# NCERT Class 9 Science Notes & Solutions – Chapter 10 – Gravitation

## Exercises Questions & Answers - Gravitation NCERT Class 9 Science

#### In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science

##### In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 134

1. State the universal law of gravitation.
Write down the answer of Q1 in detail.

2. Write the formula to find the magnitude of the gravitational force between the earth & an object on the surface of the earth.
As per universal law of gravitation ⇒ F = G (Mxm)/d2, (where G = universal gravitation constant). Let assume, Mearth = mass of the earth, mobject = mass of an object on the surface of the earth, d = R = Radius of the earth = distance between the earth & the object on the surface of the earth. Then, the magnitude of gravitational force is ⇒   F = G (Mearth x mobject)/R2.

##### In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 136

1. What do you mean by free fall?
Write down the answer of Q9 in detail.

2. What do you mean by acceleration due to gravity?
Write down the answer of Q10 in detail.

##### In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 138

1. What are the differences between the mass of an object & its weight?

Differences between mass & weight -

Mass

Weight

It is the quantity of matter that a body contains.
It is the gravitational force acting on the body.
Basically mass is the inertia of the body.
Basically weight s the measure of the gravity.
It is constant & does not vary during changing of place.
It is not constant & varies when place is changed.
Mass has only magnitude but no direction. So it is a scalar quantity.
Weight has both magnitude & direction. So it is a vector quantity.
Kilogram i.e. kg is the S.I. unit of mass.
Newton (i.e.) N is the S.I. unit of weight.

2. Why is the weight of an object on the moon 1/6 th its weight on the earth?
Write down Q17 in detail.

##### In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 141

1. Why it is difficult to hold a school bag having a strap made of a thin & strong string?
The bag exerts force on tiny & strong string. That force exerted equally by the string exerts a very less area on our body. We know that pressure is inversely proportional to its area. Therefore, force exerted on our body by the thin string of the school bag will be more. So it is tough to carry the bag.

2. What do you mean by buoyancy?
Buoyancy is the upward force exerted by a liquid on an object immersed in it. When you try to immerse an object into water, you feel an upward or upthrust force exerted on the object. The more you try to push the object deeper into water, the more pressure you will feel.

3. Why does an object float or sink when placed on the surface of water?
When the density of an object is more than the density of water, then the object will sink in water. In this case, the buoyant force acting on the object is less than the force of gravity acting on it.  On the other side, if  the density of an object is less than the density of water, then the object will float on the water. In this case, the buoyant force acting on the object is more than the force of gravity acting on it.

##### In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos.  142

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
It is indeed less than 42 kg. When you put your body on a weight machine, your body gets pushed slightly upwards. Because an upward force acts on it. Basically this force is the buoyant force. Therefore, the result showed by the weight machine is less than the actual value.

2. You have a bag of cotton & 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 & why?
An iron bar is lighter than a cotton bag of same mass. Because the surface area of a cotton bag is more than an iron bar. So the buoyant force on the cotton bag will be more than that on the iron bar.  Therefore, the cotton bag is found lighter than its actual value. When you weigh both the items, weighing machine shows same mass. But actually the mass of the cotton bag is more than that of the iron bar.

##### Exercises Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 143 - 145

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
We know gravitational force between two objects –
F = GMm/R2, where G = gravitational constant, M = mass of the 1st object, m = mass of the 2nd object, r = distance between the objects.  When the distance r is reduced to half, the new force –
F1 = GMm/(r/2)2 = 4GMm/r2 = 4F. So, the gravitational force becomes four times.

2. Gravitational force acts on all objects in proportion to their masses, Why then, a heavy object does not fall faster than a light object?
When there is no air, all objects fall on ground with constant acceleration. That acceleration due to gravity is g. The value of g is 9.8 ms-2. The gravitational force acting on an object is  –
F = GMm/R2, where G = gravitational constant, M = mass of the earth, m = mass of the object, R = radius of the earth. Also –

F = GMm/R2 = mg ,
g = GM/ R2 = constant for a particular location on the surface of the earth. It does not depend on the mass of the object. So, all objects, be it heavy or light, fall on ground with same acceleration.

###### 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on it’s surface? (Mass of the earth = 6X1024 kg, radius of the earth = 6.4×104 m).

Gravitational force = F =  G [(M x m)/R2 ]
Here,  M = mass of the earth = 6×1024 kg, R = radius of the earth = 6.4×104 m, m = mass of the object = 1 kg, G = universal gravitation constant = 6.70 x 10-11 N m2 Kg-2.
So, F  =  G [(Mxm)/R2 ] = [(6.70 x 10-11 x 6 x 1024 x 1) / (6.4×106)2 ] N. = 9.8 N .

4. The earth & 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?
The universal law of gravitation states that two objects attract each other with equal force, but in opposite direction. So the attraction force by the Moon towards the Earth & the attraction force by the Earth towards the Moon are equal & opposite. This attraction force between the Earth & the Moon is ⇒ F = GMm/d2, where G = gravitational constant, M = mass of the Earth, m = mass of the Moon, d = distance between the Earth & the Moon. Therefore, the magnitude of F is same for Earth & Moon.

