NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Q1 ) How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer :

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

\( F = G \frac{m_1 m_2}{r^2} \)

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 if the distance is reduced to half then,

r = \( \frac{1}{2} \) r

\( \therefore F = \frac{Gm_1 m_2}{ ( \frac{r}{2} )^2 } \ \)
\(\Rightarrow \ F =4 \frac{G m_1 m_2}{r^2} \ \)
\( \Rightarrow \ F = 4 F \)

Therefore once the space between the objects is reduced to half, then the force of gravitation will increase by fourfold the first force.

Q2 ) Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Answer :

All objects fall on the bottom with constant acceleration called acceleration thanks to gravity (g). It’s constant and therefore the price of ‘g’ doesn’t depend on the mass of associate object. So serious objects don’t fall quicker than light-weight objects provided there’s no air resistance.

Q3 ) 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 × 1024 kg and radius of the earth is \( 6.4 × 10^6 m. \)

Answer :

3. 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 × 1024 kg and radius of the earth is \( 6.4 × 10^6 m. \)

Q4 ) 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?

Answer :

The value of F is same for earth and the moon. Both bodies will exert the same amount of force on each other. As per universal law of gravitation, every body attracts the other body with some force and this force is same for both the bodies called gravitational force.

Q5 ) If the moon attracts the earth, why does the earth not move towards; the moon?5

Answer :

According to the universal, law of gravitation both moon and earth attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of distance between them. The force of attraction of moon’ on the earth is present, but the earth does not appear to move towards the moon as the mass of the earth is large and the distance between the moon and earth is so large, even if the earth is attracted/moves towards the ,moon it is negligible, cannot be seem.

Q6 ) 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?

Answer :

(i) If the mass of one object is doubled, the force between two objects will be doubled (increases)

As per universal law of gravitation, the force between 2 objects \( m_1 \ and \ m_2 \) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

(ii) If the distance been the objects is doubled the force between two objects will be one-fourth and if the distance will be tripled, the force will be one-ninth (1/9).

As per universal law of gravitation, the force between 2 objects \( m_1 \ and \ m_2 \) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

(iii) If the masses of both objects are doubled the force will be 4 times.

As per universal law of gravitation, the force between 2 objects \( m_1 \ and \ m_2 \) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

Q7 ) What is the importance of universal law of gravitation?

Answer :

The universal law of gravitation explains several phenomena:

(i) it explains about the force that binds the earth,

(ii) the motion of the moon around the earth,

(iii) the motion of planets around the sun, and

(iv) the tides due to the moon and the

Q8 ) What is the acceleration-of free fall?

Answer :

The acceleration of free fall is; when the Body falls due to earth’s gravitational pull, its velocity changes and is said to be accelerated due to .the earth’s gravity and it falls freely called as free fall. This acceleration is calculated to be \( 9.8 m/s^2 \)

Q9 ) What do we call the gravitational force between the earth and an object?

Answer :

The gravitational force between the earth and an object is called force due to gravity.

Q10 ) Amit buys few grams of gold at the poles 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?

[Hints: The value of g is greater at the poles than at the equator.]

Answer :

Weight of the body is given by the formula

W = mg

It depends on the value of ‘g’ i.e., acceleration due to gravity.

The weight of gold at poles = \( W_p \ = \ m × g_p \)

The weight of gold at equator = \( W_e \ = \ m × g_e \)

∵ Value of g at poles is more than the value of g at equator.

\(\therefore \) Wp > We.

\(\therefore \) the friend will not agree with the weight of gold bought because the weight at pole of the same gold is found to be more as compared to the weight at the equator.

Q11 ) Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer :

A sheet of paper has larger surface area and while falling down it has to overcome the force exerted by air/wind. current, called as air resistance. The crumpled paper has smaller surface area and it has to overcome very less amount of air current.

Q12 ) Gravitational force on the surface of the moon is only \(\frac{1}{6} \) 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?

Answer :

Mass of the object = 10 kg

Weight of the object on earth = \( W = m × g \)

\(\therefore \) \( W = 10 × 9.8 \)

\( W = 98 N \)

Weight of the object on moon =\( \frac{1}{6} \) th the weight on the earth.

As the gravitational force on the surface of the moon is only \( \frac{1}{6} \) th as strong as gravitational force on the surface of the earth.

\(\therefore \) Weight of the object on moon \( = \frac{98}{6} \ = \ 16.3 N \)

Weight on earth = 98 N

Weight on moon = 16.3 N

Q13 ) 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.

Answer :

(i) Initial velocity = 49 m/s

Final velocity = 0 m/s

\( a \ = \ g \ = \ –9.8 m/s^2 \)

We know that, \( v^2 – u^2 = 2 g s \)
\( \Rightarrow 0^2 – (49)^2 = 2 (–9.8) × s \)
\(\Rightarrow s = \frac{49 × 49}{2 × 98} \ \)
\( \Rightarrow \ s = 122.5 m \)

(ii) We know that, \( v = u + gt \)
\( \Rightarrow 0 = 49 + (–9.8) × t \)

\(\therefore \) \( t = \frac{49}{9.8} \ \)
\( \Rightarrow \ t = 5 s \)

Total time taken to return the surface of the earth by the ball is 5 s + 5 s = 10 s.

