Correct Answer - Option 4 :

\(\sqrt\frac{1} {6}\) times

__Concept:__

The** time period of a simple pendulum is**

\(T =2\pi \sqrt{\frac Lg}\)

The **frequency of oscillation (f)** is inversely proportional to the time period

\(f \propto \frac{1}{T}\)

where L = Length of the pendulum, g = acceleration due to gravity

__Calculation:__

__Given:__

The acceleration due to gravity at the moon is \(g'=\frac g6\)

The time period of the pendulum at the moon surface is:

\(T' =2\pi \sqrt{\frac L{g'}}=2\pi \sqrt {\frac {6L}g}\)

\(T'=\sqrt {6}T\)

\(f \propto \frac{1}{T}\)

\(f'=\sqrt\frac{1} {6}f\)