###### 5. If the moon attracts the earth, why does the earth not move towards the moon?

As per Newton’s 3rd law of motion, the attraction force by the Moon towards the earth is equal to the attraction force by the Earth towards the Moon. But the Moon accelerates faster than the earth. Because the mass of the Earth is much greater than the mass of the Moon. So the earth does not move towards the Moon.

6. What happens to the force between two objects if ⇒
a) the mass of one object is doubled?
b) the distance between the objects is doubled & tripled?
c) the masses of both objects are doubled
We know, the gravitational force between two objects –
F = GMm/r2, where G = gravitational constant, M = mass of the 1st  object, m = mass of the 2nd object, r = distance between the objects.
a. When the mass of any object is doubled (here mass of the 2nd object is doubled) , then new force, F1 = GMx2m/ r2 = 2GMm/ r2 = 2xF . So, the gravitational force is doubled.

b. When the distance r is doubled, the new force, F2 = GMm/ (2r)2 = GMm/ 4r2 = F/4. So, the gravitational becomes one-fourth.

c. When the mass of both objects is doubled, then new force, F3 = Gx2Mx2m/ r2 = 4GMm/ r2 = 4xF. So, the gravitational force becomes four times.

###### 7. What is the importance of universal law of gravitation?

The universal law of gravitation is very important. It describes many important phenomena in science. For example,
1. It explains the motion of planets around the Sun.
2. It also explains the motion of Moon around the Earth.
3. In addition, it describes the attraction force that binds us to the Earth.
4. It explains the occurring of tides due to the motion of the Sun & Moon.

8. What is the acceleration of free fall?
When any object falls from the upper place towards ground, we may term it as free falling under the influence of the gravitational force. Any free falling involves some changes in velocity. Because, from the time of starting of falling to the time of touching of the earth, there are some changes in velocity. Also we know that any change in velocity is associated with acceleration. So free falling  is actually an accelerated velocity. This acceleration is created by the gravity i.e. gravitational force. It is denoted by ɡ. The unit of ɡ is m/s2 or ms-2 i.e. same as that of acceleration. The value of ɡ is 9.8 ms-2.

###### 9. What do we call the gravitational force between the earth & an object?

The gravitational force between the earth & an object is the of that object. This gravitational force (F) is equal to (m x ɡ) where ɡ is acceleration due to gravity. So, the weight of an object is ⇒
W = F = m x ɡ   (m= mass of the object, ɡ = acceleration due to gravity). The S.I. Unit of weight is same as that of force and it is Newton.

10. 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].
We know, weight of an object – W = mg (m=mass of the object, g= acceleration due to gravity). Therefore,
weight at the poles – Wpole = m x gpole .
And weight at theequator – Wequator = m x gequator .
Given that, gpole > gequator . So, Wpole > Wequator . Hence, the friend will not agree. Because he will find less weight during measurement.

###### 11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

A sheet of paper is crumpled into a ball. As a result, its density increases. Therefore, resistance by the air against its motion decreases. So it fall faster than the sheet of paper.

12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight of a 10 kg object on the moon and on the earth?
Weight of an object on earth = (m x g) N.
Here , m = mass of the object = 10 kg, g = acceleration due to gravity = 9.8 ms-2.
Therefore, weight on earth = (10 x 9.8) N = 98 N and weight on moon = (1/6) x 98 N = 16.3 N.

###### 13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate i) the maximum height to which it rises and ii) total time it takes to return to the surface of the earth.

i) The ball is thrown upwards i.e. against gravitational force. So ‘g’ i.e. acceleration due to gravity acts against velocity. Then –
⇒ 2 x (-g) x s = v2 – u2      [∵ (-) g indicates that g acts against velocity ]
⇒ (-)2 x 9.8 x s = 02 – 492  [∵s = displacement, v=final velocity=0, initial velocity=49m/s ]
So, s = 492 / 2×9.8 m = 122.5 m.

ii) Time to reach maximum height –
⇒ v = u –  g x t [∵ (-) g indicates that g acts against velocity ]
⇒ t = (u –v)/g [v = final velocity = 0, initial velocity = 49 m/s]
Therefore, t = 49/9.8 s = 5 s.

Time to reach earth – when the ball starts falling after reaching maximum height, it’s initial velocity is zero and it falls due to gravitational force with acceleration g. So –
s = ut +  (½)xg x t2 [ s= 122.5 m, u = 0, t=? ]
⇒ 122.5 = 0xt + (½)x9.8xt2
t = (122.5 x 2 ) / 9.8 = 25
Therefore, t = 5 s.  So total time taken = (5 + 5) s = 10 s.

Time to reach earth – when the ball starts falling after reaching maximum height, it’s initial velocity is zero and it falls due to gravitational force with acceleration g. So –
s = ut +  (½)xg x t2 [ s= 122.5 m, u = 0, t=? ]
⇒ 122.5 = 0xt + (½)x9.8xt2
t = (122.5 x 2 ) / 9.8 = 25
Therefore, t = 5 s.  So total time taken = (5 + 5) s = 10 s.

