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(B) The time period of a spring-based clock is given by $T_S = 2 \pi \sqrt{\frac{m}{k}}$,which is independent of the acceleration due to gravity $g$.T

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$P$ will run faster than $S$

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(B) The time period of a spring-based clock is given by $T_S = 2 \pi \sqrt{\frac{m}{k}}$,which is independent of the acceleration due to gravity $g$.The time period of a pendulum-based clock is given by $T_P = 2 \pi \sqrt{\frac{l}{g}}$,which depends on $g$.Let the density of Earth be $\rho$ and its radius be $R$. The acceleration due to gravity on Earth is $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \rho \cdot \frac{4}{3} \pi R^3 = \frac{4}{3} \pi G \rho R$.For the new planet,the density $\rho' = \rho$ and the radius $R' = 2R$. The new acceleration due to gravity is $g' = \frac{4}{3} \pi G \rho (2R) = 2g$.Since $T_S$ is independent of $g$,the rate of clock $S$ remains unchanged.For clock $P$,the new time period is $T_P' = 2 \pi \sqrt{\frac{l}{g'}} = 2 \pi \sqrt{\frac{l}{2g}} = \frac{T_P}{\sqrt{2}}$.Since $T_P' < T_P$,the pendulum clock $P$ will complete its oscillations in less time,meaning it will run faster than clock $S$.

$A$ clock $S$ is based on the oscillations 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,then:

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