The weight of an object in a coal mine,at sea level,and at the top of a mountain are $W_1$,$W_2$,and $W_3$ respectively. Then:

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

$A$ body weighs $49 \, N$ on a spring balance at the North Pole. What will be its weight recorded on the same weighing machine if it is shifted to the equator (in $, N$)? (Use $g = \frac{GM}{R^2} = 9.8 \, m/s^2$ and radius of the Earth,$R = 6400 \, km$.)

During the motion of a man from the equator to the pole of the Earth,his weight will ....... $\%$ (neglect the effect of change in the radius of the Earth).

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

Two planets have radii $r_1$ and $r_2$ and their densities are $p_1$ and $p_2$ respectively. The ratio of the acceleration due to gravity on their surfaces will be: