Chapter 9: Gravitation - NCERT Solutions
In-Text Questions (Page No–102)
Q1. State the universal law of gravitation.
Answer - Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Q2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer - The formula is:
F = G × (M × m) / R²
Where:
F = Gravitational force
G = Universal gravitational constant
M = Mass of the Earth
m = Mass of the object
R = Radius of the Earth
F = G × (M × m) / R²
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)
Q1. What do you mean by free fall?
Answer - When an object falls under the influence of gravitational force only, it is said to be in free fall. During free fall, the object accelerates towards the Earth due to gravity, without any resistance from air or other forces.
Q2. What do you mean by acceleration due to gravity?
Answer - Acceleration due to gravity is the acceleration produced in a freely falling object due to the gravitational pull of the Earth. It is denoted by g and has an average value of 9.8 m/s² near the Earth's surface.
In-Text Questions (Page No–106)
Q1. What are the differences between the mass of an object and its weight?
Answer -
| Mass | Weight | |
|---|---|---|
| Definition | Mass is the amount of matter in an object. | Weight is the force with which an object is attracted towards the Earth. |
| Quantity Type | Scalar quantity | Vector quantity |
| SI Unit | Kilogram (kg) | Newton (N) |
| Value | Remains constant everywhere | Changes with location due to gravity |
| Formula | — | Weight = Mass × Acceleration due to gravity (W = mg) |
Q2. Why is the weight of an object on the moon 1/6th its weight on the earth?
Answer - The weight of an object depends on the gravitational force acting on it. The gravitational pull of the Moon is 1/6th that of the Earth. Hence, the weight of an object on the Moon is 1/6th of its weight on the Earth.
In-Text Questions (Page No–109)
Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer - It is difficult to hold a school bag with a thin strap because the pressure exerted by the strap on the shoulder is high.
Pressure = Force / Area
When the area of contact (strap) is small, the pressure increases, causing more pain and discomfort on the shoulder.
Pressure = Force / Area
When the area of contact (strap) is small, the pressure increases, causing more pain and discomfort on the shoulder.
Q2. What do you mean by buoyancy?
Answer - Buoyancy is the upward force exerted by a liquid on an object placed in it. This force acts opposite to the weight of the object and can cause the object to float or appear lighter in the fluid.
Q3. Why does an object float or sink when placed on the surface of water?
Answer - An object floats or sinks based on the relationship between its density and the density of water:
• If the object's density is less than that of water, it floats.
• If the object's density is more than that of water, it sinks.
This happens due to the balance between the object's weight and the buoyant force exerted by water.
• If the object's density is less than that of water, it floats.
• If the object's density is more than that of water, it sinks.
This happens due to the balance between the object's weight and the buoyant force exerted by water.
In-Text Questions (Page No–110)
Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer - Your actual mass is slightly more than 42 kg. A weighing machine measures the normal force, which may be slightly reduced due to the buoyant force of air acting upward. However, the difference is usually negligible in everyday conditions.
Q2. 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 the other. Can you say which one is heavier and why?
Answer - The iron bar is actually heavier. The cotton occupies more volume, so it experiences a greater upthrust (buoyant force) from the surrounding air, which slightly reduces its apparent weight. Since both show 100 kg on the machine, the real mass of the cotton is slightly less than that of the iron bar.
Exercises
Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer - The gravitational force becomes four times stronger when the distance is reduced to half.
According to the universal law of gravitation:
F = G (m₁m₂ / r²)
If r is reduced to r/2, then:
F' = G (m₁m₂ / (r/2)²) = G (m₁m₂ / (r²/4)) = 4 × (G m₁m₂ / r²)
⇒ F' = 4F
According to the universal law of gravitation:
F = G (m₁m₂ / r²)
If r is reduced to r/2, then:
F' = G (m₁m₂ / (r/2)²) = G (m₁m₂ / (r²/4)) = 4 × (G m₁m₂ / r²)
⇒ F' = 4F
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 - Though gravitational force is more on heavier objects, the acceleration due to gravity (g) is the same for all objects near the Earth’s surface. Therefore, in the absence of air resistance, all objects fall at the same rate regardless of mass.
