If the earth suddenly shrinks (without changing mass) to half of its present radius,the acceleration due to gravity will be

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

$Assertion$ : $A$ tennis ball bounces higher on hills than in plains.
$Reason$ : Acceleration due to gravity on the hill is greater than that on the surface of earth.

Two planets $A$ and $B$ have densities $\varrho_1$ and $\varrho_2$ and have radii $r_1$ and $r_2$ respectively. The ratio of acceleration due to gravity on $A$ to that of $B$ is:

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = R/3$ below the surface of the Earth,what is the new frequency of oscillation? ($R$ is the radius of the Earth)

The magnitudes of the gravitational force at distances $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then