Unit 6: Gravitation — Online MCQ Test
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Practice Unit 6: Gravitation with a free chapter-wise online MCQ test.
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....
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Unit 6: Gravitation — Important Questions & Answers
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.
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.
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.
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.
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.
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).
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.
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).
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.
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.
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.