Free Gravitation MCQs with Answers

11. Compute the gravitational force of attraction between two balls each weighing 10 kg when placed at a distance of 1 metre apart.

• A. 66 x 10 -2 N
• B. 77 x 10 -7 N
• C. 81.9 x 10 -7 N
• D. 31.4 x 10^-6 N
• E. 667.3 x 10^-11 N

Explanation: According to gravitational law: F= G m1m2/ r2 F = (6.674x 10^-11)(10)(10)(1)2 F = 667.4x 10^-11 N Option E is correct. As this question is theroretical options A, B, C and D are incorrect.

12. What will be the gravitational force of attraction between two balls each weighing 5 kg when placed at a distance of 0.33 m apart. (G = 6.673 x 10^-11 Nm2/kg2 )

• A. 9.1 x 10^-8 N
• B. 7.1 x 10^-8 N
• C. 6.1 x 10^-8 N
• D. 3.5 x 10^-8 N
• E. 1.5 x 10^-8 N

Explanation: Given: m1=5kg m2=5kg r=0.33m G=6.673×10-11 Nm2/kg2 Formula: FG=Gm1m2/r2 FG=(6.673×10-11)(5)(5)/0.332 = 1.5×10-8N

13. In motion of satellites, necessary centripetal force is provided by:

• A. Gravitational Force
• B. Coulomb's Force
• C. Magnetic Force
• D. Nuclear Force

Explanation: Satellites revolve around planets because the gravitational force of attraction between the satellites and the planets acts as a centripetal force for them. This is why option A is correct. Other forces either don't exist or are too weak to act as a centripetal force.

14. For a satellite so that it could stay over the same point on Earth's surface the orbital radius r is given by _.

• A. r2 = GMT2 / 4x2
• B. r2 = [GMT2 / 4x2 ] 1/3
• C. r = [GMT2 / 4x2 ] 1/3
• D. r = [GMT / 4x2 ] 1/3

Explanation: For a satellite, the gravitational force provides the centripetal force; therefore: Fg=Fc {where Fg is the gravitational force and Fc is the centripetal force} GMm/r2=mv2/r Make R the subject of the equation to obtain: r=[GMT2/4x2]⅓

15. A boy is standing in a lift falling freely under gravity and releases a ball from hand. As seen by the ball, the boy:

• A. Falls down
• B. Remains stationary
• C. Falls behind him.
• D. None of above

Explanation: The acceleration of the body does not depend on the mass (for freely falling). Hence both boy and ball have same acceleration. Hence their velocities at every point would be same. So "Relative" velocity is 0. Hence the boy will remain stationary as seen by the ball.

16. An object of mass 'm' is suspended in an elevator moving downward with acceleration equal to acceleration due to gravity. What is the apparent weight of object?

• A. Zero
• B. 2mg
• C. mg
• D. mg/2

Explanation: The apparent weight of the object is the force exerted on it by the elevator floor, which is equal to the net force acting on the object.Using the formula Fnet = mg - ma Where,m is the mass of the object g is the acceleration due to gravity a is the acceleration of the elevatorIn this case, the elevator is moving downward with an acceleration equal to the acceleration due to gravity, which means a = g.Therefore,Fnet = mg - ma = m(g - g) = 0.So, the apparent weight of the object is zero. Option A is the correct answer.

17. An elevator is moving upwards with constant velocity of 'v'. What is the weight of a person of a mass 'm' inside the elevator during upward motion?

• A. mg + mv
• B. mg
• C. mg ― mv
• D. Zero

Explanation: The weight of an object is defined as the force of gravity on the object and is calculated as the mass times the acceleration of gravity, w = mg. This does not change when the elevator is moving because it doesn't effect the mass of the person and the gravitational acceleration.

18. Due to some mechanical fault, a lift falls freely from the top of a multistory building. Which of the following is the apparent weight of a man inside the lift, if the mass of the man is 80 kg while the value of 'g' is 10 ms-2?

• A. 900 N
• B. Zero
• C. 800 N
• D. 700 N

Explanation: If the elevator falls freely, the downward acceleration of the man in the elevator is: a = g Now, the force exerted by the floor on the man is equal to the actual weight of the man minus the net upward force on him. F = ma - mg F = mg - mg F = 0 N Hence, the force exerted by the floor on the man is zero.

19. On the ground, the gravitational force on a satellite is W. What is the gravitational force on the satellite when at a height R/50, where R is the radius of the earth?

• A. 1.04W
• B. 1.02W
• C. 0.50W
• D. 0.96W

Explanation: Gravitational force has formula,F=GMm/R2Where G is constant, M is mass of earth ,m is mass of object and R is distance between earth's center and objectOn ground Force= GMm/R2Because distance equals to radius of earthIn given orbit,Force=GMm/(R+R/50)²Taking LCM of R+R/50, we get 51R/50GMm/(51R/50)²GMm/(2601 R²/2500) = GMmx2500/2601R²Equating both these equations:GMm x 2500/2601R²=GMm/R2All the constants will be canceled and we will be left with;2500/2601 = 0.96WSo right option will be D

20. The gravitational field strength on the surface of the Earth is g. The gravitational field strength on the surface of a planet of twice the radius and the same density is:

• A. 4g
• B. 2g
• C. g
• D. g/4

Explanation: The unit "per ohm second" (Ω⁻¹s) is indeed equivalent to the unit "farad" (F), which is the SI unit of capacitance. Both units represent different aspects of electrical properties. In the context of capacitance, the farad (F) measures the ability of a capacitor to store an electric charge for a given voltage. It's defined as one coulomb of charge per volt of potential difference. On the other hand, "per ohm second" (Ω⁻¹s) represents the unit of electrical conductance multiplied by time. Conductance is the reciprocal of resistance (measured in ohms), and when multiplied by time (seconds), it becomes equivalent to capacitance (farads). In mathematical terms: 1 F = 1 Ω⁻¹s This relationship highlights the interconnectedness of various electrical properties and units. Now, the gravitational field strength (g') on the surface of the new planet will be determined by the following equation: g' = G * (M' / R'²) where: G is the gravitational constant (a universal constant, approximately 6.674 × 10⁻¹¹ N(m/kg)²), M' is the mass of the planet (which is 8 times the mass of the Earth), R' is the radius of the planet (twice the radius of the Earth). Since the mass and radius are both larger by a factor of 8 and 2, respectively, the gravitational field strength on the surface of the new planet (g') will be: g' = 8g/4 = 2g Therefore, the gravitational field strength on the surface of a planet with twice the radius and the same density as the Earth will be two times the gravitational field strength on the surface of the Earth (2g). Refer to this video for better understanding: