CBSE Class 9 Physics Gravitation MCQs Set 10

Practice MCQs for Class 9 Physics Gravitation

Access targeted multiple-choice questions for Gravitation designed to align with the latest CBSE academic syllabus for Class 9 Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Navigate directly to the 50 objective questions for Gravitation using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The period of revolution of a certain planet in an orbit of radius R is T. Its period of revolution in an orbit of radius 4R will be
(a) 2T
(b) 2√2 T
(c) 4T
(d) 8T
Show Answer & Explanation

Answer: (d) 8T

Explanation:
1. T² ∝ R³, so T ∝ R^(3/2).
2. For 4R: 4^(3/2) = 8, so the period is 8T.

Teacher's Note:
1. Kepler's third law.

Question 2: If the mass of a body is M on the surface of the earth, then its mass on the surface of the moon will be
(a) M/6
(b) M
(c) M + 6
(d) zero
Show Answer & Explanation

Answer: (b) M

Explanation:
1. Mass is the quantity of matter in a body and does not change from place to place.
2. Only the weight changes.

Teacher's Note:
1. Weight changes with g; mass does not.

Question 3: The value of G does not depend on
(a) the nature of the interacting bodies
(b) the size of the interacting bodies
(c) the mass of the interacting bodies
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. G is a universal constant, 6.67 × 10⁻¹¹ N·m²/kg².
2. It is the same for all bodies, whatever their nature, size or mass.

Teacher's Note:
1. G is a constant of nature.

Question 4: The SI unit of the universal gravitational constant G is
(a) N²·m²/kg
(b) N·m²/kg
(c) N·m/kg
(d) N·m²/kg²
Show Answer & Explanation

Answer: (d) N·m²/kg²

Explanation:
1. From F = Gm₁m₂/r², G = Fr²/(m₁m₂).
2. Its unit is N·m²/kg².

Teacher's Note:
1. Put the units of F, r and m into the formula.

Question 5: 1 kg wt is equal to
(a) 980 dyne
(b) 9.80 dyne
(c) 98 dyne
(d) none of these
Show Answer & Explanation

Answer: (d) none of these

Explanation:
1. 1 kg wt = 9.8 N.
2. 1 N = 10⁵ dyne, so 1 kg wt = 9.8 × 10⁵ dyne = 980000 dyne.
3. None of the listed values is correct.

Teacher's Note:
1. Be careful with the factor 10⁵.

Question 6: The acceleration due to gravity is reduced to a value one-fourth of that on the surface of the earth at a height h where h = xRe (Re = radius of earth). The value of x is
(a) 1
(b) ½
(c) ⅓
(d) ¼
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. g' = g/(1 + x)² = g/4, so (1 + x)² = 4.
2. 1 + x = 2, so x = 1.

Teacher's Note:
1. Take the square root.

Question 7: If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is
(a) ½ mgR
(b) 2mgR
(c) mgR
(d) ¼ mgR
Show Answer & Explanation

Answer: (a) ½ mgR

Explanation:
1. ΔU = GMm(1/R − 1/2R) = GMm/(2R).
2. With GM = gR², ΔU = ½ mgR.

Teacher's Note:
1. Use the general formula, since h is not small.

Question 8: How does the force of gravitation between two objects change when the distance between them is reduced to half?
(a) increases 4 times
(b) increases 2 times
(c) increases 16 times
(d) increases 32 times
Show Answer & Explanation

Answer: (a) increases 4 times

Explanation:
1. F ∝ 1/r².
2. If r becomes r/2, F becomes 1/(1/2)² = 4 times.

Teacher's Note:
1. Halving the distance gives four times the force.

Question 9: A man weighs W on the surface of the earth. His weight at a height equal to R is
(a) W
(b) W/2
(c) W/4
(d) W/8
Show Answer & Explanation

Answer: (c) W/4

Explanation:
1. At height R the distance from the centre is 2R.
2. g' = g/4, so the weight is W/4.

Teacher's Note:
1. Weight ∝ 1/(distance from centre)².

Question 10: The value of g at a certain height h above the free surface of the earth is x/4, where x is the value of g at the surface of the earth. The height h is
(a) R
(b) 2R
(c) 3R
(d) 4R
Show Answer & Explanation

Answer: (a) R

Explanation:
1. (1 + h/R)² = 4, so 1 + h/R = 2.
2. h = R.

Teacher's Note:
1. Same as before.

Question 11: According to Kepler's laws, which of the following is correct?
(a) T ∝ R3/2
(b) T ∝ R³
(c) T ∝ R2/3
(d) T ∝ R²
Show Answer & Explanation

Answer: (a) T ∝ R3/2

Explanation:
1. Kepler's third law: T² ∝ R³.
2. Hence T ∝ R3/2.

Teacher's Note:
1. The square of the period is proportional to the cube of the orbit radius.

Question 12: The gravitational field due to a mass distribution is I = K/x³ in the x-direction (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is
(a) K/x
(b) K/2x
(c) K/x²
(d) K/2x²
Show Answer & Explanation

Answer: (d) K/2x²

Explanation:
1. Field I = −dV/dx, so V(x) = −∫ from ∞ to x of (K/x'³) dx'.
2. ∫ x'⁻³ dx' = −1/(2x'²), which gives V(x) = K/(2x²).

Teacher's Note:
1. Integrate the field to get the potential.

Question 13: The value of g on the earth's surface is 980 cm/s². Its value at a height of 64 km from the earth's surface is (radius of earth = 6400 km)
(a) 960.40 cm/s²
(b) 984.90 cm/s²
(c) 982.45 cm/s²
(d) 977.55 cm/s²
Show Answer & Explanation

Answer: (a) 960.40 cm/s²

Explanation:
1. g' ≈ g(1 − 2h/R) for small h.
2. 2h/R = 2 × 64/6400 = 0.02.
3. g' = 980 × 0.98 = 960.4 cm/s².

Teacher's Note:
1. For h much smaller than R use g' ≈ g(1 − 2h/R).

Question 14: Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be
(a) 3F
(b) F
(c) F/3
(d) F/9
Show Answer & Explanation

Answer: (b) F

Explanation:
1. Gravitational force does not depend on the medium between the masses.
2. So the force remains F.

Teacher's Note:
1. Gravitation cannot be shielded.

Question 15: A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h = 3R. The change in gravitational potential energy is
(a) ¼ mgR
(b) ⅔ mgR
(c) ¾ mgR
(d) ½ mgR
Show Answer & Explanation

Answer: (c) ¾ mgR

Explanation:
1. ΔU = GMm(1/R − 1/(R + 3R)) = GMm(1/R − 1/4R) = ¾ GMm/R.
2. With GM = gR², ΔU = ¾ mgR.

Teacher's Note:
1. Use the potential energy formula −GMm/r.

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Practice MCQs for Class 9 Physics Gravitation

Class 9 Physics Gravitation Objective Test Questions

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FAQs

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