Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d=\frac{R}{2}$ from the surface of earth, if its werght on the surface of earth is $200\,N$, will be $...........\,N$ ( $Given R =$ Radrus of earth)

The gravitational potential at a point above the surface of earth is $-5.12 \times 10^7 \mathrm{~J} / \mathrm{kg}$ and the acceleration due to gravity at that point is $6.4 \mathrm{~m} / \mathrm{s}^2$. Assume that the mean radius of earth to be $6400 \mathrm{~km}$. The height of this point above the earth's surface is :

A pendulum clock is set to give correct time at the sea level. This clock is moved to hill station at an altitude of $2500\, m$ above the sea level. In order to keep correct time of the hill station, the length of the pendulum

A body weighs $72 N$ on surface of the earth. When it is taken to a height of $h=2 R$, where $R$ is radius of earth, it would weigh ........ $N$

The radius of the earth is $6400\, km$ and $g = 10\,m/{\sec ^2}$. In order that a body of $5 \,kg$ weighs zero at the equator, the angular speed of the earth is

The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius)