g

c) 3.011 × 1023 number of N atoms :
Mass of 6.022 × 1023 number of N atoms = 14 g
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 5

d) 6.022 × 1023 number of N2 molecules :
Mass of 6.022 × 1023 number of N2 molecules = 28 g

Question 16.
Calculate the number of particles in each of the following. (AS 1)
a) 46 g of Na
b) 8 g of O2
c) 0.1 mole of hydrogen
Answer:
a) 46 g of Na :
Atomic weight of Na = 23
Number of particles in 23 g of Na atom = 6.022 × 1023
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 6

b) 8 g of O2 :
Molecular weight of O2 is 32.
Number of particles in 32 g of O2 molecule = 6.022 × 1023
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 7

c) 0.1 mole of hydrogen :
Atomic weight of hydrogen is 1.
Number of particles in 1 mole of hydrogen = 6.022 × 1023
Number of particles in 0.1 mole of hydrogen= \(\frac{0.1}{1}\) × 6.022 × 1023 = 6.022 × 1022

Question 17.
Convert into moles. (AS 1)
a) 12 g of O2 gas
b) 20 g of water
c) 22 g of carbon dioxide
Answer:
a) 12 g of O2 gas :
Molecular weight of O2 is 32.
∴ Number of moles of 32 g of O2 gas = 1
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 8

b) 20 g of water :
Molecular weight of water (H2O) is 18.
Number of moles of 18 g of water = 1
Number of moles of 20 g of water = \(\frac{20}{18}\) x 1 =1.11

c) 22 g of carbon dioxide :
Molecular weight of carbon dioxide (CO2) is 44.
∴ Number of moles of 44 g of CO2 = 1
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 9

Question 18.
Write the valencies of Fe in FeCl2 and FeCl3. (AS 1)
Answer:

  1. In FeCl2, the valency of Fe is 2.
  2. In FeCl3, the valency of Fe is 3.

Question 19.
Calculate the molar mass of sulphuric acid (H2SO4) and glucose (C6H12O6). (AS 1)
Answer:
a) Formula of sulphuric acid is H2SO4.
Molecular mass of H2SO4 = 2 × 1 + 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 u
∴ Molar mass of H2SO4 = 98 g

b) Formula of glucose is C6H12O6.
Molecular mass of C6H12O6 = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 u
∴ Molar mass of C6H12O6 = 180 g.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 20.
Which has more number of atoms – 100 g of sodium or 100 g of iron? Justify your answer. (Atomic mass of sodium = 23 u, atomic mass of iron = 56 u) (AS 1)
Answer:
100g of sodium has more number of atoms than 100g of iron.

Justification :
1) Atomic mass of sodium = 23 u
23 g of sodium contains 6.022 × 1023 atoms.
100 g of sodium contains = \(\frac{100}{23}\) × 6.022 × 1023 = 26.1826 × 1023 atoms of sodium.

2) Atomic mass of iron = 56 u
∴ 56 g of iron contains 6.022 × 1023 atoms.
100 g of iron contains = \(\frac{100}{56}\) × 6.022 × 1023 = 10.7535 × 1023 atoms of iron.

Question 21.
Complete the following table. (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 10
Answer:
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 11

Question 22.
Fill the following table. (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 12
Answer:
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 13

Question 23.
Make placards with symbols and valencies of the atoms of the elements separately. Each student should hold two placards, one with the symbol in the right hand and the other with the valency in the left hand. Keeping the symbols in place, students should criss-cross their valencies to form the formula of a compound.
Answer:
Student’s activity.

Question 24.
Take empty blister packs of medicines. Cut them into pieces having
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 14
Hence the formula of sodium carbonate will be Na2 CO3
Hence the forrnu A. Student’s activity.

9th Class Physical Science 4th Lesson Atoms and Molecules InText Questions and Answers

9th Class Physical Science Textbook Page No. 56

Question 1.
Does the weight of iron rod increase or decrease, on rusting?
Answer:
The weight of iron rod decreases on rusting.

Question 2.
Where does the matter charcoal go?
Answer:
The charcoal, on burning, gives off CO2 which is mixed in atmosphere. The residue is remained as ash.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 3.
Wet clothes dry after some time – where does the water go?
Answer:
Water evaporates and mixed in the atmosphere.

Question 4.
What happens to magnesium on burning it in air?
Answer:
Magnesium on burning in air gives a bright light and ash is remained. The ash is magnesium oxide.

Question 5.
What happens to sulphur on burning it in air?
Answer:
Sulphur on burning, changes its state and colour.

9th Class Physical Science Textbook Page No. 57

Question 6.
Did you observe any precipitate in the reaction?
Answer:
In the flask, a reaction takes place between lead nitrate and potassium iodide.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 7.
Do you think that a chemical reaction has taken place in the flask? Give reason.
Answer:
Yes, the contents in the flask are changed as lead iodide and potassium nitrate.

9th Class Physical Science Textbook Page No. 58

Question 8.
Do the weights of the flask and its contents change during the activity?
Answer:
The weights of the flask and its contents do not change before and after reaction.

Question 9.
What are your conclusions?
Answer:
Mass was neither created nor destroyed.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 10.
What do you observe from table – 1?
Answer:
The components of a compound are mixed at same proportions in any sample.

9th Class Physical Science Textbook Page No. 59

Question 11.
What difference do you observe in percentage of copper, carbon, and oxygen in two samples?
Answer:
The percentage of copper, carbon and oxygen are same in two samples, i.e., they are mixed at same proportions.

9th Class Physical Science Textbook Page No. 60

Question 12.
Are elements also made of atoms?
Answer:
When the particles of a substance contain only one type of atoms, that substance is called an element. In elements the smallest particle that exist may be atoms or molecules.

9th Class Physical Science Textbook Page No. 62

Question 13.
How do we write the symbols for calcium, chlorine, chromium?
Answer:
We have only 26 alphabets in English, but there are over 100 known elements. We cannot write the same symbol for carbon, calcium, chromium, etc.

9th Class Physical Science Textbook Page No. 63

Question 14.
Would you be able to recognise the elements of the table – 2, have symbols of this category?
Answer:
Yes. They are iron, gold, sodium, and potassium.

9th Class Physical Science Textbook Page No. 64

Question 15.
Observe the atomicity and fill the following table.
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 18
Answer:

Name of the element Formula Atomicity
Argon Ar Monoatomic
Helium He Monoatomic
Sodium Na Monoatomic
Iron Fe Monoatomic
Aluminium Al Monoatomic
Copper Cu Monoatomic
Hydrogen H2 Diatomic
Oxygen O2 Diatomic
Nitrogen N2 Diatomic
Chlorine Cl2 Diatomic
Ozone O3 Triatomic
Phosphorus P4 Tetratomic
Sulphur S8 Octatomic

Question 16.
What is valency?
Answer:
Every element reacts with other element according to its combining capacity, which we call as its valency.

9th Class Physical Science Textbook Page No. 66

Question 17.
Can you write the formula of carbon dioxide and carbon monoxide? Try to write formula for them as we have done in case of water molecule.
Answer:
Carbon dioxide :
The elements present are carbon and oxygen. One atom of carbon and one atom of oxygen are present in a molecule of carbon monoxide. Hence the formula of carbon monoxide is CO.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Carbon dioxide :
The elements present are carbon and oxygen. One atom of carbon and 2 atoms of oxygen are present in a molecule of carbon dioxide. Hence the formula of carbon dioxide is CO2.

9th Class Physical Science Textbook Page No. 69

Question 18.
How many molecules are there in 18 grams of water?
Answer:
6.022 × 1023 molecules are there in 18 grams of water.

Question 19.
How many atoms are there in 12 grams of carbon?
Answer:
6.022 × 1023 atoms are there in 12 grams of carbon.

9th Class Physical Science Textbook Page No. 58

Question 20.
Do you get the same result if the conical flask is not closed?
Answer:

  1. No, we cannot get the same result.
  2. When the conical flask is not closed, some gases will leave out the flask during chemical reaction.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 21.
Recall the burning of the magnesium ribbon in air. Do you think mass is conserved during this reaction?
Answer:

  1. Yes, but we cannot observe the conservation of mass.
  2. When the experiment is conducted in a closed container where there is no scope for oxygen to escape, we can observe the conservation of mass. But in this condition this experiment is not possible.

9th Class Physical Science Textbook Page No. 59

Question 22.
100 g of mercuric oxide decompose to give 92.6 g of mercury and 7.4 g of oxygen. Let us assume that 10 g of oxygen reacts completely with 125 g of mercury to give mercuric oxide. Do these values agree with the law of constant proportions?
Answer:
Proportion of oxygen = 7.4 : 10
Proportion of mercury = 92.6 : 125
\(\Rightarrow \frac{7.4}{10}=\frac{92.6}{125} \Rightarrow 0.74=0.74\)
∴ They follow law of constant proportions.

