If the radius of the earth contracts by $2\%$ and its mass remains the same,then the weight of a body at the earth's surface:

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is:

The depth at which acceleration due to gravity becomes $\frac{g}{2n}$ is ($R=$ radius of earth,$g=$ acceleration due to gravity on earth's surface,$n$ is an integer).

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

If the radius of the earth were to shrink by $1\%$ while its mass remains the same,the acceleration due to gravity on the earth's surface would