Chapter Exam

Unit 6: Gravitation — Online MCQ Test

Q: A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

Answer: A. 32 N — F = F₀ * (R / (R+h))². Given h = R/2, F = 72 * (R / (1.5R))² = 72 * (1 / 1.5)² = 72 * (2/3)² = 72 * 4/9 = 8 * 4 = 32 N.

Q: The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

Answer: A. 88 earth days — By Kepler's Third Law, T^2 ∝ r^3. T_mars / T_merc = (r_mars / r_merc)^(3/2) = 4^(3/2) = 8. T_merc = 687 / 8 ≈ 85.8, which is approximately 88 days.

Q: A body weight 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

Answer: B. 27 N — Weight at height h is W' = W * (R / (R + h))^2. Here h = R/3, so W' = 48 * (R / (R + R/3))^2 = 48 * (R / (4R/3))^2 = 48 * (3/4)^2 = 48 * 9/16 = 3 * 9 = 27 N.

Q: At what height from the surface of earth the gravitation potential and the value of g are – 5.4 × 10⁷ J kg⁻² and 6.0 ms⁻² respectively? Take the radius of earth as 6400 km:

Answer: D. 2600 km — Using V = -GM/(R+h) and g = GM/(R+h)², the ratio |V|/g gives (R+h) = 5.4 × 10⁷ / 6 = 0.9 × 10⁷ m = 9000 km. Subtracting R = 6400 km gives h = 2600 km.

Q: The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is:

Answer: A. 1 : 2√2 — Escape velocity v_e = sqrt(8/3 * G * pi * rho * R^2). Therefore, v is proportional to R * sqrt(rho). Given R_p = 2*R_e and rho_p = 2*rho_e, v_p = v_e * (R_p/R_e) * sqrt(rho_p/rho_e) = v_e * 2 * sqrt(2) = 2√2 * v_e. Ratio v_e/v_p = 1 : 2√2.

Q: When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as:

Answer: D. 1/m — By conservation of energy, (1/2)mv^2 = (k * (Ze) * (2e)) / r0. Therefore, r0 = (4kZe^2) / (mv^2). The distance of closest approach is inversely proportional to mass (1/m).

Q: The mass of a planet is one-tenth that of the Earth and its diameter is half that of the Earth. The acceleration due to gravity on that planet is:

Answer: A. 4.9 m s^-2 — g' = GM'/R'^2. Here M' = M/10 and R' = R/2, so g' = (1/10)/(1/4) g = g/2 = 4.9 m s^-2.

Q: The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is:

Answer: C. 5GMm / 6R — Altitude 2R means orbital radius r = 3R. Initial energy at Earth surface = -GMm/R. Final total energy in orbit = -GMm/(2r) = -GMm/(6R). Energy required = final - initial = GMm/R - GMm/(6R) = 5GMm/(6R).

Q: Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will

Answer: A. Move towards each other — Both the astronauts are in the condition of weightlessness. Gravitational force between them pulls them towards each other.

Q: The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then

Answer: C. d = 2 km — The change in gravity at height h (for small h) is Δg = 2hg/R. At depth d, Δg = dg/R. Equating them, d = 2h. Given h = 1 km, d = 2 km.

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Practice Unit 6: Gravitation with a free chapter-wise online MCQ test for NEET NEET (All) PHYSICS. This chapter covers: The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Kepler’s law of planetary motion. Gravitational potential energy; gravitati.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 6: Gravitation — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) PHYSICS — Unit 6: Gravitation, with answers and explanations. These are sample questions; the exam has its own separate question set.

A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

A. 32 N ✓
B. 30 N
C. 24 N
D. 48 N

Answer: A. 32 N
F = F₀ * (R / (R+h))². Given h = R/2, F = 72 * (R / (1.5R))² = 72 * (1 / 1.5)² = 72 * (2/3)² = 72 * 4/9 = 8 * 4 = 32 N.

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

A. 88 earth days ✓
B. 225 earth days
C. 172 earth days
D. 124 earth days

Answer: A. 88 earth days
By Kepler's Third Law, T^2 ∝ r^3. T_mars / T_merc = (r_mars / r_merc)^(3/2) = 4^(3/2) = 8. T_merc = 687 / 8 ≈ 85.8, which is approximately 88 days.

A body weight 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

A. 16 N
B. 27 N ✓
C. 32 N
D. 36 N

Answer: B. 27 N
Weight at height h is W' = W * (R / (R + h))^2. Here h = R/3, so W' = 48 * (R / (R + R/3))^2 = 48 * (R / (4R/3))^2 = 48 * (3/4)^2 = 48 * 9/16 = 3 * 9 = 27 N.

At what height from the surface of earth the gravitation potential and the value of g are – 5.4 × 10⁷ J kg⁻² and 6.0 ms⁻² respectively? Take the radius of earth as 6400 km:

A. 1600 km
B. 1400 km
C. 2000 km
D. 2600 km ✓

Answer: D. 2600 km
Using V = -GM/(R+h) and g = GM/(R+h)², the ratio |V|/g gives (R+h) = 5.4 × 10⁷ / 6 = 0.9 × 10⁷ m = 9000 km. Subtracting R = 6400 km gives h = 2600 km.

The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is:

A. 1 : 2√2 ✓
B. 1 : 4
C. 1 : 2
D. 1 : √2

Answer: A. 1 : 2√2
Escape velocity v_e = sqrt(8/3 * G * pi * rho * R^2). Therefore, v is proportional to R * sqrt(rho). Given R_p = 2*R_e and rho_p = 2*rho_e, v_p = v_e * (R_p/R_e) * sqrt(rho_p/rho_e) = v_e * 2 * sqrt(2) = 2√2 * v_e. Ratio v_e/v_p = 1 : 2√2.

When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as:

A. 1/m
B. 1/m^2
C. m
D. 1/m ✓

Answer: D. 1/m
By conservation of energy, (1/2)mv^2 = (k * (Ze) * (2e)) / r0. Therefore, r0 = (4kZe^2) / (mv^2). The distance of closest approach is inversely proportional to mass (1/m).

The mass of a planet is one-tenth that of the Earth and its diameter is half that of the Earth. The acceleration due to gravity on that planet is:

A. 4.9 m s^-2 ✓
B. 3.92 m s^-2
C. 19.6 m s^-2
D. 9.8 m s^-2

Answer: A. 4.9 m s^-2
g' = GM'/R'^2. Here M' = M/10 and R' = R/2, so g' = (1/10)/(1/4) g = g/2 = 4.9 m s^-2.

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is:

A. GM^2 m / R
B. GM^3 m / R
C. 5GMm / 6R ✓
D. 2GMm / 3R

Answer: C. 5GMm / 6R
Altitude 2R means orbital radius r = 3R. Initial energy at Earth surface = -GMm/R. Final total energy in orbit = -GMm/(2r) = -GMm/(6R). Energy required = final - initial = GMm/R - GMm/(6R) = 5GMm/(6R).

Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will

A. Move towards each other ✓
B. Move away from each other
C. Will become stationary
D. Keep floating at the same distance between them

Answer: A. Move towards each other
Both the astronauts are in the condition of weightlessness. Gravitational force between them pulls them towards each other.

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then

A. d = 1 km
B. d = 3/2 km
C. d = 2 km ✓
D. d = 1/2 km

Answer: C. d = 2 km
The change in gravity at height h (for small h) is Δg = 2hg/R. At depth d, Δg = dg/R. Equating them, d = 2h. Given h = 1 km, d = 2 km.

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