A pendulum has time period $T$. If it is taken on to another planet having acceleration due to gravity half and mass $ 9 $ times that of the earth then its time period on the other planet will be
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(d) $T = 2\pi \sqrt {\frac{l}{g}}$
$\Rightarrow T \propto \frac{1}{{\sqrt g }}$
$ \Rightarrow \frac{{{T_P}}}{{{T_e}}} = \sqrt {\frac{{{g_e}}}{{{g_P}}}} = \sqrt {\frac{2}{1}}$
$\Rightarrow T' = \sqrt 2 T$
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