###### 14. A stone is released from the top of a tower of height 19.6 m. Calculate it’s final velocity just before touching the ground.

2g s = v2 – u2
⇒ 2 x 9.8 x 19.6 = v2 – 02   [ ∵ s = 19.6 m, u=0, v=? ]
v2 = 19.6 x 19.6 = 384.16
So, v = 19.6 m/s.

15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g =10 m/s2, find the maximum height reached by the stone. What is the net displacement & the total distance covered by the stone?
2g s = v2 – u2
⇒ 2x 10 x s = v2 – 02   [ ∵ s = ?, g = 10 m/s2, u = 0, v = 40 m/s ]
Therefore, s = 402 / (2 x 10) m = 80 m.
Maximum height reached = 80 m. After reaching maximum height 80 m, the stone falls back to ground again covers 80 m. So total distance covered = 80 + 80  m = 160 m. Basically displacement is the distance between starting point and ending point. As in this case, the stone returns back to starting point, so net displacement is zero.

###### 16. Calculate the force of gravitation between the earth and the sun. Given data ⇒ mass of the earth = 6 x1024 kg, mass of the sun = 2 x 1030 kg, the average distance between the sun & the earth = 1.5 x 1011 m.

Gravitational force = F  =  G [(M x m)/d2]
Here,  M = mass of the earth = 6X1024 kg, d= distance between the sun & the earth = 1.5 x 1011 m, m = mass of of the sun = 2 x 1030 kg, G = universal gravitation constant = 6.673 x 10-11 N m2 Kg-2.
So, F  =  G [ ( M x m ) / d2
= [ (6.673 x 10-11 x 6 x 1024 x 2 x 1030) / (1.5 x 1011)2 ] N
= 35.58 x 1021 N = 3.56 x 1022 N.

###### 17. 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.

Let assume both the stone meet after t seconds. Distance covered by 1st stone plus distance covered by 2nd stone will be 100 m. 1st stone fall in favour of g and 2nd stone moves against g. Distance covered by 1st stone-
s1 = ut +  ½ xg x t2   [∵ u = 0]
⇒ s1 = 0xt  + ½ x 9.8 x t2 = 4.9t2
Similarly, distance covered by 2nd stone  s2 = ut –  ½ xg x t2   [∵ u = 25 m/s]
⇒ s2 = 25xt  –  ½ x 9.8 x t2 = 25t –  4.9t2.

Now ,   s1 + s2 = 100
⇒  9t2 + 25t – 4.9t2 = 100
⇒  25 t = 100
So, t = 4 s.
Now,  s1 = 4.9 x 42 m = 78.40 m and  s2 = 25×4  – 4.9×42 = 21.6 m. They meet 78.40 m from top or 21.6 m from ground.
Here we have taken g = 9.8 m/s2. If you take g = 10 m/s2, the answer will be 80 m from top or 20 m from ground.

###### 18. 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, c) it’s position after 4 s.

a) Total time taken = 6 s. So it takes 3 s to reach maximum height and  3 s to return back to the thrower. initial velocity = u = ?, final velocity = v = 0, g = 9.8 m/s2 . When stone goes up g acts against it. So,
⇒ v = u + (-g)xt ,
⇒ 0 = u – 9.8×3 ,
So, u = 29.4 m/s .

b) Maximum height it reaches = s = ut + ½ x(-g)xt2 = 29.4×3 + ½ x(-)9.8×32 = 44.1 m.

c) After 3 s it reaches maximum height, then it starts falling when initial speed = u = 0.
So, free falling in 1 s i.e. (4 s  – 3 s) = s = ut + ½ x(g)xt2 = 0x1  +  ½ x9.8×12 = 4.9 .
So after 4 s, it’s position is (44.1 – 4.9) m or 39.2 m up from the thrower.

19. In what direction does the buoyant force on an object immersed in a liquid act?
The buoyant force always acts in the upward direction. An object may sink or float in a liquid. But it always experiences buoyant force in the upward direction. That is also applicable in case of an object suspended in a liquid. The more you go deep into a liquid, the more buoyant force you experience in the upward direction.

###### 20. Why does a block of plastic released under water come up to the surface of water?

In this case, two forces act on the block of plastic immersed in water. Firstly, the gravitational force pulls the block of plastic downwards. Secondly, the buoyant force pushes that object upwards. The buoyant force is greater than the gravitational force in this case. So the block of plastic comes up to the surface on the water as soon as it is released within water.

21. The volume of a 50g substance is 20 cm3. If the density of water is 1 g cm-3, will the substance float or sink?
Density of substance  = (50/20) g cm-3 = 2.5 g cm-3 . As the density of the substance is greater than the density of the water, so the substance will sink.

###### 22. 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?

Answer: Density of sealed packet  = (500/350) = 1.42 g cm-3. Now we get that density of the sealed packet is greater than the density of water. So the substance will sink and displace water equal to it’s volume. So volume of displaced water is 350 cm3. Also given density of water is 1 g cm-3. We know-
Density =  mass/volume , so mass = density x volume = 350 x 1 = 350 g.

Gravitation Science Chapter 10 – To Be Continued – Gravitation Science Chapter 10