Looking for accurate SEBA Class 9 Science Chapter 10 (Gravitation) textual solutions for the 2026–27 academic year? Access step-by-step textbook exercise answers, Universal Law of Gravitation derivations, mass vs. weight explanations, free fall equations, and solved numerical problems designed to help you score top marks in your annual exams.
Blue question
- State the universal law of gravitation.
Answer: Every particle in the universe attracts every other particle 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.
- Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth.
Answer: Gravitational Force between Earth and an Object:
F = (G * M * m) / R^2
Where:
- F = Gravitational force (N)
- G = Universal gravitational constant ≈ 6.67 * 10^-11 N m^2 / kg^2
- M = Mass of the Earth (kg)
- m = Mass of the object (kg)
- R = Radius of the Earth (m)
- What do you mean by free fall?
Answer: Free fall is the motion of an object falling solely under the influence of gravitational force, with no other forces (such as air resistance) acting upon it.
- What do you mean by acceleration due to gravity?
Answer: Acceleration due to gravity (g) is the uniform acceleration produced in a freely falling body due to the gravitational pull of the Earth.
Near the Earth’s surface:
- g ≈ 9.8 m/s^2
- SI Unit: m/s^2
- What are the differences between the mass of an object and its weight?
Answer:
- Mass is the total quantity of matter contained in an object, whereas Weight is the force with which Earth attracts the object toward its center.
- Mass is measured in kilograms (kg), while Weight is measured in Newtons (N).
- Mass remains constant everywhere in the universe, while Weight changes depending on the local acceleration due to gravity (g).
- Mass is a scalar quantity, whereas Weight is a vector quantity (directed vertically downward).
- Why is the weight of an object on the Moon 1/6th of its weight on the Earth?
Answer: Weight depends directly on the acceleration due to gravity (g = G * M / R^2). Because the Moon has a much smaller mass and radius compared to Earth, its gravitational force is only about 1/6th that of Earth.
Weight of an object on the Moon:
W_Moon = m * g_Moon = (1/6) * (m * g_Earth) = (1/6) * W_Earth
Where:
- m = Mass of the object
- g_Moon = Acceleration due to gravity on the Moon
- g_Earth = Acceleration due to gravity on Earth
Thus, the weight of an object on the Moon is 1/6th of its weight on Earth.
- Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer: Pressure is inversely proportional to the surface area over which force is applied (Pressure = Force / Area).
A strap made of a thin string has a very small contact area with the shoulder. As a result, the weight of the bag exerts a large pressure on the shoulder, causing pain and making it difficult to hold.
- What do you mean by buoyancy?
Answer: Buoyancy (or buoyant force / upthrust) is the net upward force exerted by a fluid on an object that is partially or fully immersed in it. This upward force opposes the weight of the object.
- Why does an object float or sink when placed on the surface of water?
Answer: An object floats or sinks depending on its density relative to water:
- If the density of the object is less than the density of water: The upward buoyant force is greater than the object’s weight, so the object floats.
- If the density of the object is greater than the density of water: The upward buoyant force is less than the object’s weight, so the object sinks.
- You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer: Your mass is slightly more than 42 kg.
A weighing machine measures the normal reaction force balancing your weight minus the buoyant force exerted on your body by surrounding air. Because air exerts a small upward buoyant force on you, the reading shown on the scale is slightly less than your actual true mass.
SEBA Class 9 Science Chapter 10 Solutions
- 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 bag of cotton is actually heavier in reality.
The bag of cotton occupies a much larger volume than the dense iron bar of the same measured mass. Because of its larger volume, the cotton bag displaces a larger volume of air, experiencing a greater upward buoyant force from the air.
Since the weighing machine measures apparent weight (True Weight – Air Buoyancy), the cotton bag must have a greater true weight to show the same 100 kg reading on the machine.
Textbook exercise
1. How does the force of gravitation change if the distance between two objects is halved?
Answer: According to the universal law of gravitation, gravitational force is inversely proportional to the square of the distance between two objects:
F ∝ 1 / r^2
If the distance r is halved (r’ = r / 2):
F_new = 1 / (r / 2)^2 = 1 / (r^2 / 4) = 4 / r^2 = 4 * F
Thus, the gravitational force becomes 4 times greater when the distance is halved.
- Why does a heavy object not fall faster than a light object in a vacuum?
Answer: The gravitational force on an object is proportional to its mass (F = m * g), but acceleration is given by Newton’s second law:
a = F / m
Substituting F = m * g:
a = (m * g) / m = g
The mass (m) cancels out, meaning acceleration due to gravity (g) is completely independent of the object’s mass. In the absence of air resistance, all objects accelerate at the exact same rate regardless of their weight.
- What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface?
Answer: The gravitational force is given by:
F = (G * M * m) / R^2
Given:
- Mass of Earth (M) = 6 * 10^24 kg
- Mass of object (m) = 1 kg
- Radius of Earth (R) = 6.4 * 10^6 m
- Universal gravitational constant (G) = 6.673 * 10^-11 N m^2 / kg^2
Calculation:
F = (6.673 * 10^-11 * 6 * 10^24 * 1) / (6.4 * 10^6)^2 F = (4.0038 * 10^14) / (4.096 * 10^13) F ≈ 9.8 N
The gravitational force between the Earth and a 1 kg object is approximately 9.8 N.
