The length of a second's pendulum on the surface of earth is $1\, m$. What will be the length of a second's pendulum on the moon ?

The time period of a simple pendulum

$\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$ $\therefore \mathrm{T} \propto \sqrt{\frac{l}{g}} 2 \pi \text { is constant }$ $\therefore \frac{\mathrm{T}_{m}}{\mathrm{~T}_{e}}$ $=\sqrt{\frac{l_{m}}{g_{m}} \times \frac{g_{e}}{l_{e}}}$

where $\mathrm{T}_{e}, \mathrm{~T}_{m}$ are the periodic time on earth and moon Here, $\mathrm{T}_{e}=\mathrm{T}_{m}=2 \mathrm{~s}$

$l_{m}, l_{e}$ are the length of second's pendulum on the surface of moon and earth respectively. Now, $g_{m}, g_{e}$ are the acceleration due to gravity on moon and earth respectively.

$\therefore \frac{2}{2}=\sqrt{\frac{g_{e}}{g_{m}} \times \frac{l_{m}}{l_{e}}}$ $\therefore$ by squaring, $$ 1=\frac{g_{e}}{g_{\mathrm{m}}} \times \frac{l_{\mathrm{m}}}{l_{e}} $$ but $g_{m}=\frac{g_{e}}{6}$ and $l_{e}=1 \mathrm{~m}$ $\therefore 1=\frac{g_{e}}{g_{e}} \times \frac{l_{m}}{1}$ $\therefore 1=6 \times l_{m}$ $\therefore l_{m}=\frac{1}{6} \mathrm{~m}$

but $g_{m}=\frac{g_{e}}{6}$ and $l_{e}=1 \mathrm{~m}$