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 × 10⁶ m.)
Answer - Given:
Mass of object, m = 1 kg
Mass of Earth, M = 6 × 10²⁴ kg
Radius of Earth, r = 6.4 × 10⁶ m
G = 6.674 × 10⁻¹¹ Nm²/kg²
Using the formula: F = G (Mm / r²)
F = (6.674 × 10⁻¹¹ × 6 × 10²⁴ × 1) / (6.4 × 10⁶)²
⇒ F = (4.0044 × 10¹⁴) / (4.096 × 10¹³)
⇒ F ≈ 9.78 N
Mass of object, m = 1 kg
Mass of Earth, M = 6 × 10²⁴ kg
Radius of Earth, r = 6.4 × 10⁶ m
G = 6.674 × 10⁻¹¹ Nm²/kg²
Using the formula: F = G (Mm / r²)
F = (6.674 × 10⁻¹¹ × 6 × 10²⁴ × 1) / (6.4 × 10⁶)²
⇒ F = (4.0044 × 10¹⁴) / (4.096 × 10¹³)
⇒ F ≈ 9.78 N
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 gravitational force between the earth and the moon is equal in magnitude and opposite in direction.
This is because of Newton’s third law of motion, which states that every action has an equal and opposite reaction.
This is because of Newton’s third law of motion, which states that every action has an equal and opposite reaction.
Q5. If the moon attracts the earth, why does the earth not move towards the moon?
Answer - The earth does move slightly due to the moon’s gravitational pull, but because the mass of the earth is much greater, its motion is very small and not easily noticeable. Instead, both the earth and the moon revolve around their common center of mass.
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?
(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 doubles.
(ii) If the distance is doubled, force becomes 1/4th; if tripled, force becomes 1/9th.
(iii) If masses of both objects are doubled, the force becomes 4 times.
(ii) If the distance is doubled, force becomes 1/4th; if tripled, force becomes 1/9th.
(iii) If masses of both objects are doubled, the force becomes 4 times.
Q7. What is the importance of universal law of gravitation?
Answer - The universal law of gravitation explains the force of attraction between any two masses in the universe. It helps us understand planetary motion, tides, and explains phenomena like orbits of satellites and the structure of the solar system.
Q8. What is the acceleration of free fall?
Answer - Acceleration of free fall is the acceleration experienced by any object falling freely under gravity, denoted by 'g'. Near the Earth's surface, g ≈ 9.8 m/s².
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 the weight of the object.
Q10. 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 - No, the friend will not agree.
Weight depends on acceleration due to gravity (g). Since g is greater at the poles than at the equator, the gold weighs more at the poles and less at the equator even though the mass remains the same.
Weight depends on acceleration due to gravity (g). Since g is greater at the poles than at the equator, the gold weighs more at the poles and less at the equator even though the mass remains the same.
Q11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer - A sheet of paper falls slower because it has a larger surface area, causing more air resistance which slows it down. A crumpled ball has less surface area and experiences less air resistance, so it falls faster.
Q12. 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 - Weight on Earth = mass × gravity = 10 × 9.8 = 98 N
Weight on Moon = (1/6) × weight on Earth = 98 / 6 ≈ 16.33 N
Weight on Moon = (1/6) × weight on Earth = 98 / 6 ≈ 16.33 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.
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.
