Question

The mass and radius of a planet are double that of the earth. The time period of a pendulum on that planet which is a seconds pendulum on earth, will be

• A. $\sqrt{\frac{1}{\sqrt{2}}}s$
• B. 0.5 s
• C. $2\sqrt{2}s$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

$g=\frac{GM}{R^2}$
If, M and R are double, $g^{\prime }=\frac{G.2M}{4R^2}=g/2$
Thus $T=2\pi \sqrt{L/g^{\prime }}=2\pi \sqrt{2L/g}=\sqrt{2}seconds$