Question 23.
Discuss with your friends if the carbon dioxide that you breathe out and the carbon dioxide they breathe out are identical. Is the composition of the carbon dioxide of different sources same? (Page – 73)
Answer:
Same.
This can be justified with the help of law of constant proportions.

9th Class Physical Science Textbook Page No. 60

Question 24.
Which postulate of Dalton’s theory is the result of the law of conservation of mass?
Answer:
First postulate of Dalton’s theory i.e. “Matter consists of indivisible particles called atoms”, is the result of law of conservation of mass.

AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules

Question 25.
Which postulate of Dalton’s theory can explain the law of constant proportions?
Answer:
Third postulate of Dalton’s theory i.e. “Atoms of a given element have identical mass and chemical properties. Atoms of different elements have different masses and chemical properties”, is the result of law of constant proportions.

9th Class Physical Science 4th Lesson Atoms and Molecules Activities

Activity – 1

Question 1.
Some elements and their possible symbols are given. Correct them and give reasons for your corrections.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 15

Activity – 2

Question 2.
Write the symbols for given elements.
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 16
Answer:
AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules 17

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure? Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 3rd Lesson Is Matter Pure?

9th Class Physical Science 3rd Lesson Is Matter Pure? Textbook Questions and Answers

Improve Your Learning

Question 1.
Which separation techniques will you apply for the separation of the following? (AS 1)
a) Sodium chloridfe from its solution in water.
b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
c) Small pieces of metal in the engine oil of a car.
d) Different pigments from an extract of flower petals.
e) Butter from curd.
f) Oil from water.
g) Tea leaves from tea.
h) Iron pins from sand.
i) Wheat grains from husk.
j) Fine mud particles suspended in water.
Answer:

Mixture Separation technique
a) Sodium chloride from its solution in water Crystallization
b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride Sublimation
c) Small pieces of metal in the engine oil of a car Filtration
d) Different pigments from an extract of flower petals Chromatography
e) Butter from curd Centrifugation
f) Oil from water Separation funnel
g) Tea leaves from tea Filtration
h) Iron pins from sand Magnetism
i) Wheat grains from husk Winnowing
j) Fine mud particles suspended in water Sedimentation and decantation (or) Filtration using filter paper

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

Question 2.
Write the steps you would use for making tea. Use the words given below and write the steps for making tea. (AS 7)
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 16
Answer:

  • Take a cup of milk (solvent) in a tea kettle.
  • Add one table spoon of sugar (solute), and one table spoon of tea powder (insoluble) to the solvent.
  • Heat the tea kettle on the stove.
  • The sugar (solute) dissolves in the milk (solvent) and the tea powder remains undissolved.
  • Now filter the solution so formed.
  • The filtrate is the tea (solution).
  • The residue remained in the sieve is the insoluble component of tea powder.

Question 3.
Explain the following giving examples. (AS 1)
a) Saturated solution
b) Pure substance
c) Colloid
d) Suspension
Answer:
a) Saturated solution :
When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.

In a saturated solution, equilibrium with the undissolved solute at a certain temperature.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 1

Ex :

  1. Take 50 ml of water in a cup.
  2. Add one spoon of sugar to the cup and stir still it dissolves.
  3. Keep on adding sugar to the water in the cup and stir till no more sugar can be dissolved.
  4. The solution so formed is a saturated solution.

b) Pure substance :
A substance is pure i.e., homogeneous if the com-position doesn’t change, no matter which part of the substance we take for examination.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 2
Ex :

  1. Take a small part of pure gold biscuit as a sample.
  2. The composition is found to be same throughout it.

c) Colloid :
Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye, but is big enough to scatter light.
Ex : Milk, butter, cheese, cream, gel, etc.

d) Suspension :
Suspension is a heterogeneous mixture in which the solute particles didn’t dissolve and the particles are visible to naked eye.
Ex : Syrups, chalk powder mixed with water, etc.

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

Question 4.
Classify each of the following as a homogeneous or heterogeneous mixture. Give reasons. (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 3
Answer:

Homogeneous mixtures Heterogeneous mixtures
Soda water Soil
Air Wood
Vinegar
Filtered tea
Reason : Components in the above mixtures are uniformly distributed and we cannot see the components separately. Reason : Components in the above mixtures are not uniformly distributed.

Question 5.
How would you confirm that a colourless liquid given to you is pure water? (AS 1)
Answer:

  • Observe the smell. We should not find any smell.
  • Observe with a naked eye we should not find any suspended particles or fumes or air bubbles.
  • Pass a beam of light. It should not scatter.
  • The temperature should be normal.
    Then the given colourless liquid is pure water.

Question 6.
Which of the following materials fall in the category of a “pure substance”? Give reasons. (AS 1)
a) Ice
b) Milk
c) Iron
d) Hydrochloric acid
e) Calcium oxide
f) Mercury
g) Brick
h) Wood
i) Air
Answer:

  • Except brick and wood remaining materials given in the list can be treated as pure substances.
  • Take any small part of ice, milk, iron, hydrochloric acid, calcium oxide, mercury and air and test for their components.
  • We find that the composition is same throughout them.

Question 7.
Identify the solutions among the following mixtures. (AS 1)
a) Soil
b) Sea water
c) Air
d) Coal
e) Soda water
Answer:
The solutions are : sea water, air and soda water.

Question 8.
Which of the following will show “Tyndall effect”? How can you demonstrate “Tyndall effect” in them? (AS 1, AS 3)
a) Salt solution
b) Milk
c) Copper sulphate solution
d) Starch solution
Answer:
Milk shows Tyndall effect.

Demonstration :

  1. Prepare the milk, copper sulphate, salt and starch solutions in different beakers.
  2. Allow a beam of light through each of them.
  3. The path of the light beam is clearly visible to us through milk.
  4. The path of the light beam is not visible through remaining.
  5. This experiment will be effective if it is performed in a dark room.

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

Question 9.
Classify the following into elements, compounds and mixtures. (AS 1)
a) Sodium
b) Soil
c) Sugar solution
d) Silver
e) Calcium carbonate
f) Tin
g) Silicon
h) Coal
i) Air
j) Soap
k) Methane
l) Carbondioxide
m) Blood
Answer:

Elements Compounds Mixtures
Sodium Calcium carbonate Soil
Silver Coal Sugar solution
Tin Methane Air
Silicon Carbondioxide
Soap
Blood

Question 10.
Classify the following substances in the below given table. (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 15
Answer:

Solution Suspension Colloidal dispersion
Soda water Ink Fog
Fruit salad Nail polish Aerosol sprays
Black coffee Starch solution Boot polish
Air Brass Milk
Blood
Oil and water

Question 11.
Take a solution, a suspension, a colloidal dispersion in different beakers. Test whether each of these mixtures shows the Tyndall effect by focusing a light at the side of the container. (AS 3)
Answer:

  • Take sugar solution (solution), starch solution (suspension) and milk solution (colloidal dispersion) in three different beakers.
  • Focus a beam of light by torch or a laser beam at the side of each container and observe.
  • We can see that the path of beam of light is clearly visible through all the solutions.
  • Hence all the three solutions show ‘Tyndall effect”.

Question 12.
Draw the figures of arrangement of appatus for distillation and fractional distillation. What do you find the major difference in these apparatus? (AS 5)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 6 AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 7
The main difference between these two apparatus is that a fractionating column is fitted in between the distillation flask and the condenser.

Question 13.
Determine the mass by mass percentage concentration of a 100 g salt solution which contains 20 g salt. (AS 1)
Answer:
Mass of salt = 20 g; Mass of salt solution = 100g
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 4

Question 14.
Calculate the concentration in terms of mass by volume percentage of the solution containing 2.5 g potassium chloride in 50 ml of potassium chloride (KCl) solution. (AS 1)
Answer:
Mass of potassium chloride = 2.5 g
Volume of potassium chloride solution = 50 ml
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 5

9th Class Physical Science 3rd Lesson Is Matter Pure? InText Questions and Answers

9th Class Physical Science Textbook Page No. 40

Question 1.
Can you give few more examples of this kind?
Answer:
Some more examples of homogeneous mixtures are sugar solution, lemon squash, fruit juices, syrups and tonics used in medicine, etc.

Question 2.
Can you prove this with an experiment?
Answer:

  • Take some thick milk in a test tube.
  • Pass a beam of light from torch or a laser light.
  • We cannot observe the path of light through the solution.

