The variation of acceleration due to gravity $(g)$ with distance $(r)$ from the center of the earth is correctly represented by ... (Given $R =$ radius of earth)

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At a height of $h$ $km$ from the surface of the Earth,the gravitational potential and the value of $g$ are $-5.4 \times 10^7\, J kg^{-1}$ and $6.0\, m s^{-2}$ respectively. Take the radius of the Earth as $6400\, km$.

The height $h$ at which the weight of a body will be the same as that at the same depth $h$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

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The value of acceleration due to gravity at a depth of $ 1600 \,km $ is equal to: (Radius of Earth $ = 6400 \,km $) (in $\,ms^{-2}$)

The weight of a body on the earth is $400\,N$. Then weight of the body when taken to a depth half of the radius of the earth will be ............ $N$.

If a particle takes $t$ seconds less and acquires a velocity of $v \text{ m/s}$ more in falling through the same distance on two planets,where the accelerations due to gravity are $2g$ and $8g$ respectively,then:

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