- 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 Newton’s third law of motion and the universal law of gravitation, forces always exist in equal and opposite pairs. The Earth attracts the Moon with the exact same magnitude of force as the Moon attracts the Earth, but in the opposite direction.
- If the Moon attracts the Earth, why does the Earth not move towards the Moon?
Answer: According to Newton’s second law, acceleration is inversely proportional to mass for a given force:
a = F / m
Although the gravitational force (F) exerted by the Moon on the Earth is equal to the force exerted by the Earth on the Moon, the mass of the Earth is exceptionally large compared to the Moon. As a result, the acceleration produced in the Earth is extremely small and practically negligible, which is why the Earth is not observed moving noticeably toward the Moon.
SEBA Class 9 Science Chapter 10 Notes
- How does the force of gravitation between two objects change under the following conditions?
Answer: The general formula for gravitational force is F = (G * m1 * m2) / r^2.
(i) If the mass of one object is doubled: F_new = (G * 2 * m1 * m2) / r^2 = 2 * F
The force doubles.
(ii) If the distance between the objects is doubled or tripled:
- Doubled: F_new = 1 / (2 * r)^2 = 1 / 4 * F (The force becomes 1/4th of its original value).
- Tripled: F_new = 1 / (3 * r)^2 = 1 / 9 * F (The force becomes 1/9th of its original value).
(iii) If the masses of both objects are doubled: F_new = (G * 2 * m1 * 2 * m2) / r^2 = 4 * F
The force becomes 4 times greater.
- What is the importance of the Universal Law of Gravitation?
Answer: The universal law of gravitation successfully explains several phenomena that were previously thought to be unrelated:
- The force that binds us to the Earth.
- The motion of the Moon and artificial satellites around the Earth.
- The orbital motion of planets around the Sun.
- The occurrence of ocean tides caused by the gravitational pull of the Moon and the Sun.
- What is meant by the acceleration of free fall?
Answer: The acceleration of free fall is the uniform acceleration experienced by an object falling under the sole influence of Earth’s gravitational pull, without any air resistance or other opposing forces. Near the surface of Earth, this value is constant at approximately g ≈ 9.8 m/s^2.
- What do we call the gravitational force between the Earth and an object?
Answer: The gravitational force exerted by the Earth on an object is commonly called the weight of the object.
- Amit buys a 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? Why?
Answer: The friend will not agree with the measured weight.
Weight is given by W = m * g. Because the Earth is flattened at the poles and bulges at the equator, the distance from the center of Earth to the poles is less than to the equator. Since acceleration due to gravity (g = G * M / R^2) is inversely proportional to radius, g is slightly larger at the poles than at the equator.
As a result, gold bought at the poles will weigh noticeably less when re-weighed at the equator, even though its mass remains identical.
Class 9 Science Chapter 10 Question Answer SEBA 2026-27
- Why does a sheet of paper fall slower than one that is crumpled into a ball?
Answer: A flat sheet of paper has a much larger surface area compared to a crumpled ball of paper of equal mass. As the flat sheet falls, it encounters significantly greater air resistance.
The crumpled ball has a much smaller surface area, so it experiences less air resistance and falls faster. In a vacuum (where air resistance is absent), both would fall at the exact same rate.
- Calculate the weight of a 10 kg object on the Moon and on the Earth.
Answer: Weight is calculated using the formula W = m * g.
Weight on Earth:
- Mass (m) = 10 kg
- Acceleration due to gravity on Earth (g_e) = 9.8 m/s^2
- W_e = m * g_e = 10 * 9.8 = 98 N
Weight on Moon:
- Acceleration due to gravity on Moon (g_m) = g_e / 6 ≈ 1.63 m/s^2
- W_m = m * g_m = 10 * 1.63 ≈ 16.3 N
The weight of a 10 kg object is 98 N on Earth and approximately 16.3 N on the Moon.
- 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:
Given:
- Initial velocity (u) = 49 m/s
- Final velocity at highest point (v) = 0 m/s
- Acceleration due to gravity (g) = -9.8 m/s^2 (acting downward)
(i) Maximum Height (h):
Using equation of motion v^2 = u^2 + 2 * g * h:
0 = (49)^2 + 2 * (-9.8) * h
19.6 * h = 2401
h = 2401 / 19.6 = 122.5 m
(ii) Total Time of Flight (t_total):
Time to reach top (t): v = u + g * t
0 = 49 – 9.8 * t
t = 49 / 9.8 = 5 s
Total time of flight = Time of ascent + Time of descent = 2 * t = 2 * 5 = 10 s
- 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 m/s (released from rest)
- Height of tower (h) = 19.6 m
- Acceleration due to gravity (g) = 9.8 m/s^2
Using equation of motion v^2 = u^2 + 2 * g * h:
v^2 = 0 + 2 * 9.8 * 19.6
v^2 = 19.6 * 19.6
v = 19.6 m/s
The final velocity of the stone just before touching the ground is 19.6 m/s.