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

Question 3.
If the solution is diluted, can the path of light be visible?
Answer:

  • Take some thick milk in a test tube.
  • Dilute it by adding some water.
  • Now pass a beam of light from torch or a laser light.
  • We cannot observe the path of light through the solution.

Question 4.
What would happen if you add a little more solute to a solvent?
Answer:
The solution becomes concentrated.

Question 5.
How do you determine the percentage of the solute present in a solution?
Answer:

  • Take 100 ml of water in a beaker.
  • Take 50 g. of sugar in a plate.
  • Add a spoon of sugar to water and stir it tell the sugar dissolve in water.
  • Go on adding sugar till you reach a situation that the sugar cannot be dissolved in water.
  • Now weigh the sugar remained in the plate.
  • Subtract thin weight from 50 g. The weight so obtained is dissolved in water.
  • Hence the maximum amount of solute present in 100 ml of solvent is the percentage of solute (solubility).

9th Class Physical Science Textbook Page No. 44

Question 6.
Did you ever observe this phenomenon in the cinema halls?
Answer:
In cinema halls when we observe the projector while the movie is running, we can observe the phenomenon of “Tyndall effect”. We can see the beams of light in which dust particles also observed.

9th Class Physical Science Textbook Page No. 46

Question 7.
Is the mixture heterogeneous? Give reasons.
Answer:
The mixture of ammonium chloride and salt is a heterogeneous mixture. Even though these two are white in colour their particles do not mix.

AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure?

Question 8.
How do we separate the salt and ammonium chloride?
Answer:
We can separate the salt and ammonium chloride by the method of sublimation.

9th Class Physical Science Textbook Page No. 49

Question 9.
Can you give any examples where we use fractional distillation technique?
Answer:
We use this technique in separating the components of crude oil i.e., petrol, naphthalene, kerosene, greese, etc.

9th Class Physical Science Textbook Page No. 38

Question 10.
How does a laundry dryer squeeze out water from wet clothes?
Answer:

  • The laundry dryer contains a cylindrical vessel with holes on its walls.
  • When wet clothes are dropped in it, it is rotated with high speed with the help of an electric motor.
  • Due to centrifugation, the water from the clothes reaches to the walls of the cylinder and comes out through the holes.
  • Hence the clothes are dried up.

9th Class Physical Science Textbook Page No. 40

Question 11.
a) “All the solutions are mixtures, but not all mixtures are solutions”. Discuss about the validity of the statement and give reasons to support your argument.
Answer:

  • You take any solution like salt solution, sugar solution, air, etc. all are homogeneous mixtures.
  • Consider a mixture of sand and iron fehlings. It is not homogeneous. Hence this is not a solution.

b) Usually we think of a solution as a liquid that contains either a solid, liquid or a gas dissolved in it. But, we can have solid solutions. Can you give some examples?
Answer:
Examples of solid solutions are :

  1. Steel used in constructions (a homogeneous mixture of iron and carbon).
  2. Brass (a homogeneous mixture of zinc and copper).

9th Class Physical Science Textbook Page No. 43

Question 12.
1) Have you ever observed carefully the syrup that you take for cough? Why do you shake it before consuming?
2) Is it a suspension or colloidal solution?
Answer:

  1. The syrup used for cough will be shook before consuming because it consists some undissolved particles settled down.
  2. Hence cough syrup is a suspension.

9th Class Physical Science Textbook Page No. 45

Question 13.
Is there any difference between a true solution and colloidal solution? If you find the differences, what are those differences?
Answer:
Differences between true solutions and colloidal solutions :
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 17

Question 14.
Why do we use different separation techniques for mixtures like grain and husk as well as ammonium chloride and salt though both of them are heterogeneous mixtures? What is the basis for choosing a separation technique to separate mixtures?
Answer:
The basis for choosing a separation technique to separate mixtures is the property of a component in the mixture i.e., solubility in water, evaporation, appearance, etc.

9th Class Physical Science Textbook Page No. 47

Question 15.
Is it possible to find out adulteration of kerosene in petrol with this technique?
Answer:
The adulteration of kerosene in petrol can be found by using density meter.

9th Class Physical Science Textbook Page No. 50

Question 16.
a) Arrange the gases present in air in increas you observe?
Answer:

Gas B.P
Helium 268.93°C
Hydrogen 252.9°C
Neon 246.08°C
Nitrogen 195.8°C
Argon 185.8°C
Oxygen 183°C
Methane 164°C
Krypton 153.22°C
Xenon 108.12°C
Carbondioxide 78°C

b) Which gas forms the liquid first as the air is cooled?
Answer:
Oxygen forms the liquid first as the air is cooled.

9th Class Physical Science 3rd Lesson Is Matter Pure? Activities

Activity – 1

Question 1.
How can we separate cream from milk?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 8

  • Take some milk in a vessel.
  • Spin it with a milk churner for some time.
  • After some time you observe, separation of a paste like solid out of the milk.
  • The paste like solid is called cream.

Activity – 2

Question 2.
Explain a demonstration to identify homogeneous and heterogeneous mixtures.
Answer:

  • Take two test tubes.
  • Now add one tea spoon of salt to both the test tubes.
  • Fill one test tube with water and another with kerosene and stir them.
  • In the first test tube (water), the salt dissolves completely.
  • This is a homogeneous mixture.
  • In the second test tube (kerosene), the salt is not dissolved.
  • This is a heterogeneous mixture.

Activity – 3

Question 3.
Describe an activity to prepare saturated and unsaturated solutions.
Answer:
Preparation of saturated solution :
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 1

  1. When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.
  2. Take 50 ml of water in an empty cup.
  3. Add one spoon of sugar to the water in the cup.
  4. Stir the water until it dissolves.
  5. Keep on adding sugar to the cup and stir till no more sugar can be dissolved in it.
  6. Thus formed solution is called saturated solution.
  7. In a saturated solution, equilibrium with the undissolved solute at a certain temperature.

Preparation of unsaturated solution :
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 9

  1. If the amount of solute present in a solution is less than that in the saturated solution, is called an unsaturated solution.
  2. Now take the solution prepared by you into a beaker.
  3. Heat that solution slowly by 5 to 6°C above the room temperature.
  4. The undissolved solute dissolves.
  5. Add some more sugar to this solution.
  6. You notice that more sugar dissolves in it easily when it is heated.
  7. Thus we prepared an unsaturated solution.

Activity – 4

Question 4.
What are the factors affecting the rate of dissolving ? How do you prove them?
Answer:
Factors affecting the rate of dissolving are :

  1. Temperature of the solvent.
  2. Size of the solute particles.
  3. Stirring the solution.

Proof:

  1. Take three glass beakers and fill each of them with 100 ml of water.
  2. Add two spoons of salt to each beaker.
  3. Place the first beaker undisturbed.
  4. Stir the contents of the second beaker.
  5. Heat gently the third beaker.
  6. In all the cases, the salt dissolves but the time taken to dissolve is different.
  7. When the beaker is heated, the salt dissolved quickly.
  8. When we stir the contents, the salt dissolved but slower than heating.
  9. When we observe the undisturbed beaker, the salt dissolves but at the slowest rate.
  10. This shows that the temperature of the solvent, size of the solute particles, stirring of contents are the factors affecting the rate of dissolving.

Activity – 5

Question 5.
Describe an experiment to identify suspensions and colloids.
Answer:

  • Take some chalk powder in a test tube.
  • Take a few drops of milk in another test tube.
  • Add water to these samples and stir with a glass rod.
  • Now do the following steps and write your observations in the table given.

Step 1 :
Direct a beam of light from a torch or a laser beam on the test tubes. Observe the path of the light through the solutions.

Step 2 :
1) Leave the mixture undisturbed for some time.
2) See whether the solute settles down after some time.

Step 3 :
Filter the mixtures and observe any residue found on the filter paper.
Now read your observations :
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 10

Observations :

  1. In the chalk mixture, the particles of chalk settled at the bottom of the test tube and on filtration, we can observe a residue on the filter paper.
  2. Hence the chalk mixture is a suspension.
  3. In the milk mixture, the particles of milk are uniformly spread throughout the mixture and no residue is found on the filter paper.
  4. Hence milk mixture is a colloidal solution.

Activity – 6

Question 6.
Describe an example for the separation of mixtures by sublimation.
Describe a method of separating ammonium chloride from the mixture of ammonium chloride and common salt.
Answer:
Aim :
To separate ammonium chlo-ride from the mixture of ammonium chloride and common salt.

Materials required :
China dish, funnel, cotton, ammonium chloride, common salt and stove.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 11
Procedure:

  1. Take one table spoon of ammo-nium chloride and one table spoon of common salt and mix them.
  2. Take the mixture in a China dish.
  3. Take a glass funnel.
  4. Plug the mouth of the funnel with cotton.
  5. Invert the funnel over the dish.
  6. Heat the dish on the stove and observe the walls of the funnel.

Observations :
Initially we find vapours of ammonium chloride and then solidified ammonium chloride on the walls of the funnel.

Activity – 7

Question 7.
Describe a method to separate the dye present in ink.
(or)
Describe an example for the separation of a mixture by the process of evaporation.
Answer:
Aim :
To separate the dye present in ink by the process of evaporation.

Materials required :
Beaker, watch glass, water, ink and stove.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 12

Procedure :

  1. Take a beaker and fill it to half its volume with water.
  2. Keep 3, glass on the mouth of a beaker.
  3. Put few drops of ink on the watch glass.
  4. Heat the beaker and observe the watch glass.

Observations:

  1. We observe some fumes coming from the watch glass.
  2. Continue heating till you do not observe any further change on the watch glass.
  3. A small residue will be remained on the watch glass.

Inference :

  1. We know that ink is a mixture of a dye in water.
  2. The residue remained on the watch glass is the dye present in the ink.

Lab Activity

Question 8.
Describe paper chromatography activity to observe the colours present in a marker ink.
(OR)
How can you perform chromatography activity in your laboratory.
Answer:
Aim :
Separating the components of ink using paper chromatography.

Materials required :
Beaker, rectangular shaped filter papers, black marker (non-permanent), water, pencil and cello tape.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 13
Procedure:

  1. Draw a thick line just above the bottom of the filter paper using the marker.
  2. Pour some water into the beaker.
  3. Hang the paper strip with help of a pencil and tape in such a way that it should just touch the surface of water.
  4. Make sure that the ink line or mark does not touch the water.
  5. Allow the water to move up the paper for 5 minutes and then remove the strip from water.
  6. Let it dry.
  7. Repeat the process with green marker, a permanent marker, etc.

Observations :

  1. When black marker is used, we observe different colours like red, green, violet, black, etc. on the filter paper after drying.
  2. When green marker is used, we observe yellow, blue, green colours on the filter paper.
  3. When permanent marker is used, we cannot find any change in the mark.

Activity – 8

Question 9.
How do you separate water and kerosene from the mixture of kerosene and water?
Answer:
Aim :
To separate water and kerosene from the mixture of kerosene and water.

Materials required :
Kerosene, water, separating funnel, beakers.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 14
Procedure:

  1. Pour the mixture of kerosene and water in a separating funnel.
  2. Let it stand undisturbed for some time, so that the layers of oil and water are formed.
  3. Open the stopcock of the separating funnel and pour out the lower layer of water carefully.
  4. Close the stopcock of the separating funnel as the oil reaches the stop cock.

Principle involved :
The immiscible liquids separate out into layers depending on their densities.

Activity – 9

Question 10.
Explain the method of separation of two miscible liquids by distillation.
Answer:
Aim :
To separate two miscible liquids (water and acetone) by distillation.

Materials required :
Stand, distillation flask, thermometer, condenser, beaker, acetone and water, one holed rubber cork.
AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure 6
Procedure:

  1. Take a mixture of acetone and water in a distillation flask.
  2. Fit it with a thermometer and clamp it to stand.
  3. Attach the condenser of the flask on one side.
  4. On the other side of the condenser keep a beaker to collect distillate.
  5. Heat the mixture slowly.
  6. Keep a close watch on the thermometer.
  7. The acetone vapourizes and condenses in the condenser.
  8. The acetone can be collected from the condenser outlet.
  9. Water remains in the distillation flask.
  10. The separation technique used above is called distillation.

Activity – 10

Question 11.
How do you separate copper metal from the mixture of copper sulphate and aluminium?
Answer:

  • Take a concentrated solution of copper sulphate into a beaker.
  • Drop an aluminium foil in the beaker.
  • After some time, we observe a layer of copper deposited on the aluminium foil.
  • The solution becomes colourless.
  • A chemical reaction takes place among the copper ions present in the solution with aluminium and copper metal is separated.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 2nd Lesson Laws of Motion

9th Class Physical Science 2nd Lesson Laws of Motion Textbook Questions and Answers

Improve Your Learning

Question 1.
Explain the reasons for the following. (AS 1)
a) When a carpet is beaten with a stick, dust comes out of it.
Answer:

  1. The dust particles in the carpet are at rest.
  2. When the carpet is beaten with a stick, the state of rest of the dust particles is disturbed.
  3. Due to inertia, the dust particles comes out.

b) Luggage kept on the roof of a bus is tied with a rope.
Answer:

  1. Luggage kept on the roof of a bus is in the state of rest.
  2. As the bus moves, the luggage also moves with a velocity equal to the velocity of the bus.
  3. If the bus suddenly stops, the luggage resists to change its state of motion.
  4. Hence due to inertia it will fall down.
  5. To avoid this, the luggage is tied with a rope.

c) A pace bowler in cricket runs in from a long distance before he bowls.
Answer:

  1. When he runs in from a long distance, he gains momentum of inertia.
  2. Due to this larger inertia, larger force is applied in a short interval of time.
  3. Hence the momentum will be more.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 2.
Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why? (AS 1)
Answer:

  1. The object with mass 25 kg has more inertia.
  2. The resistance to change the state of object will be more for a body of larger mass.

Question 3.
Keep a small rectangular shaped piece of paper on the edge of a table and place an old five rupee coin on its surface vertically as shown in the figure below. Now give a quick push to the paper with your finger. How do you explain inertia with this experiment?
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 11
Answer:

  1. The coin and the paper are in inertia of rest.
  2. When we give a quick push to the paper, paper comes to inertia of motion and the coin remains in its original state i.e., inertia of rest.
  3. As a result, the paper will come out and the coin remains on the table without changing its position.

Question 4.
If a car is travelling westwards with a.constant speed of 20 m/s, what is the resultant force acting on it? (AS 1, AS 7)
Answer:

  1. A car is moving with a constant speed.
  2. Hence the net force on the car is zero both in horizontal and vertical directions.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 5.
What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s? (AS 1)
Answer:
Mass of the ball (m) = 6.0 kg
Velocity of the ball (v) = 2.2 m/s
Momentum (p) = nv = 6.0 kg × 2.2 m/s = 13.2 kg m/s (or) 13.2 N-s

Question 6.
Two people push a car for 3 sec, with a combined net force of 200 N. (AS 1)
a) Calculate the impulse provided to the car.
Answer:
Fnet = 200 N ; ∆t =3 sec
Impulse ∆p = Fnet. ∆t = 200 × 3 = 600 N – sec.

b) If the car has a mass of 1200 kg, what will be its change in velocity?
Answer:
Mass of the car (m) = 1200 kg.; Net force (Fnet) = 200 N
Time ∆t = 3 sec. ; Change in velocity ∆v =?
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 1

Question 7.
What force is required to produce an acceleration of 3 m/sec2 in an object of mass 0.7 kg? (AS 1)
Answer:
Mass of the object (m) = 0.7 kg.; Acceleration (a) = 3 m/sec²
Force required (F) = ?
F = ma = 0.7 x 3 = 2.1 N

Question 8.
A force acts for 0.2 sec on an object having mass 1.4 kg initially at rest. The force stops to act but the object moves through 4 m in the next 2 seconds, find the magnitude of the force. (AS 1)
Answer:
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 3
Velocity after 0.2 sec v = u + at = 0 + 0.2 a = 0.2a ……….. (1)
After 0.2 s, the body moves with uniform velocity, acceleration is zero, because force is removed.
∴ Velocity v = \(\frac{s}{t}=\frac{4}{2}\) =2 m/s. ………. (2)
From (1) & (2)
v = 0.2a ⇒ 2 = 0.2a
⇒ a = \(\frac{2}{0.2}\) = 10 m/s²
∴ Force applied F = ma = 1.4 kg × 10 m/s² = 14 N.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 9.
An object of mass 5 kg is moving with a velocity of 10 ms-1. A force is applied so that in 15 s, it attains a velocity of 25 ms-1. What is the force applied on the object? (AS 1)
Answer:
Mass (m) = 5 kg ; Initial velocity (u) = 10 m/s.; Time (t) = 15 s
Final velocity (v) = 25 m/s.
Acceleration a = \(=\frac{v-u}{t}=\frac{25-10}{15}=\frac{15}{15}=1 \mathrm{~m} / \mathrm{s}^{2}\).
Force applied on the object F = ma = 5 × 1=5 N.

