A) S will run faster than P
B) P will run faster than S
C) they will both run at the same rate as on the earth
D) None of the above
Correct Answer: B
Solution :
\[g=\frac{4}{3}\pi \rho GR\]. If density is same than\[g\propto R\] According to problem,\[{{R}_{p}}=2{{R}_{e}}\] \[{{g}_{p}}=2{{g}_{e}}\] For dock \[P\] (based on pendulum motion) \[T=2\pi \sqrt{\frac{l}{g}}\] Time period decreases on planet so it will run faster because\[{{g}_{p}}>{{g}_{e}}\]. For clock\[S\] (based on oscillation of spring) \[T=2\pi \sqrt{\frac{m}{k}}\] So, it does not change.
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