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The value of acceleration due to gravity at the moon is g/6, where g is the value of acceleration due to gravity at earth. The value of frequency of oscillation of simple pendulum on moon as compared to earth will be


1. \(\sqrt6\) times
2. 6 times
3. 1/6 times
4. \(\sqrt\frac{1} {6}\) times

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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:

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

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

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

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

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