Question 10.
Find the acceleration of body of mass 2 kg from the figures shown. (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 4
Answer:
1) Force 30 N is acting downwards on weight of (2 × 10) = 20 kg.
The acceleration a = \(\frac{30-20}{2}=\frac{10}{2}\) = 5 m/s²
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 5

2) m1 = 2 kg, m2 = 3 kg.
m2 pulls the body mt with a weight 3 × 10 = 30 N.
∴ Acceleration of m1 = \(\frac{30-20}{3+2}=\frac{10}{5}\)
= 2 m/s².

Question 11.
Take some identical marbles. Make a path or a track keeping your notebooks on either side so as to make a path in which marbles can move. Now use one marble to hit the other marbles. Take two, three marbles and make them to hit the other marbles. What can you explain from your observations? (AS 5)
Answer:

  1. When one marble is hit by another marble, both the marbles move with some velocity.
  2. When the marble is hit by two, three marbles, all marbles move with a velocity which is more than in the previous case.
  3. As we are hitting with more marbles, the mass increases. So that the net momentum also increases.

Question 12.
A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the maximum acceleration with which he can climb safely? (AS 1, AS 7)
Answer:
Mass m = 30 kg. ; Force F = 450 N
Acceleration a = ?
F = ma
∴ a = \(\frac{\mathrm{F}}{\mathrm{m}}=\frac{450}{30}\) =15 m/sec²
∴ The required acceleration =15 m/sec²

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 13.
An vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/sec²? (AS 1, AS 7)
Answer:
Mass of the vehicle, m = 1500 kg.; Acceleration (-a) = 1.7 m/sec²
Force, F =?
F = m (-a) = 1500 × (-1.7) = ( – ) 2550 N
∴ The force between the vehicle and road is 2550 N, in the direction opposite to that of the vehicle.

Question 14.
If a fly collides with the windshield of a fast moving bus, is the impact force experienced, same for the fly and the bus? Why? (AS 1, AS 2)
Answer:
The impact force experienced by the fly will be more, because the mass of fly is negligible when compared to the mass of the bus.

Question 15.
A truck is moving under a hopper with a constant speed of 20 m/sec. Sand falls on the truck at a rate 20 kg/s. What is the force acting on the truck due to falling of sand? (AS 1, AS 7)
Answer:
Mass of the sand falling on the truck in 1 sec = 20 kg
Constant speed of the truck = 20 m/s
Acceleration in 1 sec, a = \(\frac{\Delta v}{\Delta t}=\frac{20}{1}\) = 20 m/sec²
Force applied on the truck, F = ma = 20 kg x 20 m/sec² = 400 N

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 16.
Two rubber bands stretched to the standard length cause an object to accelerate at 2 m/sec². Suppose another object with twice the mass is pulled by four rubber bands stretched to the standard length. What is the acceleration of the second object? (AS 1)
Answer:
First object:
Let the force applied by two rubber bands = F1 Newton
Mass of the object = m1 kg; Acceleration a1 = 2 m/sec²
We know F = ma
F1 = m1 × 2
⇒ F1 = 2m1 …….(1)

Second object:
The force applied by 4 rubber bands = 2F1 Newton
Mass of the object = 2m1 kg ; Acceleration a2 =?
We know F = ma
2F1 = 2m1. a2
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 6
∴ Acceleration of the second object = 2 m/sec²

Question 17.
Illustrate an example of each of the three laws of motion. (AS 1)
Answer:
First law of motion :
A body continues its state of rest or of uniform motion unless a net force acts on it.
Ex:

  1. When the bus which is at rest begins to move suddenly, the person standing in the bus falls backward.
  2. When you are travelling in bus, the sudden stop of the bus makes you fall forward.

Second law of motion :
The rate of change of momentum of a body is directly proportional to the net force acting on it and it takes place in the direction of net force.
Ex : Place a ball on the veranda and push it gently. Then the ball accelerates from rest. Thus, we can say that force is an action which produces acceleration.

Third law of motion :
If one object exerts a force on the other object, the second object exerts a force on the first one with equal magnitude but in opposite direction.
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 2
Ex :

  1. When birds fly, they push the air downwards with their wings, and the air pushes back the bird in opposite upward direction.
  2. When a fish swims in water, the fish pushes the water back and the water pushes the fish with equal force but in opposite direction.
  3. A rocket accelerates by expelling gas at high velocity. The reaction force of the gas on the rocket accelerates the rocket in a direction opposite to the expelled gases.

Question 18.
Two ice-skaters initially at rest, push of each other. If one skater whose mass is 60 kg has a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg? (AS 1, AS 7)
Answer:
Mass of first skater m1 = 60 kg.;
Velocity of first skater v1 = 2 m/s.
Mass of second skater m2 = 40 kg.; Velocity of second skater v2 =?
As the two skaters push each other, the resultant momentum will become zero.
The resultant momentum m1v1 + m2v2 = 0
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 12
∴ Velocity of second skater is 3 m/s, but in the direction opposite to the first skater.

Question 19.
A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is (AS 7)
a) Accelerated
b) Uniform
c) Retarded
d) Circular motion
Answer:
a) Accelerated

Question 20.
A horse continues to apply a force in order to move a cart with a constant speed. Explain. (AS 1)
Answer:

  1. The cart moves when the force (in the form of pulling) is applied by the horse.
  2. As the horse and cart are moving, the net momentum will be zero at any instance of time.
  3. Hence when the horse comes to rest, the cart also comes to rest.
  4. To avoid this and to move the cart with a constant speed, the horse must apply force continuously.

Question 21.
A force of 5 N produces an acceleration of 8 m/sec² on a mass m, and an acceleration of 24 m/sec² on a mass m2. What acceleration would the same force provide If both the masses are tied together? (AS 1)
Answer:
For the first mass (m1)
Force F = 5 N
Acceleration a = 8 m/s²
We know, F = ma
5 = m, . 8
m1= \(\frac{5}{8}\) kg

For the second mass (m2)
Force F = 5 N
Acceleration a = 24 m/sec²
We know, F = ma
5 = m2.24
m2 = \(\frac{5}{24}\) kg
When both the masses are tied together and the same force is applied, then
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 7

Question 22.
A hammer of mass 400 g, moving at 30 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? (AS 1)
Answer:
Mass of the hammer (m) = 400 g = 0.4 kg. ; Velocity of the hammer (v) = 30 m/s.
Momentum (∆p) = 30 × 0.4 N – s.
The nail stops the hammer with in a time 0.01 s.
∴ ∆t = 0.01 s.
The stopping force of the nail on the hammer.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 23.
System is shown in figure. Assume there is no friction. Find the acceleration of the liloc-ks-and tension in the string. Take g = 10 m/s² (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 8
Answer:
m1 = 3 kg, m2 = 3 kg.
1) Acceleration ’a’ and tension T on m1 are shown in figure.
2) Acceleration a and tension T will be as shown in figure.
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 9
Normal force = weight 3s – T = 3a ………… (2)
∴ only tension applied
Tension T = 3 kg × a = 3a
From (1) & (2)
3g – 3a = 3a ⇒ 3g = 6a ⇒ a = \(\frac{3 g}{6}=\frac{3 \times 10}{6}\) = 5 m/s².
Tension T = 3a = 3 × 5 = 15 N.

Question 24.
Three identical blocks, each of mass 10 kg, are pulled as shown on the hoii ;ontal frictionless surface. If the tension (F) in the rope is 30 N, what is the acceleration oi each block? And what are the tensions in the other ropes? (Neglect the masses of the ropes) (AS 1)
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 10
Answer:
Three blocks, each of mass 10 kg are pulled by a rope.
∴ Total mass = 10 + 10 + 10 = 30 kg
Force applied by the rope, F = 30 N
∴ Acceleration of each block, a = \(\frac{F}{m}=\frac{30}{30}\) = 1 m/sec².

Tension in first rope (T1)
First rope pulls only one block whose mass is 10 kg with an acceleration 1 m/sec².
∴ F = ma = 10 kg × 1 m/sec² = 10 N

Tension in second rope (T2)
Second rope pulls two blocks, each of mass 10 kg.
∴ Total mass = 10 + 10 = 20 kg.
Acceleration, a = 1 m/sec²
Force, F = ma = 20 × 1 = 20 N

Question 25.
A ball of mass’m’ moves perpendicularly to a wall with a speed v, strikes it and rebounds with the same speed in the opposite direction. What is the direction and magnitude of the average force acting on the ball due to the wall? (AS 7)
Answer:
According to the Newton s third law of motion,
Force exerted by ball on the wall = – (Force exerts by the wall on the ball)
∴ FB.W = – FW.B
Force exerted by ball :
Mass of ball = m, speed = v, FBW = ma = \(\frac{\mathrm{m} \cdot \mathrm{v}}{\mathrm{t}}\)
As the wall is at rest and exerts some force on the ball of mass m, then it moves in the other direction with the same speed.
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 13

Question 26.
Divya observed a horse pulling a cart. She thought that cart also pulls the horse with same force in opposite direction. As per third law of motion, the cart should not move forward. But her observation of moving cart raised some questions in her mind. Can you guess what questions are raised in her mind? (AS 2)
Answer:

  • According to Newton’s third law, when horse pulls a cart, the cart also pulls the horse with same force but in opposite direction. So the cart has to stop. Why is the cart moving?
  • What is the effect of friction of ground on the cart and horse?
  • Does the cart continue to move even if the horse stops pulling it?
  • What makes the cart to move continuously?
  • Does it become an isolated system?
  • What is the action and reaction in this system?

