A clock S is based on the oscillation of a sp | vedclass

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(B) The acceleration due to gravity is given by $g = \frac{4}{3}\pi \rho GR$. Since the density $\rho$ is the same,$g \propto R$.Given that the radius

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$P$ will run faster than $S$

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(B) The acceleration due to gravity is given by $g = \frac{4}{3}\pi \rho GR$. Since the density $\rho$ is the same,$g \propto R$.Given that the radius of the planet is twice that of Earth $(R_p = 2R_e)$,the acceleration due to gravity on the planet is $g_p = 2g_e$.For clock $P$ (pendulum clock),the time period is $T_p = 2\pi \sqrt{\frac{l}{g}}$. Since $g_p > g_e$,the time period $T_p$ decreases,meaning the clock runs faster.For clock $S$ (spring clock),the time period is $T_s = 2\pi \sqrt{\frac{m}{k}}$. Since $T_s$ is independent of $g$,the time period remains unchanged.Therefore,clock $P$ will run faster than clock $S$.

$A$ clock $S$ is based on the oscillation of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,which of the following is true?

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