Question

The length of a seconds pendulum on the surface of earth is $1m$. Its length on the surface of the moon is

• A. $\frac{1}{6}m$
• B. $1m$
• C. $\frac{1}{36}m$
• D. $36m$

Answer 1

Answer: (A)

Time period of seconds pendulum is $2s$
$\therefore 2=2\pi \sqrt{\frac{L}{g}}$
On earth, $L=1m$, $g=g_e$
$\therefore 2=2\pi \sqrt{\frac{1}{g_e}}$ $\dots \left(i\right)$
On moon, $L=L^{\prime },g=\frac{g_e}{6}$
$\therefore 2=2\pi \sqrt{\frac{L^{\prime }}{g_e/6}}=2\pi \sqrt{\frac{6L^{\prime }}{ge}}\dots \left(ii\right)$
From(i) and (ii), we get
$6L^{\prime }=1$ or $L^{\prime }=\frac{1}{6}m$