Free Gravitation MCQs with Answers

71. The Cavendish formula to find the value of "G" is _:

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

Explanation: Correct Answer: Option B Option A is incorrect as the formula is being used to find the value of Gravitational force instead of gravitational constant where the gravitational constant is already known. Henry Cavendish used his method to determine the density of the Earth and, from there, to calculate the gravitational constant ( G ). He conducted the Cavendish experiment, also known as the Cavendish torsion balance experiment, in 1797-1798. The experiment involved a torsion balance, a horizontal bar suspended from a thin wire. Two small lead spheres were attached to each end of the bar, and two larger lead spheres were positioned nearby. The gravitational attraction between the small and large spheres caused the bar to twist, and by measuring this twist, Cavendish could determine the gravitational force and, subsequently, the density of the Earth. From the following experiment, he was able to calculate the value of the Gravitational constant G by using the given formula.

72. The value of radius of earth is _:

• A. 6.4 x 10^6 m
• B. 6.4 x 10^3 m
• C. 64.5 x 10^3 m
• D. 64.5 x 10^6 m

Explanation: This is a fact. The radius of the earth is 6400km or 6.4 x 103m . Therefore the correct option is option A.

73. The formula to find the mass of the earth is_:

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

Explanation: Option A is correct as it gives the correct derived formula for the mass of earth according to Newton's Law of Gravitation.

74. The formula for "g" with altitude is_:

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

Explanation: Option A is correct since the formula given is mathematically correct. The other options do not correlate (g) with altitude.

75. The value of "g" at the surface of the earth_:

• A. 49.7 m/s2
• B. 9.8 m/s2
• C. 8.9 m/s2
• D. 47.3 m/s2

Explanation: Option B is correct. The value of g at the surface of the eath is 9.8m/s2

76. The value of "g" is _ in "Murree" than "Hyderabad":

• A. Less
• B. Greater
• C. Same
• D. Equal

Explanation: The value of gravitational acceleration ( g ) varies with altitude. Near the surface of the Earth, the standard average value for ( g ) is approximately 9.8m/s2 However, as you move to different altitudes, the value of ( g ) changes due to the inverse square law of gravitation. Since Murree is at a higher altitude than Hyderabad, the value of g is comparatively lesser due to the inverse relationship between g and the distance from the center of the earth Rearth. As you move to higher altitudes, the distance Rearth from the center of the Earth increases, leading to a decrease in gravitational acceleration. This means that gravitational acceleration is slightly weaker at higher altitudes. The effect is more pronounced at greater distances from the Earth's surface, such as in space. However, for most practical purposes on or near the Earth's surface, ( g ) is often considered approximately constant at 9.8 m/s2.

77. At the center of the earth, the value of "g" becomes_:

• A. Greater
• B. Decreases
• C. Increases
• D. Zero

Explanation: At the center of the Earth, the gravitational acceleration ( g ) theoretically becomes zero. This is because, at the center of a spherically symmetric mass distribution, the gravitational forces from all directions cancel each other out. As you move towards the center, the mass above you pulls you in one direction, but the mass in the opposite direction also pulls you. At the center, these forces balance, resulting in a net gravitational force of zero. The formula for gravitational acceleration near the surface of a massive sphere (like Earth) is given by: g= GMearth/R2earth where: - g is the gravitational acceleration, - G is the gravitational constant 6.67430 x 10-11 Nm2/Kg2 - Mearth is the mass of the Earth, and - Rearth is the radius of the Earth. As you approach the center Rearth=0 g theoretically approaches zero. However, in practice, the Earth is not a perfect sphere, and there are other factors (such as variations in density and composition) that can affect the gravitational field. The above explanation assumes a simplified model of a spherically symmetric and homogeneous Earth.

78. In reference frame if a = 0, then weight of the body is equal to the force of_:

• A. Mass
• B. Weight
• C. Gravity
• D. None of these

Explanation: In a reference frame where acceleration (a) is zero, an object is either at rest or moving at a constant velocity. In this situation, the net force acting on the object is also zero, according to Newton's first law of motion. If the net force is zero, then the weight of the body (which is the force due to gravity acting on the object) is equal to the force of tension (if the object is hanging or connected by a string) or any other supporting force that counteracts the gravitational force. Mathematically, this is expressed as: Weight = Tension or, more generally: Weight =Supporting Force In the absence of acceleration, the forces are balanced, and the object is in a state of equilibrium. This concept is consistent with Newton's first law, which states that an object at rest will stay at rest, and an object in motion will stay in motion with a constant velocity unless acted upon by a net external force.

79. When a satellite moves in a circular orbit around the earth, the necessary _force is provided by the force of gravity:

• A. Centripetal
• B. Artificial
• C. Mass
• D. None of these

Explanation: When a satellite moves in a circular orbit around the Earth, the necessary centripetal force is provided by the force of gravity. The gravitational force between the satellite and the Earth acts as the centripetal force required to keep the satellite in its circular path. The centripetal force is directed toward the center of the circular orbit and is necessary to counteract the tendency of the satellite to move in a straight line tangentially to its orbit. In the case of a satellite orbiting the Earth, gravity is what provides this centripetal force, keeping the satellite in a stable orbit. The mathematical relationship between the centripetal force Fc, gravitational force Fg, and other parameters can be expressed using the following equation: Fc= GMsatelliteMearth/R2 where: - Fc is the centripetal force, - G is the gravitational constant, - Msatellite is the mass of the satellite, - Mearth is the mass of the Earth, - r is the radius of the satellite's orbit.

80. If the mass of the earth becomes double, then value of "G" _:

• A. Also doubles
• B. Remains same
• C. Four times
• D. Two times

Explanation: The value of the gravitational constant (G) is not affected by changes in the mass of the Earth or any other celestial body. The gravitational constant is a fundamental constant of nature and is considered to be universal and constant throughout the universe. If the mass of the Earth becomes double, it will affect the force of gravity between two objects on Earth, but (G) remains constant. In other words, (G) is a fundamental constant that doesn't change based on the masses involved in the gravitational interaction.