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The value of g at a certain height above the surface of the earth is 16% of its value on the surface. The height is R=6300 km,

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Important Questions on Gravitation

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Assuming the earth as a sphere of uniform density, the acceleration due to gravity half way towards the centre of the earth will be,
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If the change in the value of g at a depth d below the surface of the earth is equal to that on the surface at a latitude of angle ϕ, then,
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If the earth of radius R, while rotating with angular velocity ω, becomes stand still, what will be the effect on the weight of a body of mass m at a latitude of 45°?
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The mass of the moon is 18 of the mass of the earth but the gravitational pull is 16 of that of the earth. It is due to the fact that
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A body weighs 63 N on the surface of the Earth. At a height h above the surface of Earth, its weight is 28 N while at a depth h below the surface of the Earth, the weight is 31.5 N. The value of h is
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If the earth were assumed to have uniform density and spherical symmetry, then the value of g in m s-2 halfway towards the centre of earth would be g=10 m s-2
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The radius of the Earth shrinks by 1%, its mass remaining the same. The percentage change in the value of g is
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If the value of the gravitational acceleration at the height h be 1% of its value at the surface of the earth, then h is equal to (given Re=6400 km)