What is the change in mass of a body,when taken $64 \ km$ below the surface of the earth? $[$Take radius of the earth as $6400 \ km]$

The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

Consider a planet in some solar system which has a mass double the mass of Earth and density equal to the average density of Earth. If the weight of an object on Earth is $W$,then the weight of the same object on that planet will be

Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$,the error in weighing is

$A$ planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing $W$ on earth will weigh on that planet:

If the mass of the Earth decreases by $25 \%$ and its radius increases by $50 \%$,then the acceleration due to gravity at its surface decreases by nearly ......... $\%$