The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{1}{6}$ th as much as at present is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

The mass of the earth is $81$ times that of the moon and the radius of the earth is $3.5$ times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to

$A$ planet of radius $R = \frac{1}{10} \times$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $\lambda = 10^{-3} \ kg \ m^{-1}$ into it. If the wire is not touching anywhere,the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $= 6 \times 10^6 \ m$ and the acceleration due to gravity on Earth $g_e = 10 \ m \ s^{-2}$) (in $N$)

The weights of an object are measured in a coal mine of depth $h_1$,then at sea level of height $h_2=0$,and lastly at the top of a mountain of height $h_3$ as $W_1, W_2$,and $W_3$ respectively. Which one of the following relations is correct? [$h_1 \ll R, h_3 \ll R, R=$ radius of the earth]