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].

$A$ certain planet completes one rotation about its axis in time $T$. The weight of an object placed at the equator on the planet's surface is a fraction $f$ ($f$ is close to unity) of its weight recorded at a latitude of $60^{\circ}$. The density of the planet (assumed to be a uniform perfect sphere) is given by

Match Column-$I$ with Column-$II$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

The radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

The figure shows the variation of acceleration due to gravity with distance from the center of a uniform spherical planet of radius $R$. What is $r_2 - r_1$?

Select the correct statement out of the following: