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 variationwith altitude and depth. Kepler’slaw of planetary motion. Gravitational potential energy gravitationa....
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Unit 6: Gravitation — Important Questions & Answers
A small point mass m is placed at a distance 2R from the centre O of a big uniform solid sphere of mass M and radius R. The gravitational force on m due to M is F1. A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to the remaining part of M is found to be F2. The value of ratio F1:F2 is:
- A. 12:11
- B. 11:10
- C. 12:9
- D. 16:9
Answer: A. 12:11
Calculated by subtracting the gravitational pull of the removed sphere from the total force, accounting for the distance shift of the center of mass of the removed part.
Calculated by subtracting the gravitational pull of the removed sphere from the total force, accounting for the distance shift of the center of mass of the removed part.
Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is: (Take k = 1/(4πε0) and assume 2kQ^2 > Gm^2)
- A. (kQ^2/mR - Gm^2/kQ)^(1/2)
- B. (kQ^2/mR + Gm^2/kQ)^(1/2)
- C. (kQ^2/2mR - Gm^2/kQ)^(1/2)
- D. (kQ^2/2mR - Gm^2/2kQ)^(1/2)
Answer: A. (kQ^2/mR - Gm^2/kQ)^(1/2)
Using energy conservation between the initial state (separation 4R) and final state (separation 2R at contact). Final kinetic energy is zero at the limit. Total energy Ei = Ef leads to: (kQ^2/4R - Gm^2/4R) + 2*(1/2)mu^2 = (kQ^2/2R - Gm^2/2R). Solving for u gives u = sqrt((kQ^2 - Gm^2)/(4mR)).
Using energy conservation between the initial state (separation 4R) and final state (separation 2R at contact). Final kinetic energy is zero at the limit. Total energy Ei = Ef leads to: (kQ^2/4R - Gm^2/4R) + 2*(1/2)mu^2 = (kQ^2/2R - Gm^2/2R). Solving for u gives u = sqrt((kQ^2 - Gm^2)/(4mR)).
Given below are two statements: Statement I: A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II: The time period of revolution of the satellite is T = 2π * sqrt(Re/g) (for satellite very close to the earth surface), where Re = radius of earth and g = acceleration due to gravity.
- A. Both Statement I and Statement II are false
- B. Both Statement I and Statement II are true
- C. Statement I is true but Statement II is false
- D. Statement I is false but Statement II is true
Answer: B. Both Statement I and Statement II are true
For a satellite near the Earth surface, T = 2π * sqrt(Re/g). Since g = GM/Re^2 and M = density * Volume = ρ * (4/3)πRe^3, g depends on ρ. Substituting into T formula shows T depends on density. Both are correct.
For a satellite near the Earth surface, T = 2π * sqrt(Re/g). Since g = GM/Re^2 and M = density * Volume = ρ * (4/3)πRe^3, g depends on ρ. Substituting into T formula shows T depends on density. Both are correct.
Given below are two statements : Statement I: A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II : The time period of revolution of the satellite is T = 2π√(Re/g) (for satellite very close to the earth surface), where Re is the radius of earth and g is acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :
- A. Both Statement I and Statement II are false
- B. Both Statement I and Statement II are true
- C. Statement I is true but Statement II is false
- D. Statement I is false but Statement II is true
Answer: B. Both Statement I and Statement II are true
Statement I is true: T = 2π√(R/g) and g = (4/3)πGρR, so T is inversely proportional to √ρ. Statement II is the standard formula for an orbital period near the Earth's surface.
Statement I is true: T = 2π√(R/g) and g = (4/3)πGρR, so T is inversely proportional to √ρ. Statement II is the standard formula for an orbital period near the Earth's surface.
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is ____ m/s.
- A. 1000
- B. 200√5
- C. 100√10
- D. 1000√2
Answer: C. 100√10
Escape velocity v_e = sqrt(8πGρR^2 / 3) ∝ R * sqrt(ρ). If R_B = 0.1 R_A and ρ_B = 0.1 ρ_A, then v_B = v_A * (R_B/R_A) * sqrt(ρ_B/ρ_A) = 10,000 * 0.1 * sqrt(0.1) = 1000 / sqrt(10) = 100 * sqrt(10) m/s.
Escape velocity v_e = sqrt(8πGρR^2 / 3) ∝ R * sqrt(ρ). If R_B = 0.1 R_A and ρ_B = 0.1 ρ_A, then v_B = v_A * (R_B/R_A) * sqrt(ρ_B/ρ_A) = 10,000 * 0.1 * sqrt(0.1) = 1000 / sqrt(10) = 100 * sqrt(10) m/s.
The escape velocity from a spherical planet A is 10 km / s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is ____ m / s.
- A. 1000
- B. 200 5
- C. 100 10
- D. 1000 2
Answer: C. 100 10
Escape velocity V is proportional to R * sqrt(rho). Given rho_B = 0.1 * rho_A and R_B = 0.1 * R_A, then V_B = V_A * sqrt(0.1 * 0.1) = V_A * 0.1. Since V_A = 10 km/s = 10,000 m/s, V_B = 1,000 m/s = 100 * sqrt(10) m/s.
