The weight of a body of mass $m$ decreases by $1\%$ when it is raised to a height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,the change in its weight is:

The height at which the weight of the body becomes $\frac{1}{16}$ of its weight on the surface of the earth of radius $R$ is: (in $R$)

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating,the duration of the day would be approximately ........ minutes
(Take: $g = 10 \, m/s^2$,the radius of earth,$R = 6400 \times 10^3 \, m$,Take $\pi = 3.14$)

The depth $d$ below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth).

$A$ uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_o$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_o}{4}$ is,

If the acceleration due to gravity at the Earth is $g$,the mass of the Earth is $80$ times that of the Moon,and the radius of the Earth is $4$ times that of the Moon,then the value of the acceleration due to gravity at the surface of the Moon will be: