# NCERT Solutions Class 9 Science Chapter 10 Gravitation

NCERT Solution MCQ Quiz Notes Important Questions MCQ Quiz – 1 MCQ Quiz – 2**In-Text Questions Solutions**

**In Text-Questions Page 134**

**State the universal law of gravitation.****Ans:-**According to universal law of gravitation, every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of distance between them. The direction of force is along the line joining the two particles.

Consider two objects of masses*m*_{1}and*m*_{2}and let the distance between their centres be*r.*The gravitational force of attraction (*F*) acting between them is given by the universal law of gravitation as

where, G is the universal gravitation constant**Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.****Ans:-**The formula for the magnitude of gravitational force between the earth and an object on its surface is,

where F is the gravitational force. G is the gravitational constant. M_{e}is the mass of the earth. m is the mass of the object on the surface of the earth. R_{e}is the radius of the earth.

**In Text-Questions Page 136**

**What do you mean by free fall?****Ans:-**When an object falls towards the earth due to gravitational force only then the object is said to be in free fall.**What do you mean by acceleration due to gravity?****Ans:-**The acceleration with which an object fall freely towards the earth is known as acceleration due to gravity. It is denoted by g and its value is 9.8 m s^{-2}.

**In Text-Questions Page 138**

**What are the differences between the mass of an object and its weight?****Ans:-**

Mass | Weight | |

1. | The quantity of matter present in a body is called its mass. | The force with which the earth attracts a body towards its center is called its weight. |

2. | S.I. unit is kg. | S.I. unit is Newton. |

3. | It remains constant everywhere. | Its value changes from place to place. |

4. | It is measured by common balance. | It is measured by spring balance. |

2. ** Why is the weight of an object on the moon 1/6 its weight on the earth?****Ans:-** The weight of an object depends on the value of acceleration due to’gravity g. The value of g on earth is 6 times more than that of moon because, the mass and radius of the earth is more than the mass and radius of the moon.

We have, g = GM/R^{2} and W = mg

Weight of a body of mass m on earth is

W_{e} = mg_{e} = mGM_{e}/R_{e}^{2}

Weight of a body of mass m on moon is

W_{m} = mg_{m} = mGM_{m}/R_{m}^{2} or, W_{m}/W_{e} = M_{m}/R_{m}^{2} x R_{e}^{2}/M_{e}

As mass of the moon M_{m} is 1/100 times the mass of earth M_{e} and radius of moon R_{m} is 1/4 times the radius of earth R_{e}.

∴ W_{m}/W_{e} = M_{m}/M_{e} x (R_{e}/R_{m})^{2} = 1/100 x (4)^{2} ≈ 1/6

**In Text-Questions Page 1**41

**Why is it difficult to hold a school bag having a strap made of a thin and strong string?****Ans:-**It is difficult to hold a school bag having a thin strap because the pressure exerted on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller the surface area, the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulders is very large.**What do you mean by buoyancy?****Ans:-**The upward force exerted by a fluid on a substance when immersed in it is called buoyancy.**Why does an object float or sink when placed on the surface of water?****Ans:-**An object float or sink when placed on the surface of the water:

→If the density of object is more than the density of fluid, it will sink.

→If the density of object is less than the density of fluid, it will float.

**In Text-Questions Page** **142**

**You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?****Ans:-**When weighing our body, it is acting by an upward force. The buoyant force is this upward force. As a result, the body is pushed up slightly, resulting in the weighing machine showing less reading than the actual value. So, my mass will be more than 42 kg.**You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?****Ans:-**The bag of cotton is heavier than the bar of iron. The cotton bag has a larger air thrust than the iron bar. The weighing machine therefore indicates a smaller cotton bag weight than its actual weight.

**Chapter End Exercises**

**How does the force of gravitation between two objects change when the distance between them is reduced to half ?****Ans:-**Consider, the universal law of gravitation

According to that law, the force of attraction between two bodies is

Where, m_{1}and m_{2}are the masses of the two bodies.

G is the gravitational constant.

r is the distance between the two bodies.

Given that the distance is reduced to half then,

r = ½r

Therefore,

F = 4F

Therefore, once the space between the objects is reduced to half, then the force of gravitation will increase by fourfold the first force.**Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?****Ans:-**Acceleration due to gravity is independent of mass. So, all objects fall at the same rate under the gravitational force only. That’s why a heavy object does not fall faster than a light object.**What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10**^{24}kg and radius of the earth is 6.4 × 10^{6}m.)**Ans:-**From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by

Here,

m_{1}= mass of earth = 6 x 10^{24}Kg

m_{2}= mass of the body = 1 Kg

r = distance between the two bodies

Radius of the earth = 6.4 x 10^{6}m

G = 6.67 x 10^{-11}Nm^{2}/Kg^{2}

By substituting all the values in equation:

F = 9.8 N

This shows that Earth exerts a force of a 9.8 N on a body of mass 1 Kg.**The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?****Ans:-**According to the universal law of gravitation, two objects attract each other with equal force but in opposite directions. The earth attracts the moon with an equal force with which the moon attracts the earth.**If the moon attracts the earth, why does the earth not move towards the moon?****Ans:-**The earth and the moon experience equal gravitational forces from each other. However, the mass of the earth is much larger than the mass of the moon. Hence it accelerates at a rate lesser than the acceleration rate of the moon towards the earth. For this reason, the earth does not move towards the moon.**What happens to the force between two objects, if**

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?**Ans:-**(i) the mass of one object is doubled

(ii) the distance between the objects is doubled

Thus, the force between the two objects becomes one-fourth.

Now, the distance is tripled .

Thus, the force becomes one-ninth when the distance is tripled.

(iii) the masses of both objects are doubled?

If both masses are doubled, then the force is

F = 4F, Force will therefore be four times greater than its actual value.**What is the importance of universal law of gravitation?****Ans:-**The universal law of gravitation successfully explained some phenomena like:

➩ the force that binds us to the earth.

➩ the motion of the moon around the earth.

➩ the motion of the planets around the sun.

➩ the tides due to the moon and the sun.**What is the acceleration of free fall?****Ans:-**Acceleration of the free fall is the acceleration produced when a body falls under the influence of the gravitational force only. It is also known as acceleration due to gravity. It is denoted by g and it’s value on the surface of the earth is 9.8 m/s^{2}.**What do we call the gravitational force between the earth and an object?****Ans:-**The gravitational force between the earth and an object is called object’s weight.**Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]****Ans:-**Weight of a body on the earth is given by:

w = mg

where,

m = mass of the body

g = acceleration due to gravity

The value of g is greater at poles than equator. Therefore, gold at the equator weighs less than at the poles. Amit’s friend will not agree with the weight of the gold bought.**Why will a sheet of paper fall slower than one that is crumpled into a ball?****Ans:-**A sheet of paper has a lot of area as compared to a crumpled paper ball. A sheet of paper must face a lot of air resistance. As result a sheet of paper falls slower than the crumpled ball.**Gravitational force on the surface of the moon is only ⅙ as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?****Ans:-**Acceleration due to earth’s gravity = g_{e}or g = 9.8 m/s^{2}

Object weight m = 10 kg

Acceleration due to moon gravity = g_{m}

Weight on the earth= W_{e}

Weight on the moon = W_{m}

Weight = mass x gravity

g_{m}= (1/6) g_{e}(given)

So W_{m}= m g_{m}= m x (1/6) g_{e}

W_{m}= 10 x (1/6) x 9.8 = 16.34 N

W_{e}= m x g_{e}= 10 x 9.8

W_{e}= 98N**A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:**

(i) the maximum height to which it rises,

(ii) the total time it takes to return to the surface of the earth.**Ans:-**Initial velocity u = 49m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/s^{2}(thus negative as ball is thrown up).

