The period of seconds pendulum on a planet, whose mass and radius are three times that of earth, is

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Aicharya · Accepted Answer

$\begin{aligned} \frac{g^{\prime}}{g} &=\frac{M^{\prime}}{M} \cdot \frac{R^{2}}{R^{\prime 2}} \ &=3 \cdot \frac{1}{3^{2}}=\frac{1}{3} \ \frac{T^{\prime}}{T} &=\sqrt{\frac{g}{g^{\prime}}}=\sqrt{3} \ 	herefore T^{\prime} &=\sqrt{3} T=\sqrt{3} \cdot 2 	ext { second } \end{aligned}$

The period of seconds pendulum on a planet, whose mass and radius are three times that of earth, is

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