Class 9 Science Chapter 9 question answer Gravitation

Ncert Solutions for Class 9 Science Chapter 9: Gravitation class 9 questions and answers

Textbook	Ncert
Class	Class 9
Subject	Science
Chapter	Chapter 9
Chapter Name	Gravitation class 9 ncert solutions
Category	Ncert Solutions
Medium	English

Are you looking for Class 9 Science Chapter 9 question answer ? Now you can download Gravitation class 9 questions and answers pdf from here.

In Text Questions Page No: 102

Question 1: State the universal law of gravitation.

Answer 1: The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Question 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 2: The formula to find the magnitude of the gravitational force between the Earth and an object on its surface is:

\(F = \frac{G \cdot M \cdot m}{R^2}\)

Where:

• ( F ) = gravitational force
• ( G ) = universal gravitational constant
• ( M ) = mass of the Earth
• ( m ) = mass of the object
• ( R ) = radius of the Earth

In Text Questions Page No: 104

Question 1: What do you mean by free fall?

Answer 1: Earth’s gravity attracts each object to its center. When an object is dropped from a certain height, under the influence of gravitational force it begins to fall to the surface of Earth. Such an object movement is called free fall.

Question 2: What do you mean by acceleration due to gravity?

Answer 2: When an object falls freely towards the surface of earth from a certain height, then its velocity changes. This change in velocity produces acceleration in the object which is known as acceleration due to gravity denoted bu letter g . The value of acceleration due to gravity is g= 9.8 m/s².

In Text Questions Page No: 106

Question 1: What are the differences between the mass of an object and its weight?

Answer 1: Here are the key differences between mass and weight:

Mass	Weight
Mass is the amount of matter in an object.	Weight is the force with which Earth attracts the object.
It is a scalar quantity.	It is a vector quantity.
It is measured in kilograms (kg).	It is measured in newtons (N).
Mass remains constant everywhere.	Weight changes with location due to gravity.
Measured using a beam balance.	Measured using a spring balance.

Question 2: Why is the weight of an object on the moon 1/6th its weight on the earth?

Answer 2: The moon has a radius of 1/4 and a mass of 1/100 that of Earth. Because of this, the moon’s gravitational pull is around one-sixth that of Earth’s. The mass and diameter of the moon determine its gravitational pull. As a result, an object’s weight on the moon is one-sixth that of its weight on Earth. The moon has a different radius (R) than the Earth and is significantly less substantial than the planet.

In Text Questions Page No: 109

Question 1: Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Answer 1: It is difficult to hold a school bag with a thin and strong strap because the pressure on the shoulder increases. Pressure is equal to force divided by area. Since the strap is thin, it has less surface area, so the same weight (force) creates more pressure, making it feel more painful and uncomfortable.

Question 2: What do you mean by buoyancy?

Answer 2: When a body is put in a liquid or gas, it experiences an upward force. Due to this upward force, substances appear lighter when immersed wholly or partially in a liquid. The tendency of a fluid (liquid or gas) to exert an upward force on the object is called buoyancy. The upward force that acts on the object is called buoyant force.

Question 3: Why does an object float or sink when placed on the surface of water?

Answer 3: 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.

In Text Questions Page No: 110

Question 1: You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Answer 1: The weighing machine actually measures the weight of the body as the acceleration due to gravity ‘g’ is acting on the body. Hence the mass reading of 42 kg given by a weighing machine is same as the actual mass of the body. As mass is the quantity of inertia, it remains the same.

Question 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 2: The cotton bag weighs more than an iron bar, which is the right response. The cotton bag weighs more than the iron bar. The air thrust on the cotton bag is greater than that of the iron bar. As a result, the weighing machine gives the cotton bag a lower weight reading than it does. The explanation is True weight = (apparent weight + up thrust) Since the density of the cotton bag is lower than that of the iron bar, it has a larger volume than the latter. Because of the air, the cotton bag experiences greater upthrust. As a result, the true weight of the cotton bag is greater than the true weight of the iron bar when air is present.

Exercises

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

Answer 1: When the distance between two objects is reduced to half, the force of gravitation becomes four times greater.

This is because gravitational force is inversely proportional to the square of the distance between the objects:

\(F \propto \frac{1}{r^2}\)

So, if the distance r becomes \( \frac{r}{2} \), then:

\(F’ = \frac{1}{\left(\frac{r}{2}\right)^2} = \frac{1}{\frac{r^2}{4}} = 4 \times \frac{1}{r^2}\)

Hence, the force becomes 4 times stronger.

Question 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?