Answer - Given:
Initial velocity, u = 49 m/s
Final velocity at max height, v = 0 m/s
Acceleration, a = -9.8 m/s² (gravity acts downwards)
(i) Using v² = u² + 2as:
0 = (49)² + 2 × (-9.8) × s
=> 0 = 2401 - 19.6s
=> 19.6s = 2401
=> s = 122.5 m
(ii) Time to reach max height, t = (v - u)/a = (0 - 49)/(-9.8) = 5 s
Total time to return = 2 × 5 = 10 s
Initial velocity, u = 49 m/s
Final velocity at max height, v = 0 m/s
Acceleration, a = -9.8 m/s² (gravity acts downwards)
(i) Using v² = u² + 2as:
0 = (49)² + 2 × (-9.8) × s
=> 0 = 2401 - 19.6s
=> 19.6s = 2401
=> s = 122.5 m
(ii) Time to reach max height, t = (v - u)/a = (0 - 49)/(-9.8) = 5 s
Total time to return = 2 × 5 = 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:
Initial velocity, u = 0
Distance, s = 19.6 m
Acceleration, a = 9.8 m/s²
Using v² = u² + 2as:
v² = 0 + 2 × 9.8 × 19.6 = 384.16
=> v = √384.16 ≈ 19.6 m/s
Initial velocity, u = 0
Distance, s = 19.6 m
Acceleration, a = 9.8 m/s²
Using v² = u² + 2as:
v² = 0 + 2 × 9.8 × 19.6 = 384.16
=> v = √384.16 ≈ 19.6 m/s
Q15. 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 - Given:
Initial velocity, u = 40 m/s
Final velocity at max height, v = 0 m/s
Acceleration, a = -10 m/s²
Maximum height, s:
Using v² = u² + 2as
0 = 1600 + 2 × (-10) × s
=> 0 = 1600 - 20s
=> 20s = 1600
=> s = 80 m
Net displacement = 0 (since it returns to the starting point)
Total distance covered = Upwards distance + Downwards distance = 80 + 80 = 160 m
Initial velocity, u = 40 m/s
Final velocity at max height, v = 0 m/s
Acceleration, a = -10 m/s²
Maximum height, s:
Using v² = u² + 2as
0 = 1600 + 2 × (-10) × s
=> 0 = 1600 - 20s
=> 20s = 1600
=> s = 80 m
Net displacement = 0 (since it returns to the starting point)
Total distance covered = Upwards distance + Downwards distance = 80 + 80 = 160 m
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:
Mass of Earth, m₁ = 6 × 10²⁴ kg
Mass of Sun, m₂ = 2 × 10³⁰ kg
Distance, r = 1.5 × 10¹¹ m
Gravitational constant, G = 6.67 × 10⁻¹¹ N·m²/kg²
Using Newton's law of gravitation:
F = G × (m₁ × m₂) / r²
=> F = 6.67 × 10⁻¹¹ × (6 × 10²⁴ × 2 × 10³⁰) / (1.5 × 10¹¹)²
=> F = 6.67 × 10⁻¹¹ × 12 × 10⁵⁴ / 2.25 × 10²²
=> F = (6.67 × 12 / 2.25) × 10^(−11 + 54 − 22)
=> F = 35.56 × 10²¹ = 3.56 × 10²² N
Mass of Earth, m₁ = 6 × 10²⁴ kg
Mass of Sun, m₂ = 2 × 10³⁰ kg
Distance, r = 1.5 × 10¹¹ m
Gravitational constant, G = 6.67 × 10⁻¹¹ N·m²/kg²
Using Newton's law of gravitation:
F = G × (m₁ × m₂) / r²
=> F = 6.67 × 10⁻¹¹ × (6 × 10²⁴ × 2 × 10³⁰) / (1.5 × 10¹¹)²
=> F = 6.67 × 10⁻¹¹ × 12 × 10⁵⁴ / 2.25 × 10²²
=> F = (6.67 × 12 / 2.25) × 10^(−11 + 54 − 22)
=> F = 35.56 × 10²¹ = 3.56 × 10²² 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 - Given:
Height of tower, h = 100 m
Initial velocity of stone from ground, u = 25 m/s (upwards)
Acceleration due to gravity, g = 9.8 m/s²
Let the stones meet after time t seconds from the start.
