Q: The mass and diameter of a planet are twice those of earth . What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on the earth ?

(a) $\sqrt{2} sec$

(b) $2 \sqrt{2} sec$

(c) $\frac{1}{\sqrt{2}} sec$

(d) $\frac{1}{2 \sqrt{2}} sec$

**Solution :** T_{e} = 2 sec (Having seconds pendulum)

$\displaystyle T = 2 \pi \sqrt{\frac{l}{g}}$

$\displaystyle g = \frac{G M}{R^2} $

$\displaystyle T = 2 \pi \sqrt{\frac{l}{\frac{G M}{R^2} }}$

$\displaystyle T = 2 \pi \sqrt{\frac{l R^2}{G M}}$

$\displaystyle \frac{T_p}{T_e} = \sqrt{\frac{R_p^2}{R_e^2} \times \frac{M_e}{M_p}}$

$\displaystyle \frac{T_p}{T_e} = \sqrt{\frac{4 R_e^2}{R_e^2} \times \frac{M_e}{2 M_e}}$

$\displaystyle \frac{T_p}{T_e} = \sqrt{2} $

$\displaystyle T_p = \sqrt{2} T_e$

$\displaystyle T_p = \sqrt{2} \times 2$

$\displaystyle T_p = 2 \sqrt{2} sec $

Correct option is (b)