Solution:
Since \(g \propto \rho R\), on a planet with twice the radius and same density, \(g' = 2g\). Pendulum time period decreases (\(T_p = 2\pi\sqrt{l/g}\)), so clock P ticks faster. Spring clock is unaffected.
Since \(g \propto \rho R\), on a planet with twice the radius and same density, \(g' = 2g\). Pendulum time period decreases (\(T_p = 2\pi\sqrt{l/g}\)), so clock P ticks faster. Spring clock is unaffected.
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