Class 9th Science Chapter-9 Exercises

NCERT Class 9 Science – Gravitation

Exercises  | Complete Answers

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Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer:

Gravitational force is inversely proportional to the square of the distance between two objects.

$$F\propto \frac{1}{d^2}$$

If distance is reduced to half:

$$d\to \frac{d}{2}$$
$$F\propto \frac{1}{(d\mathrm{/}2{)}^2}=\frac{4}{d^2}$$

The gravitational force becomes 4 times greater.

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Question 2. Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object?

Answer:

According to Newton’s second law:

$$a=\frac{F}{m}$$

Though a heavy object experiences a greater gravitational force, its mass is also greater.
These two effects cancel each other.

As a result, all objects fall with the same acceleration (g), irrespective of their mass (ignoring air resistance).

Hence, a heavy object does not fall faster than a light object.

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Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface?

Given:

• Mass of earth, $M=6\times {10}^{24}kg$

• Mass of object, $m=1kg$

• Radius of earth, $R=6.4\times {10}^{6}m$

• Gravitational constant,
  $G=6.7\times {10}^{-11}N{m}^{2}k{g}^{-2}$

Formula:

$$F=G\frac{Mm}{R^2}$$

Calculation:

$$F=\frac{6.7\times {10}^{-11}\times 6\times {10}^{24}\times 1}{(6.4\times {10}^6{)}^2}$$
$$F=9.8N$$

Answer: The gravitational force is 9.8 N.

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Question 4. 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, smaller, or the same as the force with which the moon attracts the earth? Why?

Answer:

The earth attracts the moon with the same force as the moon attracts the earth.

Reason:
According to Newton’s third law of motion, every action has an equal and opposite reaction.

Hence, the forces are equal in magnitude but opposite in direction.

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

Answer:

Although the moon attracts the earth, the mass of the earth is very large compared to the moon.

From Newton’s second law:

$$a=\frac{F}{m}$$

Due to its huge mass, the acceleration of the earth is extremely small, so its motion is not noticeable.

Therefore, the earth does not move noticeably towards the moon.

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Question 6. What happens to the force between two objects if:

(i) the mass of one object is doubled?

$$F\propto m$$

Force becomes double.

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

$$F\propto \frac{1}{d^2}$$

• Distance doubled →

  $$F=\frac{1}{4}\text{ times}$$

• Distance tripled →

  $$F=\frac{1}{9}\text{ times}$$

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(iii) the masses of both objects are doubled?

$$F\propto M\times m$$
$$(2M)\times (2m)=4Mm$$

Force becomes 4 times.

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

Answer:
The universal law of gravitation explains many natural phenomena, such as:

• The force that binds objects to the earth

• The motion of the moon around the earth

• The motion of planets around the Sun

• The occurrence of tides due to the moon and the Sun

Thus, it shows that the same gravitational force acts everywhere in the universe.

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Question 8. What is the acceleration of free fall?

Answer:
The acceleration of an object when it falls towards the earth under the influence of gravity alone is called acceleration of free fall.
It is denoted by g and its value near the earth’s surface is:

$$g=9.8m{s}^{-2}$$

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Question 9. 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 the weight of the object.

$$\text{Weight}=m\times g$$

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Question 10. Amit buys a few grams of gold at the poles and gives it to his friend at the equator. Will the friend agree with the weight? If not, why?

Answer:
No, the friend will not agree with the weight.

Reason:

• The value of g is greater at the poles than at the equator.

• Since weight = m × g, the gold will weigh less at the equator.

• Although the mass remains the same, the weight decreases at the equator.

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

Answer:
A sheet of paper has a larger surface area, so it experiences more air resistance.
A crumpled paper has a smaller surface area, so air resistance is less.

Therefore, the sheet of paper falls slower than the crumpled ball.

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Question 12. Gravitational force on the moon is $\frac{1}{6}$ of that on the earth. Find the weight of a 10 kg object on the earth and on the moon.

Given:

• Mass, $m=10kg$

• $g_{earth}=9.8m{s}^{-2}$

Weight on Earth

$$W_e=m\times g=10\times 9.8=98N$$

Weight on Moon

$$W_m=\frac{1}{6}\times W_e=\frac{1}{6}\times 98$$
$$W_m=16.3N(\text{approx.})$$

Answer:

• Weight on Earth = 98 N

• Weight on Moon = ≈ 16.3 N

Question 13. A ball is thrown vertically upwards with a velocity of 49 m/s.

Find
(i) maximum height,
(ii) total time to return to the earth.

