Question

A seconds pendulum is moved to moon where acceleration due to gravity is $1/6$ times that of the earth, the length of the second pendulum on moon w.r.t length on earth would be:

• A. $6\text{times}$
• B. $12\text{times}$
• C. $\frac{1}{6}\text{times}$
• D. $\frac{1}{12}\text{times}$

Answer 1

The correct option is C $\frac{1}{6}\text{times}$

For second's pendulum at the surface of earth
$\Rightarrow T=2\pi \sqrt{\frac{l_e}{g_e}}...\left(i\right)$
For second's pendulum at the surface of moon
$\Rightarrow T=2\pi \sqrt{\frac{l_m}{g_m}}...\left(ii\right)$
The time period of second's pendulum on earth and moon will be equal i.e $T=2\text{s}$
On equating Eq. $\left(i\right)$ and $\left(ii\right)$,
$\sqrt{\frac{l_e}{g_e}}=\sqrt{\frac{l_m}{g_m}}\Rightarrow l_m=\left(\frac{g_m}{g_e}\right)l_e$
Since, $g_m=\frac{g_e}{6}\Rightarrow l_m=\frac{l_e}{6}$
Hence length of pendulum on moon will be $\frac{1}{6}\text{times}$ length of pendulum on earth.