Question

The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is $T$, the time period on the planet is

• A. $\sqrt{2T}$
• B. $\frac{1}{\sqrt{2T}}$
• C. $\sqrt{3T}$
• D. $\frac{1}{\sqrt{3T}}$

Answer 1

Answer: (A)

$\sqrt{2T}$

Time period is given by-

$T=2\pi \sqrt{\frac{R^2\times l}{GM}}$
$T_{planet}=2\pi \sqrt{\frac{\left(2R^2\right)\times l}{G\times \left(2M\right)}}$
$=\sqrt{2}\times 2\pi \sqrt{\frac{R^{2}l}{GM}}$
$=\sqrt{2}\times T$