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Q. A clock S is based on oscillation of a spring and a clock P 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

Rajasthan PMTRajasthan PMT 2010

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.