The mass and diameter of a planet are twice that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is a seconds pendulum on earth.

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$\sqrt{2} \sec$

2 sec

$\frac{1}{\sqrt{2}} \sec$

$2 \sqrt{2} \sec$

answer is D.

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Detailed Solution

$\begin{array}{l} g = \frac{GM}{R^{2}}; g_{p} = \frac{{GM}_{p}}{R_{p}^{2}}; \\ R_{P} = 2 R; M_{p} = 2 m \\ \frac{g_{p}}{g} = \frac{{GM}_{p}}{R_{p}^{2}} \times \frac{R^{2}}{GM} = \frac{2 M}{4 R^{2}} \times \frac{R^{2}}{M} \\ \frac{g_{p}}{g} = \frac{1}{2} g_{p} = g / 2 \end{array} \begin{array}{l} \frac{T}{T_{p}} = \sqrt{\frac{g_{p}}{g}} = \sqrt{\frac{g}{2} / g} = \frac{1}{\sqrt{2}} \\ \frac{2}{T_{p}} = \frac{1}{\sqrt{2}} \Rightarrow T_{p} = 2 \sqrt{2} \sec \end{array}$

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