A pendulum clock that keeps correct time on the earth is taken to the moon it will run (it is given that $g_{Moon} = g_{Earth}/6$ )
Medium
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(d) $T = 2\pi \sqrt {\frac{l}{g}} $==>$\frac{{{T_e}}}{{{T_m}}} = \sqrt {\frac{{{g_m}}}{{{g_e}}}} = \sqrt {\frac{{{g_e}/6}}{{{g_e}}}} = \frac{1}{{\sqrt 6 }}$
$ \Rightarrow $ ${T_m} = \sqrt 6 {T_e}$ i.e. clock becomes slower.
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