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A mass falls from a helght $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of earth it is found that $t =2 T$. The entre setup is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as $t'$ and $T'$.
$\mathrm{t}^{\prime}=\sqrt{2} \mathrm{T}^{\prime}$
$\mathrm{t}^{\prime} > 2 \mathrm{T}^{\prime}$
$\mathrm{t}^{\prime} < 2 \mathrm{T}^{\prime}$
$\mathrm{t}^{\prime}=2 \mathrm{T}^{\prime}$
Solution
Time of flight $=\sqrt{\frac{2 h}{g}} \propto \frac{1}{\sqrt{g}}$
Time perlod of pendulum $=2 \pi \sqrt{\frac{l}{g}} \propto \frac{1}{\sqrt{g}}$
Ratio of time of fillght and time period of pendulum is independent of $g$. Hence $t'=2 T'$