By third equation of motion,

v^{2}= u^{2}– 2gs

Substitute all the values in the above equation;

Total time T = Time to ascend (Ta) + Time to descend (Td)

V = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5s

Also, Td = 5s

Therefore,

T = Ta + Td

T = 5 + 5

T = 10s**A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.****Ans:-**Initial velocity = u = 0

Tower height = total distance = 19.6m

g = 9.8 m/s^{2}

Consider third equation of motion

v^{2}= u^{2}+ 2gs

v^{2 }= 0 + 2 × 9.8 × 19.6

v^{2}= 384.16

v = √(384.16)

v = 19.6m/s**A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s**^{2}, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?**Ans:-**Initial velocity u = 40m/s

g = 10 m/s^{2}

Max height final velocity = 0

Consider third equation of motion

v^{2}= u^{2}– 2gs [negative as the object goes up]

0 = (40)^{2}– 2 x 10 x s

s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

Total displacement = 0 (The first point is the same as the last point)**Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10**^{24}kg and of the Sun = 2 × 10^{30}kg. The average distance between the two is 1.5 × 10^{11}m.**Ans:-**Mass of the sun m_{s}= 2 × 10^{30}kg

Mass of the earth m_{e}= 6 × 10^{24}kg

Gravitation constant G = 6.67 x 10^{-11}N m^{2}/ kg^{2}

Average distance r = 1.5 × 10^{11}m

Consider Universal law of Gravitation

**A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.****Ans:-**(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

From equations (a) and (b)

5t^{2}= 100 -25t + 5t^{2}

t = (100/25) = 4sec.

After 4sec, two stones will meet

From (a) x = 5t^{2}= 5 x 4 x 4 = 80m.

Putting the value of x in (100-x) = (100-80) = 20m.

This means that after 4sec, 2 stones meet a distance of 20 m from the ground.**A ball thrown up vertically returns to the thrower after 6 s.****Find****(a) the velocity with which it was thrown up,**

(b) the maximum height it reaches, and

(c) its position after 4 s.**Ans:-**g = 10m/s^{2}

Total time T = 6sec

T_{a}= T_{d}= 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u – gt_{a}

u = v + gt_{a}

= 0 + 10 x 3

= 30m/s

The velocity with which stone was thrown up is 30m/s.

(b) From second equation of motion

The maximum height stone reaches is 45m.

(c) In 3sec, it reaches the maximum height.

Distance travelled in another 1sec = s’

The distance travelled in another 1sec = 5m.

Therefore in 4sec, the position of point p (45 – 5)

= 40m from the ground.**In what direction does the buoyant force on an object immersed in a liquid act?****Ans:-**The buoyant force on an object that is immersed in a liquid will be in a vertically upward direction.**Why does a block of plastic released under water come up to the surface of water?**The density of plastic is a smaller amount than that of water, therefore the force of buoyancy on plastic block are going to be bigger than the load of plastic block displaced. Hence, the acceleration of plastic block are going to be in upward direction, and comes up to the surface of water.

Ans:-**The volume of 50 g of a substance is 20 cm**To find the Density of the substance the formula is^{3}. If the density of water is 1 g cm^{-3}, will the substance float or sink?

Ans:-

Density = (Mass/Volume)

Density = (50/20) = 2.5g/cm^{3}

Density of water = 1g/cm^{3}

Density of the substance is greater than density of water. So the substance will sink.**The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm–3? What will be the mass of the water displaced by this packet?****Ans:-**Density of sealed packet = 500/350 = 1.42 g/cm^{3}

Density of sealed packet is greater than density of water

Therefore the packet will sink.

Considering Archimedes Principle,

Displaced water volume = Force exerted on the sealed packet.

Volume of water displaced = 350cm^{3}

Therefore displaced water mass = ρ x V

= 1 × 350

Mass of displaced water = 350g.

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