SEBA Class 9 Science Chapter 10 Numericals
- 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:
- Initial velocity (u) = 40 m/s
- Final velocity at peak (v) = 0 m/s
- Acceleration due to gravity (g) = -10 m/s^2
Maximum Height (h):
v^2 = u^2 + 2 * g * h
0 = (40)^2 + 2 * (-10) * h
20 * h = 1600
h = 1600 / 20 = 80 m
Net Displacement and Total Distance:
- Net Displacement = 0 m (since the stone returns to its initial starting point).
- Total Distance Covered = Distance upward + Distance downward = 80 m + 80 m = 160 m.
- Calculate the force of gravitation between the Earth and the Sun.
Answer:
Given:
- Mass of Earth (m1) = 6 * 10^24 kg
- Mass of Sun (m2) = 2 * 10^30 kg
- Average distance between Earth and Sun (r) = 1.5 * 10^11 m
- Gravitational constant (G) = 6.673 * 10^-11 N m^2 / kg^2
Formula:
F = (G * m1 * m2) / r^2
Calculation:
F = (6.673 * 10^-11 * 6 * 10^24 * 2 * 10^30) / (1.5 * 10^11)^2
F = (8.0076 * 10^44) / (2.25 * 10^22)
F ≈ 3.56 * 10^22 N
The gravitational force between Earth and the Sun is approximately 3.56 * 10^22 N.
- A stone is dropped 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:
- Total height (H) = 100 m
- Acceleration due to gravity (g) = 9.8 m/s^2
Let the two stones meet after time t.
For the stone dropped from top (Stone 1):
- u1 = 0 m/s
- Distance traveled downward (s1) = u1 * t + (1/2) * g * t^2 = 0 + 4.9 * t^2 = 4.9 * t^2
For the stone thrown upward from ground (Stone 2):
- u2 = 25 m/s
- Distance traveled upward (s2) = u2 * t – (1/2) * g * t^2 = 25 * t – 4.9 * t^2
Since total height is 100 m:
s1 + s2 = 100
(4.9 * t^2) + (25 * t – 4.9 * t^2) = 100
25 * t = 100
t = 4 s
Position where they meet:
Distance from ground (s2) = 25 * 4 – 4.9 * (4)^2 = 100 – 78.4 = 21.6 m (or 20 m if using g = 10 m/s^2).
The two stones will meet after 4 seconds at a height of 21.6 m from the ground.
- 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:
- Total time of flight = 6 s
- Time taken to reach maximum height (t) = 6 / 2 = 3 s
- Final velocity at maximum height (v) = 0 m/s
- Acceleration due to gravity (g) = -9.8 m/s^2
(a) Initial Velocity (u):
v = u + g * t
0 = u – (9.8 * 3)
u = 29.4 m/s
(b) Maximum Height (h):
h = u * t + (1/2) * g * t^2
h = 29.4 * 3 – (0.5 * 9.8 * 3^2)
h = 88.2 – 44.1 = 44.1 m
(c) Position after 4 s:
s = u * t + (1/2) * g * t^2
s = (29.4 * 4) – (0.5 * 9.8 * 4^2)
s = 117.6 – 78.4 = 39.2 m from the ground
The ball is at a height of 39.2 m from the ground after 4 seconds (moving downwards).
- In which direction does the buoyant force on an object immersed in a liquid act?
Answer: The buoyant force (upthrust) acting on an object immersed in a liquid always acts in the vertically upward direction, directly opposing the downward pull of gravity.
- Why does a block of plastic released under water come up to the surface of water?
Answer: The density of plastic is less than the density of water. As a result, the upward buoyant force (upthrust) exerted by the water on the submerged plastic block is greater than the downward gravitational force (weight) of the block.
Because the net force acting on the block is directed upward, it rises to the surface.
Gravitation Class 9 SEBA Textual Exercise Solutions
21. You have an object of mass 50 g and volume 20 cm³. Will it float or sink in water?
Answer: An object floats or sinks depending on its density relative to water (density of water = 1 g/cm³).
Density of object (ρ) = mass / volume
ρ = 50 g / 20 cm³ = 2.5 g/cm³
Since the density of the object (2.5 g/cm³) is greater than the density of water (1 g/cm³), the object will sink.
- The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water having a density of 1 g/cm³? What will be the mass of the water displaced by this packet?
Answer:
Density of the packet = mass / volume
Density = 500 g / 350 cm³ ≈ 1.43 g/cm³
Since the density of the packet (1.43 g/cm³) is greater than the density of water (1 g/cm³), the packet will sink.
Mass of water displaced:
Since the packet sinks completely, it displaces a volume of water equal to its own volume (350 cm³).
Mass of displaced water = volume * density of water
Mass of displaced water = 350 cm³ * 1 g/cm³ = 350 g
🔬 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 9 Science Chapter 10 (Gravitation) Textual Exercise Solutions updated for the 2026–27 academic session. All formula derivations, weight calculations on the Moon, acceleration due to gravity ($g$) numericals, and buoyancy explanations strictly follow the latest revised Assam Board (SEBA) textbook.
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