Question

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

[take $g_{Moon}=\frac{1}{6}{g}_{Earth}$ ]

• A. at correct rate
• B. 5 times faster
• C. $\sqrt{6}$ times faster
• D. $\sqrt{6}$ times slower

Answer 1

The correct option is D $\sqrt{6}$ times slower

Time period of pendulum clock is

$T=2\pi \sqrt{\frac{L}{g}}$

Time on earth=$T_1$

$T_1=2\pi \sqrt{\frac{L}{g_{earth}}}$

Time on moon=$T_2$

$T_2=2\pi \sqrt{\frac{L}{g_{moon}}}$

$T_2=2\pi \sqrt{\frac{L}{\frac{1}{6}{g}_{earth}}}$

$T_2=\sqrt{6}{T}_1$

Hence it will run $\sqrt{6}$ times slower