Answer 2: 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.

Question 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 × 10⁶ m).

Answer 3: To find the magnitude of the gravitational force, we use the formula:

\(F = \frac{G \cdot M \cdot m}{R^2}\)

Where:

• G = 6.674×10⁻¹¹ Nm²/kg² (gravitational constant)
• M = 6×10²⁴ (mass of the Earth)
• m = 1kg (mass of the object)
• R = 6.4×10⁶ (radius of the Earth)

Plugging in the values:

\(F = \frac{6.674 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2}\)

\(F = \frac{4.0044 \times 10^{14}}{4.096 \times 10^{13}} \approx 9.78 \, \text{N}\)

The gravitational force is approximately 9.8 N.

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 or smaller or the same as the force with which the moon attracts the earth? Why?

Answer 4: According to the Universal law of gravitation, two objects attract each other and according to Newton’s third law of motion, the force of attraction between two objects is the same but acts in the opposite direction. Thus, the earth attracts the moon with the same force as the moon exerts on earth but the force acts in the opposite direction.

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

Answer 5: Force of attraction due to gravitation provides the necessary centripetal force to the moon to enable it to move round the earth in almost a circular path. Acceleration is inversely proportional to mass of the Earth which is very large; therefore, acceleration produced in the earth is so small that it cannot be observed. Therefore, we cannot see the earth moving towards the moon.

Question 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 6: (i) The mass of one object is doubled: Gravitational force is directly proportional to mass. So, the force will also double.

(ii) The distance between the objects is doubled and tripled: Gravitational force is inversely proportional to the square of the distance.

• If distance is doubled, force becomes:
  \(\frac{1}{2^2} = \frac{1}{4}\)
  → Force becomes one-fourth.
• If distance is tripled, force becomes:
  \(\frac{1}{3^2} = \frac{1}{9}\)
  → Force becomes one-ninth.

(iii) The masses of both objects are doubled:
Gravitational force is proportional to the product of the two masses.
\(F \propto (2m_1) \times (2m_2) = 4m_1m_2\)
→ Force becomes 4 times greater.

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

Answer 7: Importance of Universal Law of Gravitation:

• Explains the motion of planets around the Sun.
• Helps understand the moon’s motion around the Earth.
• Responsible for tides due to the moon’s gravity.
• Explains how objects fall towards the Earth.
• Helps in calculating weight of objects on different planets.

Question 8: What is the acceleration of free fall?

Answer 8: Acceleration of free fall is the acceleration produced when a body falls under the influence of the force of gravitation of the earth alone. It is denoted by g and its value on the surface of the earth is 9.8 ms^-2.

Question 9: What do we call the gravitational force between the earth and an object?

Answer 9: 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.

Question 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 10: The weight of a body on the earth’s surface; W = mg (where m = mass of the body and g = acceleration due to gravity) The value of g is larger at the poles when compared to the equator. So gold can weigh less at the equator as compared to the poles. Therefore, Amit’s friend won’t believe the load of the gold bought.

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

Answer 11: When a sheet of paper is crumbled into a ball, then its surface area becomes much lesser than the surface area of a plain non-crumpled sheet of paper. Hence, the upward force exerted by air on the sheet is greater as compared to the one exerted on the ball. Hence the sheet falls slower as compared to a paper ball.

Question 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 newtons of a 10 kg object on the moon and on the Earth?

Answer 12: To find the weight, use the formula:
Weight = mass × g

On Earth:

• Mass = 10 kg
• g = 9.8 m/s²

Weight on Earth = 10 × 9.8 = 98N​

On Moon:

• Gravity on Moon = \( \frac{1}{6} \) of Earth
• \( g_{\text{moon}} = \frac{9.8}{6} \approx 1.63 \, \text{m/s}^2 \)

Weight on Moon = 10 × 1.63 = 16.3N

Question 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 13:

• (i) Given data:
• 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).
• By third equation of motion,
• 2gH = v² – u²
• 2 × (- 9.8) × H = 0 – (49)²
• – 19.6 H = – 2401
• H = 122.5 m

• (ii) Total time T = Time to ascend (T_a) + Time to descend (T_d)
• v = u + gt
• 0 = 49 + (-9.8) x T_a
• Ta = (49/9.8) = 5 s
• Also, T_d = 5 s
• Therefore T = T_a + T_d
• T = 5 + 5
• T = 10 s

Question 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 14: u = 0; s = 19.6 m; a = 9.8 ms^-2; v = ?
Using v² – u² = 2as, we have
v² – (0)² = 2 (9.8) (19.6) = (19.6)²
v = 19.6 ms^-1

