Question

Diameter and mass of a planet is double that earth. Then time period of a pendulum at surface of planet is how much times of time period at earth surface

• A. $\frac{1}{\sqrt{2}}$ times
• B. $\sqrt{2}$ times
• C. $equal$
• D. $noneofthese$

Answer 1

The correct option is B $\sqrt{2}$ times

Time period of simple pendulum. $t=2\pi \sqrt{\frac{l}{g}}$

For earth: $g=\frac{G{M}_E}{R}$

For planet: $M_P=2M_E{R}_P=2R$

Thus $g_p=\frac{G\left(2M_E\right)}{(2R{)}^2}⟹g_p=\frac{g}{2}$

Now, $\frac{t_p}{t_E}=\sqrt{\frac{g}{g/2}}$

$⟹t_p=\sqrt{2}{t}_E$