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 27.
How do you appreciate Galileo’s thought of “any moving body continues in the stale only until some external force acts on it”, which is contradiction to the Aristotle’s belief of “any moving body naturally comes to rest”? (AS 6)
Answer:

  • Science is dynamic.
  • All theories can change time to time so that the science and technology will be developed.
  • Aristotle’s belief proved to be wrong only by the experiments conducted by Galileo.
  • So anybody can challenge the existing theories with proper experimentation.
  • Aristotle’s and Galileo’s contradictory thought lead Newton to propose most popular laws of motion.
  • Newton’s third law of motion is the basic principle in rocket launching.
  • Nowadays we are enjoying the results of satellites launched by rockets.
  • Hence comfortable life is the effort on experiments, theories, and calculations made by scientists with a zeal to invent new.

9th Class Physical Science 2nd Lesson Laws of Motion InText Questions and Answers

9th Class Physical Science Textbook Page No. 24

Question 1.
Do all the bodies have the same inertia?
Answer:
The inertia of all bodies is not same. It depends on the mass of the object.

Question 2.
What factors can decide the inertia of a body?
Answer:
Mass is the factor, which decides the inertia of a body.

9th Class Physical Science Textbook Page No. 26

Question 3.
Is the acceleration increased when net force is increased?
Answer:
Yes, the acceleration increased, as we increased the net force without changing mass of the object.

9th Class Physical Science Textbook Page No. 29

Question 4.
What do you notice from the readings in the spring balances?
Answer:
The two spring balances stretch up to a certain limit equally.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 5.
Are the readings of two spring balances the same?
Answer:
Yes, the readings of two spring balances are equal.

Question 6.
Are we able to make the spring balances to show different readings by pulling them simultaneously in opposite directions? Why not?
Answer:
When same force is applied in both the directions, we are unable to make the spring balances to show different readings because the action and reaction are same in magnitude and opposite in direction. When we use two forces with different magnitudes, then the spring balances can show different readings.

9th Class Physical Science Textbook Page No. 30

Question 7.
Does the rocket exert a force on the gas expelled from it?
Answer:
The rocket also exerts a force on the gas expelled from it.

9th Class Physical Science Textbook Page No. 32

Question 8.
Why does a pole vault jumper land on thick mats of foam?
Answer:
A thick mat of foam reduces the force of impact of the jumper, so that he doesn’t have any damage to his body.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 9.
Is it safe to jump on sand rather than a cement floor? Why?
Answer:

  • It is safe to jump on sand rather than a cement floor.
  • A soft and more cushioned landing surface provides a greater stopping distance because of the longer time taken to stop.

9th Class Physical Science Textbook Page No. 24

Question 10.
You may have seen the trick where a tablecloth is jerked from a table, leaving the dishes that were on the cloth nearly in their original positions.
a) What do you need to perform this successfully?
Answer:
We need a table, a cloth and some massive objects to perform this activity. The performer drag the cloth from the table very skillfully.

b) Which cloth should we use? Is it cloth made of thick cotton or thin silk?
Answer:
We have to use a thin silk cloth to perform this activity.

c) Should the dishes possess large mass or small mass?
Answer:
The dishes must possess large mass. We should not use lighter objects like plastic cups, etc.

d) Is it better to pull the cloth with a large force or pull it with a gentle and steady force?
Answer:
The cloth must be pulled with a gentle force, but with a sudden jerk.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 11.
What is the velocity of a small object that has separated from a rocket moving in free space with velocity 10 km/s?
Answer:
When a small object is separated from another object which is moving with a certain velocity. The small object also moves with a velocity equal to that of the object from which it is separated. Hence, the speed of the small object is 10 km/s.

9th Class Physical Science Textbook Page No. 27

Question 12.
Observe the following diagram.
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 23
What is the upper limit of weight that a strong man of mass 80 kg can lift as shown in figure?
Answer:
Total force mg = N + T (N = normal force, T = tension)
As the person is standing on the floor, the normal force N = 0.
80 = 0 + T
∴ T = 80
∴ The upper limit of the weight that the person in the figure can lift is 80 kg.

Question 13.
What is the momentum of a ceiling fan when it is rotating?
Answer:
Ceiling fan when it is rotating, possesses angular momentum.
Angular momentum L = mvr or mr²ω.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 14.
Is it possible to move in a curved path in the absence of a net force?
Answer:
A body comes into curved path, when centripetal force real force acts on it. Immediately after coming into curved path, an imaginary force which acts away from the centre i.e., centrifugal force comes into existence. These two forces are equal in magnitude and opposite in direction. Hence the net force is zero. So it is possible to move in a curved path in the absence of a net force.

Question 15.
Prove that the tension throughout the string is uniform when the mass of string is considered to be zero.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 24
Let a body of mass m is suspended through a string. The weight of the object mg acts downwards. Now tension in the string T = mg + msg when ms is the mass of the string.
Here ms is considered as zero.
Hence TA= mg + 0 = mg; TB = mg ; TC = mg ; TD = mg
∴ The tension throughout the string is uniform when the mass of string is considered to be zero.

9th Class Physical Science Textbook Page No. 31

Question 16.
The force exerted by the earth on the ball is 8 N. What is the force on the earth by the ball?
Answer:

  • The force exerted by the earth on the ball is 8 N.
  • The force exerted by the ball on the earth is – 8 N.

“According to Newton’s third law, if a body A exerted a force p on another body B, the B exerts a force -p on A, the two forces acting along the same line”.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 17.
A block is placed on the horizontal surface. There are two forces acting on the block. One, the downward pull of gravity and other a normal force acting on it. Are these forces equal and opposite? Do they form action – reaction pair? Discuss with your friends.
Answer:
These two forces form action-reaction pair.

Question 18.
Why is it difficult for a fire fighter to hold a hose that ejects large amount of water at high speed?
Answer:
A large amount of water with high speed ejects from the hose of a fire engine, produces a large force in forward direction. According to action-reaction, the hose moves back with the same force. But the fire fighter has to resist that reaction force. Hence it becomes very difficult for him.

9th Class Physical Science Textbook Page No. 33

Question 19.
A meteorite burns in the atmosphere before it reaches the earth’s surface. What happens to its momentum?
Answer:
The momentum of the meteorite becomes zero. It doesn’t touch the ground as it burns in the atmosphere. So no mass of the meteorite hits the ground.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion

Question 20.
As you throw a heavy ball upward, is there any change in the normal force on your feet?
Answer:
The normal force on the feet changes its direction and acts in upward direction. As a result we raise our foot while throwing the ball.

Question 21.
When a coconut falls from a tree and strikes the ground without bouncing. What happens to its momentum?
Answer:
Its momentum doesn’t change but its impact force will be very less because it is not bouncing.

Question 22.
Air bags are used in’the cars for safety. Why?
Answer:
When a car hits another vehicle, the air bags immediately comes in between the persons in the car and the wind shield of the car, to prevent damage to life of passengers.

9th Class Physical Science 2nd Lesson Laws of Motion Activities

Activity – 1

Question 1.
Explain the motion of a pen cap kept on a thick paper ring.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 14

  1. Make a circular strip from a thick paper.
  2. Balance the hoop on the centre of the mouth of the bottle.
  3. Now balance a pen cap on the paper hoop aligning it on the centre of the bottle’s mouth.
  4. Give the paper hoop a sharp push with your finger as fast as you can.
  5. We observe that the pen cap suddenly falls into the bottle.
  6. As we push paper hoop, we applied force on the paper hoop. So it changed its state from rest to motion.
  7. Pen cap cannot change its state of rest.
  8. Due to gravitational force, the pen cap falls into the bottle.

Activity – 2

Question 2.
Explain the motion of the carrom coins hit by a striker.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 15

  1. Make a stack of carrom coins on the carrom board.
  2. Give a sharp hit at the bottom of the stack with striker.
  3. We can find that the bottom coin will be removed from the stack.
  4. The other coins in the stack will slide down.
  5. When we apply force on the bottom coin, the coin will move, due to change in the state of rest.
  6. The stack of remaining coins does not fall vertically due to inertia.

Activity – 3

Question 3.
Show that the object with larger mass has greater inertia.
Answer:
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 16

  1. Take two rectangular wooden blocks with different masses.
  2. Place them on a straight line drawn on a floor.
  3. Give the same push at the same time to both the blocks with the help of a wooden scale.
  4. We observe that the block with small mass will accelerate more and goes farther.
  5. The block with large mass accelerates less and moves shorter, due to high inertia.
  6. This shows that the bodies of higher mass have high inertia.

Activity – 4

Question 4.
Show that the larger the net force greater the acceleration.
Answer:

  1. Gently push a block of ice on a smooth surface and observe how the object speeds up, in other words how it accelerates.
  2. Now increase the net force and observe change in its speed.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 17
Observation : The acceleration increases.

Conclusion : If the net force is larger, then the accelerations greater.

Activity – 5

Question 5.
Show that the larger the mass smaller the acceleration.
Answer:

  1. Apply a force on an ice block.
  2. It undergoes some acceleration.
  3. Now take a block of ice with greater mass.
  4. Then apply almost the same force on the ice block which has greater mass.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 18
Observations :

  1. In both cases the object accelerates.
  2. But we can observe in the second case, it will not speed up as quickly as before.

Conclusion : If the mass is larger, then the acceleration is smaller.

Activity – 6

Question 6.
Pulling two spring balances.
Answer:
Let’s take two spring balances of equal calibrations. Connect the two spring balances as shown in figure. Pull the spring balances in opposite directions as shown in figure.
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 19
Observation :
There is no change in the reading of spring balances. We are not able to make the spring balances to show different readings by pulling them simultaneously in opposite directions.

Conclusion :
According to third law of motion, when an object exerts a force on the other object, the second object also exerts a force on the first one which is equal in magnitude but opposite in direction.

The two opposing forces are known as action and reaction pair. Newton’s third law explains what happens when one object exerts a force on another object.

Activity – 7

Question 7.
Describe the preparation of a balloon rocket. What is the principle involved in it?
Answer:
AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 20
Preparation of a balloon rocket :

  1. Inflate a balloon and press its neck with fingers to prevent air escaping from it.
  2. Pass a thread through a straw and tape the balloon on the straw.
  3. Hold one end of the thread and ask your friend to hold the other end of the thread.
  4. Now release air from balloon by removing fingers from the neck of the balloon.
  5. The balloon moves like a rocket towards the other end

Principle involved in it:

  1. Newton’s third law of motion is the principle.
  2. As the air in the balloon moves backward, the balloon moves forward.

Lab Activity

Question 8.
Describe an activity to study the action and reaction forces acting on two different objects.
Answer:
Aim : To show the action and reaction forces acting on two different objects.

Material required : Test tube, rubber cork cap, Bunsen burner, laboratory stand and thread.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 21
Procedure :

  1. Take a test tube and put a small amount of water in it.
  2. Place a rubber cork cap at its mouth to close it.
  3. Now suspend the test tube horizontally to a stand with the help of two strings.
  4. Heat the test tube with a bunsen burner until water vapourize and the rubber cork cap blows out.

Observations :

  1. Observe the movement of test tube when cork cap blows out.
  2. As the cork cap blows out in forward direction, the test tube recoils back.
  3. We can observe the velocities of cork cap and recoil of test tube to be same.

Activity – 8

Question 9.
Show that the impulse will be less on a soft and cushioned surface.
Answer:

  • Take two eggs.
  • Drop them from a certain height, so that one egg falls on a concrete floor and the other on a cushioned pillow.
  • We observe that the egg that falls on a concrete floor will break.

AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion 22

  • The reason is large force acts on the egg for short interval of time.
    ∆p = Fnet∆t
  • The egg which falls on a cushioned pillow doesn’t break, because a smaller force acts on the egg for a longer time.
    ∆p= Fnet ∆t
  • This shows that the impulse (∆P) will be less on a soft and cushioned surface.

Note : Even if the ∆p is the same in both cases, the magnitude of the net force (Fnet</sub) acting on the egg determines whether the egg will break or not.

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 6C An Icon of Civil Rights Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 6C An Icon of Civil Rights

9th Class English Chapter 6C An Icon of Civil Rights Textbook Questions and Answers

Answer the following questions.

Question 1.
The speaker talks about “creative battle” in the beginning of his speech. What does he mean by this phrase?
Answer:
Martin Luther King (Jr) describes the Civil Rights Movement of the blacks in the USA as a ‘Creative battle’. This is to show that the battle is going to ‘create’ a new world. The battle uses ‘good’ will and ‘good’ intention as weapons. It follows ‘good’ methods like ‘truth’, ‘non-violence’ and ‘love’. Its aim is to promote universal brotherhood. Hence it is a creative battle.

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

Question 2.
What is Martin Luther King’s speech about? List the issues he is talking about.
Answer:
Martin Luther King’s speech is about justice and equality. It is about universal brotherhood. It is about food to every body, education to every mind and dignity and respect for every spirit. It is about truth, love and peace. It is delivered as Nobel Prize acceptance speech.

Question 3.
Do you think that this is an emotive speech ? If yes, pick out the expressions that show it is an emotive speech.
Answer:
Yes, it is an emotive speech. Every part, in fact, is an example to prove the point. Yet, here are some striking expressions : 22 million Negroes are engaged in a creative battle ; our children, crying for brotherhood, were answered with fire hoses beleaguered and unrelenting struggle ………

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

Question 4.
What sort of future does the speaker visualize for the Americans and the mankind in general?
Answer:
Martin Luther King (Jr) is full of hope. He visualises a bright future for Americans and humanity. He dreams of a widening and lengthening super highway. Blacks and whites will travel along it in a cooperative and brotherly mood. That will lead them to an ideal land. There everyone gets food, education, justice, equality and dignity. Love, truth and peace will rule supreme.

Vocabulary

I. Given below are the words taken from the reading passage listed as key words. Match the word with the meaning as used in the text.

Key word Choice words
afflict affect, touch, cause pain
beleaguered experienced criticism, shattered, humiliated
retaliation violation, reformation, revenge
tortuous complicated, unclear, straight
prostrate lie flat, roll on, unmoved
turmoil certainty, great confusion, trouble
curator representative, person in charge, physician

Answer:

afflict (v) cause pain
beleaguered (adj) experienced criticism
retaliation (n) revenge
tortuous (adj) complicated
prostrate (adj) lie flat
turmoil (n) great confusion
curator(n) person in charge

II. Read the following expressions taken from the reading passage.
1. blazing light of truth
2. wounded justice
3. majestic scorn

Do they have any specific meaning?
Why does the speaker use such expressions?

The above phrases are figurative expressions. They mean a word or a phrase used in a different way from its usual meaning in order to create a particular mental image or effect to add interest to a speech or a writing. Here the two words that convey opposite meaning are combined together to get a positive meaning.

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

Now read the passage once again and pick out the figurative expressions.

Find out the meanings of all the expressions including the ones given above.
Answer:

1. blazing light of truth bright light generated by pure truth
2. wounded justice justice destroyed ; injustice pervading everywhere
3. majestic scorn contempt or hatred of highest order
4. sterile passivity unproductive inactiveness
5. creative psalm of brotherhood the feeling of being brothers is like a song praising God
6. super highway of justice a smooth path of justice

Writing

I. You have listened to the speech delivered by Subhash Chandra Bose and read the speech by Martin Luther King Jr.
Let’s analyze their speeches.

Discuss the following questions in groups.
1) How do they begin their speeches?
2) Do you find any logical sequence of ideas in their speeches?
3) What sort of language do they use? (Persuasive, argumentative, emotive)
4) Do you notice any quotations, examples? (to support their argument)
5) Do they use any linkers for cohesion?
6) Do they maintain unity of ideas/thoughts for coherence?
7) What expressions do they use to conclude their speeches?
Answer:

  1. They begin their speeches with one’s obligations to family and country and to the Civil Rights Movement in the USA.
  2. Yes, there is absolute logical sequence of ideas in their speeches.
  3. They use argumentative and emotive language.
  4. Yes, I do notice quite a good number of examples.
  5. Yes, they use linkers liberally for the purpose of cohesion.
  6. Yes, they do maintain unity of ideas for coherence.
  7. They conclude their speeches with expressions of hope, freedom, justice, and equality for all!

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

II. Prepare a speech on the following occasion in your school.
You can use some of the quotations given in the box.

Independence Day
AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights 1

Answer:
Respected teachers, elders and guests and my dear friends! A very brilliant morning and warm greetings of our greatest day of days to you all.

