The length of the second pendulum on the surface of earth is $1\, m$. The length of seconds pendulum on the surface of moon, where g is 1/6th value of $g$ on the surface of earth, is
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(a) $T = 2\pi \sqrt {\frac{l}{g}} $
==> $\sqrt {\frac{l}{g}} $= constant
$ \Rightarrow l \propto g;$
==> $\frac{{{l_m}}}{1} = \frac{1}{6}\frac{g}{g}$
$\Rightarrow {l_m} = \frac{1}{6}\,m$
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