Given:
Initial velocity $u=49m{s}^{-1}$
Acceleration $a=-9.8m{s}^{-2}$ (upward motion)
Final velocity at top $v=0$

(i) Maximum height

$$v^2=u^2+2as$$
$$0=(49{)}^2+2(-9.8)s$$
$$s=\frac{{49}^2}{2\times 9.8}=\frac{2401}{19.6}=122.5m$$

Answer: Maximum height = 122.5 m

(ii) Total time of flight

Time to reach top:

$$v=u+at\Rightarrow 0=49-9.8t\Rightarrow t=5s$$

Total time $=2\times 5=10s$

Answer: Total time = 10 s

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Question 14. A stone is released from the top of a tower of height 19.6 m.

Find the final velocity just before touching the ground.

Given:
$u=0$, $s=19.6m$, $a=9.8m{s}^{-2}$

$$v^2=u^2+2as=0+2\times 9.8\times 19.6$$
$$v^2=384.16\Rightarrow v=19.6m{s}^{-1}$$

Answer: Final velocity = 19.6 m s⁻¹ (downwards)

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Question 15. A stone is thrown vertically upward with $u=40m\mathrm{/}s$.

Take $g=10m\mathrm{/}{s}^2$. Find
(i) maximum height,
(ii) net displacement,
(iii) total distance covered.

Given: $u=40m\mathrm{/}s$, $a=-10m\mathrm{/}{s}^2$, $v=0$ at top.

(i) Maximum height

$$0={40}^2+2(-10)s\Rightarrow s=\frac{1600}{20}=80m$$

Maximum height = 80 m

(ii) Net displacement

The stone returns to the starting point.

Net displacement = 0 m

(iii) Total distance covered

Upward distance $=80m$
Downward distance $=80m$

Total distance = 160 m

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Question 16. Force of gravitation between the earth and the Sun

Given:
$M=6\times {10}^{24}kg$
$m=2\times {10}^{30}kg$
$d=1.5\times {10}^{11}m$
$G=6.7\times {10}^{-11}N{m}^{2}k{g}^{-2}$

Formula:

$$F=G\frac{Mm}{d^2}$$

Calculation:

$$F=\frac{6.7\times {10}^{-11}\times 6\times {10}^{24}\times 2\times {10}^{30}}{(1.5\times {10}^{11}{)}^2}$$

$$F=\frac{8.04\times {10}^{44}}{2.25\times {10}^{22}}=3.57\times {10}^{22}N$$

Answer:

$$\boxed{{\displaystyle F=3.57\times {10}^{22}N}}$$

Question 17. Two stones are released simultaneously

• Stone A: dropped from top of a 100 m tower

• Stone B: projected upwards from ground with 25 m s⁻¹
  Find when and where they meet.

Take: $g=10m{s}^{-2}$

For stone A (downward):

$$s_1=\frac{1}{2}g{t}^2=5t^2$$

For stone B (upward):

$$s_2=ut-\frac{1}{2}g{t}^2=25t-5t^2$$

They meet when:

$$s_1+s_2=100$$

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

$$25t=100\Rightarrow t=4s$$

Position:

$$s_1=5(4{)}^2=80m$$

Answer:

• Time = 4 s

• Meeting point = 80 m below the top (or 20 m above ground)

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Question 18. A ball returns to the thrower after 6 s

Find
(a) initial velocity
(b) maximum height
(c) position after 4 s

Given: Total time = 6 s → time to reach top = 3 s
$g=9.8m{s}^{-2}$

(a) Initial velocity

$$u=gt=9.8\times 3=29.4m{s}^{-1}$$

(b) Maximum height

$$h=ut-\frac{1}{2}g{t}^2$$

$$h=(29.4\times 3)-\frac{1}{2}(9.8)(3^2)$$

$$h=88.2-44.1=44.1m$$

(c) Position after 4 s

$$s=ut-\frac{1}{2}g{t}^2$$

$$s=(29.4\times 4)-\frac{1}{2}(9.8)(16)$$

$$s=117.6-78.4=39.2m$$

Answers:

• (a) 29.4 m s⁻¹

• (b) 44.1 m

• (c) 39.2 m above the point of projection

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19. In what direction does the buoyant force act?

Answer:
The buoyant force acts vertically upward, opposite to the direction of gravity.

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

Answer:
The buoyant force acting upward is greater than the weight of the plastic block.
Hence, the net force is upward and the block rises to the surface.

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Question 21

The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm⁻³, will the substance float or sink?

Answer:

First, calculate the density of the substance.

$$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$$

$$=\frac{50}{20}=2.5gc{m}^{-3}$$

The density of the substance (2.5 g cm⁻³) is greater than the density of water (1 g cm⁻³).

Therefore, the substance will sink in water.

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Question 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:

Step 1: Calculate the density of the packet

$$\text{Density}=\frac{500}{350}=1.43gc{m}^{-3}$$

Since the density of the packet (1.43 g cm⁻³) is greater than the density of water (1 g cm⁻³):

The packet will sink in water.