Answer:
\[T=2\pi
\,\sqrt{l/g}\] on the surface of
moon,
\[T=2\pi
\,\sqrt{\frac{lm}{gm}}\,=\,2\pi \,\sqrt{\frac{6lm}{g}}\] \[\,(\because
\,{{g}_{m}}\,=\frac{g}{6})\]
\[\therefore
\] \[6\,l\,m=l\]
or \[lm\,=\,\frac{1}{6}=\,\frac{1m}{6}\,=0.17\,m\]
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