Distance travelled by falling stone after time t:
s₁ = (1/2)gt² = 4.9t² (downwards)
Distance travelled by upward stone after time t:
s₂ = ut - (1/2)gt² = 25t - 4.9t² (upwards)
Since total distance between them is 100 m:
s₁ + s₂ = 100
4.9t² + 25t - 4.9t² = 100
=> 25t = 100
=> t = 4 seconds
Distance from the ground where they meet:
s₂ = 25 × 4 - 4.9 × 16 = 100 - 78.4 = 21.6 m
Height of tower, h = 100 m
Initial velocity of stone from ground, u = 25 m/s (upwards)
Acceleration due to gravity, g = 9.8 m/s²
Let the stones meet after time t seconds from the start.
Distance travelled by falling stone after time t:
s₁ = (1/2)gt² = 4.9t² (downwards)
Distance travelled by upward stone after time t:
s₂ = ut - (1/2)gt² = 25t - 4.9t² (upwards)
Since total distance between them is 100 m:
s₁ + s₂ = 100
4.9t² + 25t - 4.9t² = 100
=> 25t = 100
=> t = 4 seconds
Distance from the ground where they meet:
s₂ = 25 × 4 - 4.9 × 16 = 100 - 78.4 = 21.6 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.
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answer - Given:
Total time of flight, T = 6 s
Acceleration due to gravity, g = 9.8 m/s²
(a) Velocity with which ball was thrown up:
Time to reach max height = T/2 = 3 s
Using v = u - gt, and final velocity at max height v = 0:
0 = u - 9.8 × 3
=> u = 29.4 m/s
(b) Maximum height reached:
Using s = ut - (1/2)gt²
s = 29.4 × 3 - 0.5 × 9.8 × 9 = 88.2 - 44.1 = 44.1 m
(c) Position after 4 seconds:
Time after throw, t = 4 s
s = ut - (1/2)gt² = 29.4 × 4 - 0.5 × 9.8 × 16 = 117.6 - 78.4 = 39.2 m above the thrower
Total time of flight, T = 6 s
Acceleration due to gravity, g = 9.8 m/s²
(a) Velocity with which ball was thrown up:
Time to reach max height = T/2 = 3 s
Using v = u - gt, and final velocity at max height v = 0:
0 = u - 9.8 × 3
=> u = 29.4 m/s
(b) Maximum height reached:
Using s = ut - (1/2)gt²
s = 29.4 × 3 - 0.5 × 9.8 × 9 = 88.2 - 44.1 = 44.1 m
(c) Position after 4 seconds:
Time after throw, t = 4 s
s = ut - (1/2)gt² = 29.4 × 4 - 0.5 × 9.8 × 16 = 117.6 - 78.4 = 39.2 m above the thrower
Q19. In what direction does the buoyant force on an object immersed in a liquid act?
Answer - The buoyant force on an object immersed in a liquid acts vertically upward, opposite to the direction of gravity.
Q20. Why does a block of plastic released under water come up to the surface of water?
Answer - A block of plastic comes up to the surface because its density is less than the density of water. Hence, the upward buoyant force is greater than the weight of the plastic, making it rise.
Q21. The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance float or sink?
Answer - Density of substance = mass/volume = 50 g / 20 cm³ = 2.5 g/cm³
Since 2.5 g/cm³ > 1 g/cm³ (density of water), the substance will sink in water.
Since 2.5 g/cm³ > 1 g/cm³ (density of water), the substance will sink in water.
Q22. 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/cm³? What will be the mass of the water displaced by this packet?
Answer - Density of packet = mass/volume = 500 g / 350 cm³ ≈ 1.43 g/cm³
Since 1.43 g/cm³ > 1 g/cm³, the packet will sink.
Mass of water displaced = density of water × volume displaced = 1 g/cm³ × 350 cm³ = 350 g
Since 1.43 g/cm³ > 1 g/cm³, the packet will sink.
Mass of water displaced = density of water × volume displaced = 1 g/cm³ × 350 cm³ = 350 g