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The period of seconds pendulum on a planet, whose mass and radius are three times
that of earth, is
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that of earth, is
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The correct answer is:
$2 \sqrt{3}$ second
$\begin{aligned} \frac{g^{\prime}}{g} &=\frac{M^{\prime}}{M} \cdot \frac{R^{2}}{R^{\prime 2}} \\ &=3 \cdot \frac{1}{3^{2}}=\frac{1}{3} \\ \frac{T^{\prime}}{T} &=\sqrt{\frac{g}{g^{\prime}}}=\sqrt{3} \\ \therefore T^{\prime} &=\sqrt{3} T=\sqrt{3} \cdot 2 \text { second } \end{aligned}$
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