NCERT Solutions Class 9 Science Chapter 10 Gravitation

NCERT Solutions Class 9 Science

The NCERT Solutions in English Language for Class 9 Science Chapter – 10 (Gravitation) has been provided here to help the students in solving the questions from this exercise.

Chapter – 10 (Gravitation)

Questions

1. State the universal law of gravitation.
Answer – Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.
F ∝ $\frac{Mm}{d^2}$ or F = $\frac{GMm}{d^2}$

2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer – Let M_E be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:
$F = \frac{G m_{1} m_{2}}{r^{2}}$

Questions

1. What do you mean by free fall?
Answer – Whenever an object falls toward earth under the force of gravity one and no other force is present, the motion of object is said to be “free fall”.

2. What do you mean by acceleration due to gravity?
Answer – The acceleration of free fall is the acceleration due to gravity. We can also say the acceleration of an object due to gravitational force of earth acting on it is known as acceleration due to gravity.
The value of the acceleration due to gravity on Earth is, g = $\frac{9.8m}{s^2}$

Questions

1. What are the differences between the mass of an object and its weight?
Answer – The differences between the mass of an object and its weight are tabulated below.

Mass	Weight
Mass is the quantity of matter contained in the body.	Weight is the force of gravity acting on the body.
It is the measure of inertia of the body.	It is the measure of gravity.
It only has magnitude.	It has magnitude as well as direction.
Mass is a constant quantity.	Weight is not a constant quantity. It is different at different places.
Its SI unit is kilogram (kg).	Its SI unit is the same as the SI unit of force, i.e., Newton (N).

2. Why is the weight of an object on the moon 1/6th its weight on the earth?
Answer – The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitional attraction on the moon is about one sixth when compared to earth. Hence, the the weight of an object on the moon 1/6th its weight on the earth.

Questions

1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer – We know that lesser is the surface area of an object the more is the pressure it exerts. Hence, it is difficult to hold a school bag having a strap made of thin and strong string, because the thin string has very less contacting area in case of a school bag having a strap made of thin and strong string which increases pressure which is un comfortable to carry the school bag.

2. What do you mean by buoyancy?
Answer – The upward force possessed by a liquid on an object that’s immersed in it is referred to as buoyancy.

3. Why does an object float or sink when placed on the surface of water?
Answer – An object floats or sinks when placed on the surface of water because of two reasons.
(i) If its density is greater than that of water, an object sinks in water.
(ii) If its density is less than that of water, an object floats in water.

Questions

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer – When you weigh your body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

2. 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?
Answer -The cotton bag is heavier than the iron bar. The cotton bag experiences larger up thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

Excercises

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer – Consider the Universal law of gravitation,
According to that law, the force of attraction between two bodies is
F = $\frac{(Gm_1m_2)}{r^2}$
Where,
m₁ and m₂ 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 = $\frac{1}{2}r$
Therefore,
F = $\frac{(Gm_1m_2)}{r^2}$

F = $\frac{(Gm_1m_2)}{\left (\frac{r }{2}\right )^2}$

F = $\frac{(4Gm_1m_2)}{r^2}$
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.

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 – All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

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 × 10⁶m.)

Answer – According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:

F = $\frac{(Gm_1m_2)}{r^2}$
Here
m₁ = Mass of Earth = 6.0 × 10²⁴ kg
m₂ = Mass of the body = 1 kg
r = Distance between the two bodies
Radius of Earth = 6.4 × 10⁶ m
G = Universal gravitational constant = 6.6 × 10^-11 Nm²kg^-2By substituting all the values in the equation
F = $\frac{(Gm_1m_2)}{r^2}$

F = $\frac{6.67\times10^{-11} \times 6.0\times 10^{24}\times 1}{(6.4\times 10^6)^2}$
F = 9.8 N
This Shows that Earth exerts a force of 9.8 N on a body of mass 1 kg. the body will exert an equal force of attraction of 9.8 N on the Earth.

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 or smaller or the same as the force with which the moon attracts the earth? Why?
Answer – 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 mcn)n attracts the earth.

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

6. 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).
(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).
(iii) If the masses of both objects are doubled the force will be 4 times. As F ∝ $\frac{m_1m_2}{r^2}$

7. What is the importance of universal law of gravitation?
Answer – The universal law of gravitation explains many phenomena that were believed to be unconnected:
(i) The motion of the moon round the earth
(ii) The responsibility of gravity on the weight of the body which keeps us on the ground
(iii) The tides because of the moon and therefore the Sun
(iv) The motion of planets round the Sun

8. What is the acceleration of free fall?
Answer – When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 m s^-2, which is constant for all objects (irrespective of their masses).

9. What do we call the gravitational force between the earth and an object?
Answer – The gravitation force between the earth and an object is called weight. Weight is equal to the product of acceleration due to the gravity and mass of the object.

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.]
Answer – 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 at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer – When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls 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 in newton’s of a 10 kg object on the moon and on the earth?
Answer – Weight of an object on the moon $= \frac{1}{6} \times$ Weight of an object on the Earth Also, Weight = Mass x Acceleration Acceleration due to gravity, g = 9.8 m/s² .
Therefore, weight of a 10 kg object on the Earth = 10 $= \frac{1}{6} \times$ 9.8 = 98 N And, weight of the same object on the moon $= \frac{1}{6} \times 98 = 16.3 N$

13. A ball is thrown vertically upwards with a velocity of 49 m/s.
Calculate
(i) The maximum height to which it rises,
(ii) The total time it takes to return to the surface of the earth.
Answer –
Initial velocity u = 49 m/s
Final speed v at maximum height = 0
Acceleration due to earth gravity g = -9.8 m/s² (thus negative as ball is thrown up).
(i) the maximum height to which it rises,
By third equation of motion,
2gH = v² – u²2 × (- 9.8) × H = 0 – (49)²– 19.6 H = – 2401
H = 122.5 m

(ii) the total time it takes to return to the surface of the earth.
Total time T = Time to ascend (T_a) + Time to descend (T_d)
v = u + gt
0 = 49 + (-9.8) x T_aTa = (49/9.8) = 5 s
Also, T_d = 5 s
Therefore T = T_a + T_dT = 5 + 5
T = 10 s

14. 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 –
Initial velocity u = 0
Tower height = total distance = 19.6m
g = 9.8 m/s²Consider third equation of motion
v² = u² + 2gs
v²= 0 + 2 × 9.8 × 19.6
v² = 384.16
v = √(384.16)
v = 19.6m/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 and the total distance covered by the stone?
Answer –
Initial velocity u = 40m/s
g = 10 m/s²Max height final velocity = 0
Consider third equation of motion
v² = u² – 2gs [negative as the object goes up]
0 = (40)² – 2 × 10 × s
s = (40 × 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)

16. 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 –
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/ kg²Average distance r = 1.5 × 10¹¹ m
Consider Universal law of Gravitation
F = $\frac{(Gm_1m_2)}{r^2}$

F = $\frac{6.67\times 10^{-11} \times 6\times 10^{24}\times 2\times 10^{30}}{(1.5\times 10^{11})^2}$
= 3.56 × 10²²N

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.
Answer – Let the two stones meet after a time t.
(i) When the stone from the top of the tower is thrown,
Initial velocity u’ = 0
Distance travelled = x
Time taken = t
Therefore,
s = ut + $\frac{1}{2}$ gt²

x = 0 + $\frac{1}{2}$ gt²
x = $\frac{1}{2}$ × 9.8 × t²
x = 4.9 t²——————- (i)

(ii) When the stone is thrown upwards,
Initial velocity u = 25 m/s
Distance travelled = (100 – x)
Time taken = t
s = ut – $\frac{1}{2}$ gt²

(100 – x) = 25t – $\frac{1}{2}$ × 10 t²x = 100 – 25t + 5t² —————– (ii)
From equations (i) and (ii)
5t² = 100 -25t + 5t²t = (100/25) = 4sec.
After 4sec, two stones will meet
From (i)
x = 5t² = 5 × 4 × 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.

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, and
(c) Its position after 4s.
Answer –
u = ?
v = 0
g = 10 m/s²
Total time = 6 s (to go up and down)
∴ Time for upward journey = $\frac{6}{2}$ = 3 s

(a) Final velocity at maximum height v = 0
From first equation of motion:-
v = u – gt_au = 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
s = ut_a – $\frac{1}{2}$ g(t_a)²
= 30 × 3 – $\frac{1}{2}$ × 10 × (3)²
= 90 – 45
= 45 m
The maximum height stone reaches is 45m.

(c) In 3sec, it reaches the maximum height.
Distance travelled in another 1sec = s’

s = ut_a – $\frac{1}{2}$ g(t_a)²

s =0 – $\frac{1}{2}$ (-10)(1)²
= 0 + 5
= 5 m
The distance travelled in another 1sec = 5m.
Therefore in 4sec, the position of point p (45 – 5)
= 40m from the ground.

19. In what direction does the buoyant force on an object immersed in a liquid act?
Answer – An object immersed in a liquid experiences buoyant force in the upward direction.

20. Why a block of plastic when released under water come up to the surface of water?
Answer – The density of plastic is lesser than that of water. Therefore, the force of buoyancy on plastic block will be greater than the weight of plastic block. Hence, the acceleration of plastic block is going to be in the upward direction. So, the plastic block comes up to the surface of water.

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 – To find the Density of the substance the formula is
Density = (Mass/Volume)
Density = (50/20) = 2.5g/cm³Density of water = 1g/cm³Density of the substance is greater than density of 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/cm³Density 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 = 350cm³Therefore displaced water mass = ρ x V
= 1 × 350
Mass of displaced water = 350g.

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NCERT Solutions Class 9 Science Chapter 10 Gravitation

NCERT Solutions Class 9 Science

Chapter – 10 (Gravitation)

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