Q14 ) A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground?

Answer :

Given, u = 0 m/s

\( h = s = 19.6 m \)

g = 9.8 m/s2 (falling down)

We know that, \( v^2 – u^2 = 2 g s \)

\( \Rightarrow v^2 – (0)^2 = 2 x 9.8 x 19.6 \)

\(\Rightarrow \) v = 19.6 m/s

The final velocity just before touching the ground is 19.6 m/s.

Q15 ) 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?

Answer :

Given data:

Initial velocity u = 40 m/s

\( g = 10 m/s^2 \)

Max height final velocity = 0

We know that, \( v^2 = u^2 – 2 g s \)
[negative as the object goes up]

\( \Rightarrow 0 = (40)^2 – 2 x 10 x s \)
\( \Rightarrow s = \frac{(40 x 40)}{20} \)

Maximum height s = 80 m

Total Distance = s + s = 80 + 80

Total Distance = 160 m

Total displacement = 0 (The first point is the same as the last point)

Q16 ) Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Answer :

Given data:

Mass of the sun m_s = 2 × 10³⁰ kg

Mass of the earth m_e = 6 × 10²⁴ kg

Gravitation constant G = 6.67 x 10^-11 N m²/ kg²

Average distance r = 1.5 × 10¹¹ m

From Universal law of Gravitation

\( F = \frac{Gm_1 m_2 }{d^2} \ \)
\( \Rightarrow \ F = \frac{ 6.67 × 10^{-11} × 6 × 10^{24} × 2 × 10^{30} }{(1.5 × 10^{11})^2 } \ \)
\( \Rightarrow \ F =3.56 × 10^{23} N \)

Q17 ) 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.

Answer :

(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

We know that,
\( s = ut + \frac{1}{2}gt^2 \ \)
\( \Rightarrow \ x = 0 + \frac{1}{2}gt^2 \ \)
\( \Rightarrow \ x = 5t^2 \) ......- I

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

We know that,
\( s = ut + \frac{1}{2}gt^2 \ \)
\(\Rightarrow \ 100-x = 25t + \frac{1}{2} × 10 × t^2 \ \)
\( \Rightarrow \ x = 100 - 25t + 5t^2 \) ...... - II

From equations I and II

\( 5t^2 = 100 -25t + 5t^2 \)

\( \Rightarrow t = \frac{100}{25} = 4 sec \)

So, after 4 sec, two stones will meet

Now, from I

\( x = 5 t^2 = 5 × 4 × 4 = 80 m \)

Putting the value of x in (100-x), we get

\( x = 100-80 = 20 m \)

Q18 )

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.

Answer :

Given data:

\( g = 10 m/s^2 \)

Total time T = 6 sec

\( T_a = T_d = 3 sec \)

(a) Final velocity at maximum height v = 0

We know that, \( v = u – gt_a \ \)
\( \ u = v + gt_a \ \)
\( \Rightarrow \ u = 0 + 10 × 3 \)

\(\Rightarrow u = 30 m/s \)

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

(b) Now, \( s = ut_a - \frac{1}{2}g(t_a)^2 \ \)
\( \Rightarrow \ s = 30 × 3 - \frac{1}{2} × 3^2 \ \)
\(\Rightarrow \ s = 45 m \)

The maximum height stone reaches is 45 m.

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

Distance travelled in another 1 sec = s’

So, \( s = ut_a - \frac{1}{2}g(t_a)^2 \ \)
\( \Rightarrow \ s = 0 + 10 × 1 × 1 \ \)
\( \Rightarrow \ s = 5 m \)

The distance travelled in another 1 sec = 5 m.

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

= 40 m from the ground.

Q19 ) In what direction does the buoyant force on an object immersed in a liquid act?

Answer :

The buoyant force on an object that is immersed in a liquid will be in a vertically upward direction.

Q20 ) Why a block of plastic does released under water come up to the surface of water?

Answer :

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.

Q21 ) The volume of 50 g of a substance is \( 20 cm^3 \). If the density of water is \( 1 g/cm^3 \) , will the substance float or sink?

Answer :

We know that, \( Density \ = \ \frac{Mass}{Volume} \)

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

Density of water = \( 1 g/cm^3 \)

Density of the substance is greater than density of water. So the substance will float.

Q22 ) The volume of a 500 g sealed packet is \( 350 cm^3 \) . 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 = \( \frac{500}{350} \ = \ 1.42 g/cm^3 \)

Density of sealed packet is greater than density of water

\(\therefore \) the packet will sink.

From Archimedes Principle, we know that

Displaced water volume = Force exerted on the sealed packet.

Volume of water displaced \( = 350 cm^3 \)

\(\Rightarrow \) displaced water mass = ? x V

\(\Rightarrow \) displaced water mass = 1 × 350

\(\therefore \) Mass of displaced water = 350g.