The period of oscillation of a second's pendulum on the planet whose mass and radius are twice that of earth…
The period of oscillation of a second's pendulum on the planet whose mass and radius are twice that of earth will
$\sqrt{2} \mathrm{~s}$
$\frac{1}{\sqrt{2}} \mathrm{~s}$
$\frac{1}{2} \mathrm{~s}$
$2 \sqrt{2} \mathrm{~s}$
Solution
Gravitational acceleration on the surface of planet of mass $M$ and radius $R$ is given by,
$g=\frac{G M}{R^2}$
Gravitational acceleration on the surface of that planet with twice mass and radius will be,
$g_p=\frac{G(2 M)}{(2 R)^2}=\frac{g_e}{2}$
The period of oscillation of a second's pendulum on earth,
$T_e=2 \pi \sqrt{\frac{l}{g_e}}=2 \mathrm{~s}$
The period of oscillation of a second's pendulum on planet,
$T_p=2 \pi \sqrt{\frac{l}{g_p}}=2 \pi \sqrt{\frac{2 l}{g_e}}=2 \sqrt{2} \mathrm{~s}$
.