Question 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 15:

• Initial velocity (u) = 40m/s
• g = −10m/s²
• Final velocity at maximum height (v) = 0 
• Third equation of motion
• v² = u² + 2gh
• 0 = 40² + 2 × (−10) × h
• 1600 = 20h
• 80m = h
• Max Distance = 80m
• Total distance travelled by the stone = 80 + 80 = 160m
• Because the stone was thrown upwards, but it came back.
• Hence, total displacement = 0

Question 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 16: We use the universal law of gravitation:

\(F = \frac{G \cdot M_1 \cdot M_2}{r^2}\)

Where:

• G = 6.67 x 10^-11 Nm²/ kg²
• m₁ = 6 × 10²⁴ kg (mass of Earth)
• m₂ = 2 × 10³⁰ kg (mass of Sun)
• r = 1.5 × 10¹¹ m (distance between Earth and Sun)

Substitute values:

\(F = \frac{6.674 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}\)

\(F = \frac{6.674 \times 6 \times 2 \times 10^{43}}{2.25 \times 10^{22}} = \frac{80.088 \times 10^{43}}{2.25 \times 10^{22}}\)

F ≈ 3.56×10²²N

The gravitational force between the Earth and the Sun is approximately 3.56×10²²N.

Question 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 17: We are given:

• Height of the tower = 100 m
• Stone A: Falls from the top (initial velocity u₁ = 0
• Stone B: Thrown upwards from ground with velocity u₂ = 25 m/s
• Acceleration due to gravity g = 9.8m/s²

Let the two stones meet after ( t ) seconds.
Let them meet at a height ( h ) from the ground.

Displacement of stone A (falling from the top):

\(s_1 = \frac{1}{2} g t^2 = \frac{1}{2} \times 9.8 \times t^2 = 4.9 t^2\)
So, its position from the ground is:
\(100 – s_1 = 100 – 4.9 t^2\)

Displacement of stone B (going upward):

\(s_2 = u t – \frac{1}{2} g t^2 = 25t – 4.9t^2\)

Since they meet at the same height:

100 − 4.9t² = 25t − 4.9t²

Cancel ( 4.9t² ) from both sides:

\(100 = 25t \Rightarrow t = \frac{100}{25} = 4 \, \text{seconds}\)

Now find the height from the ground where they meet:

Use s₂ = 25t – 4.9t²

s₂ ​= 25 × 4 − 4.9 × 16 = 100 − 78.4 = 21.6m

• They meet after 4 seconds
• At a height of 21.6 meters from the ground.

Question 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 4 s.

Answer 18: a. Final velocity (v) = 0

Time (t) = 6s

g = 10m/s² 

The first equation of motion

v = u + gt

0 = u + (−9.8) × 3 (negative sign due to opposite direction)

u = 30m/s 

Hence, the ball is thrown upwards with a velocity of 30m/s.

b. Initial velocity (u) = 30m/s

Final velocity (v) = 0

Time (t) = 3s

the second equation of motion

h = ut + 1/2 gt²

h = 30×3+1/2×(−10) x (3)²

h = 90 − 45

h = 45m

Hence, the maximum height attained by the ball is 45m.

c. Position of the ball after 4 s Distance travelled downwards in 1 sec the second equation of motion

h = ut + 1/2 gt²

h = 0×1+1/2×10×(1)2

h = 5m 

Distance covered in 1 second = 5m

Distance covered in 4 seconds = 45 − 5

= 40 m

Hence, after 4 seconds, the ball will be 40m above the ground.

Question 19: In what direction does the buoyant force on an object immersed in a liquid act?

Answer 19: An object immersed in a liquid experiences buoyant force in the upward direction.

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

Answer 20: 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.

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 21: To find the Density of the substance the formula is

Density = (Mass/Volume)

Density = (50/20) = 2.5g/cm³

Density of water = 1g/cm³

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

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 22: To determine if the packet will float or sink, we compare its density with that of water.

Find the density of the packet

\(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{500 \, \text{g}}{350 \, \text{cm}^3} \approx 1.43 \, \text{g/cm}^3\)

Since 1.43 g/cm³ > 1 g/cm³, the packet is denser than water, so it will sink.

Mass of water displaced

When an object is fully submerged, it displaces water equal to its volume.

Volume displaced = 350cm³
Density of water = 1g/cm³
Mass of water displaced = Volume × Density = 350 × 1 = 350g

• The packet will sink in water.
• Mass of water displaced = 350 g.