We breathe this cool morning the inspiring air of freedom. Today we stand in the protective shelter of our tricolour flag as it flies high in the sky and proclaims to the world the great achievement of Indians! But the flag also reminds us of the selfless sacrifice thousands of Indians made to gift us this invaluable independence. The flag, at the same time, advises us to be mindful of our responsibilities.

As we celebrate this great event in the lap of goddess of learning, shouldn’t we pledge to follow the path our national leaders have so brilliantly illuminated for us? As the present students and future citizens, shall we plan for our future and our mother India’s future too?

As you all know well, planning for future demands analysis of present. Are we, then, rellay free in its comprehensive sense? Are we, for example, free from fear? For, as Aristotle has well said, “He who has overcome his fears will truly be free”. Are we really aware of and following Eleanor Roosevelt when he says, “Freedom makes a huge requirement of every human being. With freedom comes responsibility. For the person who is unwilling to grow up, the person who does not want to carry his own weight, this is a frightening prospect?” Are we really respecting freedom of expression? Do we follow

S.G. Talientyre’s mantra in this regard? The mantra is : “I disapprove of what you say, but I will defend to the death your right to say it.” Are we free from hatred and jealousy? Shouldn’t we put into practice Martin Luther King’s (Jr) wise advice – Let us not seek to satisfy our thirst for freedom by drinking from the cup of bitterness and hatred – even after independence? Are we allowing our innate nature to move ahead in its own way? Do we prove Virginia Woolf right in the saying – Lock up your libraries if you like, but there is no gate, no lock, no bolt that you can set upon the freedom of my mind?

The present picture is full of challenges. Illiteracy, poverty, corruption, pollution, fast depleting natural resources – the list is too long to complete here. It haunts us day and night. Let us all work together to solve these burning problems. Let us empower ourselves first to be able to work for Mother India! Let us hope to see Mother India full of smiles! Thank you all for your encouraging attention. Jai Hind I

Project Work

Collect information about the great leaders who fought for the freedom of our country. Arrange the information in the table given below:

Discuss in groups and write down the questions you will need to get the information.

On the basis of the information collected in the table below, write a brief biographical sketch of any one of them and present it before others in class.
AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights 2
AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights 3
Answer:
A BRIEF BIOGRAPHICAL SKETCH OF SAROJINI NAIDU

Sarojini Devi was born on 13th February 1879 in Hyderabad to Aghoranath Chattopadhyaya and Varada Sundari Devi.

She passed matriculation when she was just 12. Then she went to England for higher studies in King’s college in London and later at Cambridge.

She married Mutyala Govindarajulu Naidu in 1898. Kandukuri Veeresalingam acted as the priest on the occasion.

Gopala Krishna Gokhale influenced and guided Sarojini Naidu into Freedom Struggle. She met Gandhiji in 1915 and Nehru in 1916 for the first time. She took a very active part in Congress activities and Independence movements. She inspired many men with her poetic recitals which won for her the title ‘Nightingale of India’, Recognizing her important role, Congress leaders made her the party president in 1925. But the British government imprisoned herforthe same role.

After India became independent, she was made the governor of United Provinces (now Uttar Pradesh), the first Indian woman to become so. People remember her as a patriot, poet, philospher and lover of peace. She remains a source of inspiration to many persons-particularly women.

An Icon of Civil Rights Summary in English

Martin Luther King (Junior) is one of the most influential and inspiring Afro-American leader. His Nobel Prize acceptance speech is noted for its excellent content and effective expression. He accepts the Nobel Prize on behalf of the Civil Rights Movement in America. He remembers the suffering of blacks in the USA. He praises the Indians for showing the path of non-violence. He believes in the power of truth and love. He is hopeful of bright future for humanity. He has faith in man’s ability to lay a super highway of justice on which all people will cooperatively go ahead and create a world where every body gets food, mind finds education and spirit receives dignity, equality and brotherhood. He says faith gives us courage and confidence to complete the mission of establishing universal brotherhood that has roots in love, truth and peace.

An Icon of Civil Rights Glossary

icon (n) : a famous person ; a symbol of an idea

majesty, highness, excellencies (nouns) : terms showing high respect to people of top positions

accept(v) : take; receive

determination (n) : firmness in continuing to do something despite difficulties

scorn (n) : contempt; hatred

reign (n) : rule (Note that the letter ‘g’ is silent.)

mindful (adj) : aware ; remembering

sanctuary (n) : a place that is safe

segregation (n) : separation ; isolation

debilitating (v+ing) : weakening

grinding (v+ing) (adj) : never ending ; never improving

afflicts (v) : causes suffering

rung (n) : step

AP Board 9th Class English Solutions Chapter 6C An Icon of Civil Rights

beleaguered (adj) : experiencing a lot of criticism

unrelenting (v+ing : adj) : not stopping and not becoming less severe

contemplation (n) : serious thinking

antithetical (adj) : opposite

sterile (adj) : not capable of producing

elegy (n) : a poem or song about sad feelings

psalm (n) : a song praising god (Note that the letters ‘p’ and T are silent.)

revenge(n) : causing suffering to others because the caused suffering

aggression (n) : attack

retaliation (n) : action against someone because they harmed

tortuous (adj) : complicated, long, full of bends

alliances (n-plural) : groups working together to achieve what they want

audacious (adj) : daring ; willing to take risks

despair (n) : the feeling of losing all hope

flotsam and jetsam (phrase-noun) : useless things, persons with no job, no home

cynical (adj) : not believing that something good will happen

triumphant (adj) : successful; victorious

prostrate (adj) : lying on the ground facing downwards

altars (n-plural) : places of worships

redemptive (adj) : that saves from evil

dreary (adj) : that which makes one dull, bored

soared (v-past tense) : rose high ; went up

righteousness (n) : morally good behaviour

heirloom (n) : a valuable object from forefathers

AP Board 9th Class English Solutions Chapter 6B Freedom

AP Board 9th Class English Solutions Chapter 6B Freedom

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 6B Freedom Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 6B Freedom

9th Class English Chapter 6B Freedom Textbook Questions and Answers

Answer the following questions.

Question 1.
What sort of freedom does the poet wish to have?
Answer:
The poet wants freedom from fear and physical, intellectual, psychological, emotional, social, and financial weaknesses.

Question 2.
The poet talks about fear in the opening lines of the poem. What kind of fear is he talking about?
Answer:
The poet refers to fear of country’s future, development, social and economic equality, etc.

AP Board 9th Class English Solutions Chapter 6B Freedom

Question 3.
What does the expression “truth’s adventurous paths” mean?
Answer:
“Truth’s adventurous paths” means the road taken by persons speaking only truth is full of challenges and problems.

Question 4.
What does the poet mean by ‘figures’ in the poem? What sort of figures are they? What does the poet want them to be?
Answer:
‘Figures’ here are people. They are obedient and patient. The poet wants them to have their own individuality.

Question 5.
“Where figures wait with patience and obedience for the master of show.” What does the poet mean by this?
Answer:
Persons here wait for their leader. They just follow their master without any original thinking.

AP Board 9th Class English Solutions Chapter 6B Freedom

Question 6.
What does the ‘shackles of slumber’ mean? How does it arrest the progression of life?
Answer:
‘Shackles of slumber’ means chains of laziness, inaction and indifference. When one is a prisoner of laziness, life does not progress.

Question 7.
Do you think we are all free from fear? What kind of fears are haunting our motherland now?
Answer:
No, we are not at all free from fears. The fears of over population, over pollution, lawlessness, insecurity to women and the weak are some of the striking ones haunting our mother land.

Question 8.
What does freedom mean to you? Is it freedom from hunger? Is it freedom from physical attack? Is it freedom from illiteracy? is it freedom from social oppression? What else?
Answer:
Freedom from desire, selfishness, illiteracy, dirt and disease, hatred, poverty, etc. is what freedom means to me.

Freedom Summary in English

The poem ‘Freedom’ by Rabindranath Tagore is both philosophical and psychological. He wants a special kind of freedom for the motherland. It is freedom from fear. It is freedom from old age and related weaknesses. It is freedom from lack of vision and slumber. It is freedom from mistrust. It is freedom from uncertain destiny, cruel power and lack of individuality. It is freedom from meaningless movements, senseless statistics and heartless artificiality.

Freedom Glossary

claim (v) : demand ; achieve

beckoning (v+ing) : showing signs to call

shackles (n-plural) : chains

slumber (n) : sleep

mistrusting (v+ing) : having no faith

AP Board 9th Class English Solutions Chapter 6B Freedom

anarchy(n) : a situation without order, control or government

helm (n) : steering wheel or handle

puppet (n) : a person whose actions are controlled by others

mimicry (n) : copying; imitation