Question

Radius of planet A is twice that of planet B and the density of A is one third that of B. The ratio of the
acceleration due to gravity at the surface of A to that at the surface of Bis
a) 2:3
6) 3:2
c) 3:4
d) 4:3​

Answer 1

We're given,

$\dfrac {R_A}{R_B}=2\quad;\quad\dfrac {\rho_A}{\rho_B}=\dfrac {1}{3}$

We know the expression for the acceleration due to gravity on the surface of a planet of mass M and radius R.

$g=\dfrac {GM}{R^2}$

But we know that density of the planet,

$\rho=\dfrac {M}{V}\quad\iff\quad M=\rho V$

Then,

$g=\dfrac {\rho VG}{R^2}$

Since the planet is spherical in shape, its volume will be,

$V=\dfrac {4}{3}\pi R^3$

Thus,

$g=\dfrac {\frac {4}{3}\pi\rho R^3G}{R^2}\\\\\\g=\dfrac {4}{3}\pi\rho RG\\\\\\\implies\ g\propto R\rho\\\\\\\therefore\ \dfrac {g_A}{g_B}=\dfrac {R_A\rho_A}{R_B\rho_B}=2\times\dfrac {1}{3}=\dfrac{2}{3}$

Hence the ratio is 2:3.

Thus (a) is the answer.