The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].

If the density of the earth is doubled keeping the radius constant,find the new acceleration due to gravity (in $m/s^2$)? $(g = 9.8 \ m/s^2)$

$A$ point mass of $10 \,kg$ is placed at the centre of the Earth. The weight of the point mass is

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):

$A$ body weighs $500 \, N$ on the surface of the earth. How much would it weigh halfway below the surface of the earth (in $, N$)?

Given below are two statements:
Statement $I:$ Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.
Statement $II:$ Acceleration due to earth's gravity is same at a height '$h$' and depth '$d$' from earth's surface,if $h = d$.
In the light of above statements,choose the most appropriate answer from the options given below: