Free Gravitation MCQs with Answers

81. If the mass of earth becomes four times, then the value of "g" _:

• A. Also becomes four times
• B. One fourth of the mass of the earth
• C. Twice the mass of the earth
• D. None of these

Explanation: The acceleration due to gravity (g) is directly proportional to the mass of the Earth and inversely proporional to the square of the distance from the center of the earth. If the mass of the Earth were to become four times its current value, and assuming the radius remains constant, the value of (g) would also become four times smaller.

82. The formula to find the volume of earth is_:

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Explanation: Option C is correct as it gives the correct formula used to calculate the volume of the earth. Earth is a sphere and to find the volume of a sphere we use the formula 4/3 pi r3

83. The value of g is _:

• A. Greatest at the equator
• B. Least at the equator
• C. Least at the poles
• D. Same all over the earth surface

Explanation: The value of ( g ), the acceleration due to gravity, is least at the equator compared to other latitudes. This is primarily due to two factors: rotation and shape. 1. Centrifugal Force: The Earth is not a perfect sphere; it's slightly flattened at the poles and bulging at the equator due to its rotation. The centrifugal force generated by this rotation is highest at the equator, where the Earth's rotation is fastest. This centrifugal force partially counteracts the gravitational force, resulting in a net decrease in the effective gravity at the equator. 2. Shape of the Earth: The Earth's equatorial diameter is larger than its polar diameter. As you move towards the poles, you get closer to the center of the Earth, where gravity is slightly stronger. At the equator, you are farther from the Earth's center, and gravity is slightly weaker. The combination of these factors leads to a decrease in the effective acceleration due to gravity at the equator compared to other latitudes. The actual value of ( g ) at the equator is approximately 9.78 m/s², while at the poles, it's about 9.83 m/s².

84. The weight of a body is minimum at:

• A. The equator
• B. The poles
• C. A place in between pole and equator
• D. None of these

Explanation: W= mg where g= acceleration due to gravity. At the equator, the weight (mg) of a body is minimum because the Earth's rotation generates a centrifugal force that partially offsets the gravitational force This reduces the effective acceleration due to gravity (g), resulting in a smaller (mg) value compared to other latitudes.

85. As we go below the surface of the earth the value of g:

• A. Decreases
• B. Increases
• C. Becomes zero
• D. Remains unchanged

Explanation: Certainly, as you go below the surface of the Earth, the value of ( g ) initially decreases due to the diminishing influence of mass above. However, as you go deeper, the increasing contribution of mass below you, particularly from the dense core, causes ( g ) to start increasing. Therefore, ( g ) decreases near the surface and then tends to increase with greater depth.

86. At what depth below the earth surface the value of g reduces to one-half of its value on the earth's surface:

• A. Re/4
• B. Re/2
• C. Re
• D. 2Re

Explanation: The correct option is B. The solution is given below:

87. The artificial gravity is created in a satellite to overcome the state of weightlessness, experienced by the astronauts in it, by:

• A. Decreasing the orbital speed of the satellite
• B. In creasing the orbital speed of the satellite
• C. Causing the satellite to rotate about its own axis
• D. None of the above

Explanation: Artificial gravity in a satellite to overcome the state of weightlessness experienced by astronauts can be achieved by creating centripetal acceleration through the rotation of the satellite. This involves spinning the satellite, generating centrifugal force, and simulating gravity for objects and astronauts inside. The formula for centripetal acceleration (\(a_c\)) is used to determine the necessary rotational parameters, and adjusting the rotation rate and radius of rotation can create artificial gravity equivalent to Earth's gravity.

88. The artificial gravity is created in the satellite to:

• A. Increase the orbital speed of the satellite
• B. Keep the satellite in its orbit
• C. Keep the orbital radius of the satellite constant
• D. Overcome the weightlessness experienced by astronaut

Explanation: Artificial gravity in a satellite to overcome the state of weightlessness experienced by astronauts can be achieved by creating centripetal acceleration through the rotation of the satellite. This involves spinning the satellite, generating centrifugal force, and simulating gravity for objects and astronauts inside. The formula for centripetal acceleration (\(a_c\)) is used to determine the necessary rotational parameters, and adjusting the rotation rate and radius of rotation can create artificial gravity equivalent to Earth's gravity.

89. Weight rather than mass must be used when calculating:

• A. The gravitational force between two bodies
• B. Specific heat capacity of a body
• C. The stress in a wire due to a load hanging from it
• D. The moment of inertia of a body

Explanation: When calculating the stress in a wire, weight is used instead of mass because stress is a force per unit area, and weight is a force. Stress is defined as the force ( F ) applied to a material divided by the cross-sectional area ( A ) over which the force is applied. In the case of a load hanging from a wire, the force ( F ) is the weight of the object, which is given by W = mg , where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity. Using weight in stress calculations is appropriate because stress is a measure of the internal forces within a material, and the force acting on the material is the weight of the hanging load. This approach is common in engineering and materials science, where understanding the impact of external forces on materials is crucial.

90. A body of mass m is projected from the earth's surface at the point of launch, the acceleration of free fall is g and the radius of earth is Re. To escape from the gravitational field of the earth, the speed of the body (i.e. escape velocity) must be at least:

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Explanation: Option C is correct since it is the correct formula.