Question

A pendulum clock that keeps correct time on the earth is taken to the moon. It will run

• A. At correct rate
• B. 6 times faster
• C. $\sqrt{6}$ times faster
• D. $\sqrt{6}$ times slower

Answer 1

The correct option is D

$\sqrt{6}$ times slower

The length of pendulum will be same but g on moon is$\frac{1}{6}$th of g on earth.

$g^{\prime }=\frac{g}{6}$

so$T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}$

$=2\pi \sqrt{\frac{1}{\left(\frac{g}{6}\right)}}$

$T^{\prime }=\sqrt{6}T$

Time period or the time taken by pendulum to complete one full to and fro motion has increased i.e., the pendulum is rolling slower as its taking more time.