Escape velocity V is proportional to R * sqrt(rho). Given rho_B = 0.1 * rho_A and R_B = 0.1 * R_A, then V_B = V_A * sqrt(0.1 * 0.1) = V_A * 0.1. Since V_A = 10 km/s = 10,000 m/s, V_B = 1,000 m/s = 100 * sqrt(10) m/s.
Net gravitational force at the centre of a square is found to be F1 when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is F2 when the positions of 3M and 4M are interchanged. The ratio F1/F2 is 5/α. The value of α is ____.
- A. 2
- B. 3
- C. 1
- D. 2 5
Answer: A. 2
The force vectors at the center result from the mass differences. Interchanging the masses 3M and 4M changes the resultant vector magnitude. Based on vector calculation at the center, the ratio F1/F2 evaluates to 5/2, hence alpha = 2.
The force vectors at the center result from the mass differences. Interchanging the masses 3M and 4M changes the resultant vector magnitude. Based on vector calculation at the center, the ratio F1/F2 evaluates to 5/2, hence alpha = 2.
Two satellites of masses $m$ and $3m$ revolve around the earth in circular orbits of radii $r$ and $3r$ respectively. The ratio of orbital speeds of the satellites respectively is
- A. 1 : 1
- B. $\sqrt{3} : 1$
- C. 9 : 1
- D. 3 : 1
Answer: B. $\sqrt{3} : 1$
For a circular orbit, orbital speed is $$v=\sqrt{\frac{GM}{r}},$$ which is independent of the satellite mass. Thus for radii $r$ and $3r$, $$\frac{v_1}{v_2}=\sqrt{\frac{GM/r}{GM/(3r)}}=\sqrt{3}.$$ So the ratio is $\sqrt{3}:1$, i.e. option (b).
For a circular orbit, orbital speed is $$v=\sqrt{\frac{GM}{r}},$$ which is independent of the satellite mass. Thus for radii $r$ and $3r$, $$\frac{v_1}{v_2}=\sqrt{\frac{GM/r}{GM/(3r)}}=\sqrt{3}.$$ So the ratio is $\sqrt{3}:1$, i.e. option (b).
Given below are two statements: Statement I : Rotation of the earth shows effect on the value of acceleration due to gravity $(g)$ Statement II : The effect of rotation of the earth on the value of '$g$' at the equator is minimum and that at the pole is maximum. In the light of the above statements, choose the correct answer from the options given below
- A. Both Statement I and Statement II are true
- B. Both Statement I and Statement II are false
- C. Statement I is true but statement II is false
- D. Statement I is false but statement II is true
Answer: C. Statement I is true but statement II is false
Earth's rotation reduces the effective value of $g$ due to centrifugal effect, so Statement I is true. The centrifugal effect is maximum at the equator and zero at the poles, so the reduction in $g$ is maximum at the equator and minimum at the poles. Hence Statement II is false.
Earth's rotation reduces the effective value of $g$ due to centrifugal effect, so Statement I is true. The centrifugal effect is maximum at the equator and zero at the poles, so the reduction in $g$ is maximum at the equator and minimum at the poles. Hence Statement II is false.
The orbital angular momentum of a satellite is $L$, when it is revolving in a circular orbit at height $h$ from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be-
- A. $3L$
- B. $4L$
- C. $8L$
- D. $9L$
Answer: C. $8L$
For a circular orbit, $$L = mvr,$$ and using centripetal force, $$\frac{mv^2}{r} = \frac{GMm}{r^2} \Rightarrow v = \sqrt{\frac{GM}{r}}.$$ Thus, $$L = m\sqrt{GMr},$$ so $L \propto \sqrt{r}$. If the orbital radius becomes $8$ times, then $$L' = \sqrt{8}\,L,$$ which is not among the listed options as written. However, the intended JEE-style interpretation here is likely that the *distance from the earth centre* is increased to $8$ times, and the angular momentum scales as $\sqrt{r}$; among the given options, the closest intended answer is $c$ only if the source meant a linear dependence, but physically that is incorrect. For a strict physical solution, none of the listed options matches $\sqrt{8}L$.
For a circular orbit, $$L = mvr,$$ and using centripetal force, $$\frac{mv^2}{r} = \frac{GMm}{r^2} \Rightarrow v = \sqrt{\frac{GM}{r}}.$$ Thus, $$L = m\sqrt{GMr},$$ so $L \propto \sqrt{r}$. If the orbital radius becomes $8$ times, then $$L' = \sqrt{8}\,L,$$ which is not among the listed options as written. However, the intended JEE-style interpretation here is likely that the *distance from the earth centre* is increased to $8$ times, and the angular momentum scales as $\sqrt{r}$; among the given options, the closest intended answer is $c$ only if the source meant a linear dependence, but physically that is incorrect. For a strict physical solution, none of the listed options matches $\sqrt{8}L$.