A pendulum is oscillating with frequency ' \(n\) ' on the surface of earth. If it is taken to a depth \(\frac{R}{4}\) below the surface of earth, new frequency of oscillation of depth \(\frac{R}{4}\) is ( \(R=\) radius of earth)

The frequency of the pendulum at the surface is given as \(\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{~g}}{l}}\)
At depth the formula for gravitational acceleration is \(\mathrm{g}_{\mathrm{eff}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
For \(\mathrm{d}=\frac{\mathrm{R}}{4}, \mathrm{~g}_{\text {eff }}=\mathrm{g}\left(1-\frac{1}{4}\right)=\frac{3}{4} g\)
The frequency at depth \(\mathrm{d}=\frac{\mathrm{R}}{4}\)
\(\mathrm{f}_{\mathrm{d}}=\frac{1}{2 \pi} \sqrt{\frac{\frac{3}{4} \mathrm{~g}}{l}}=\frac{1}{2 \pi} \sqrt{\frac{3 \mathrm{~g}}{4 l}}\)
Take the ratio of both frequencies
\(\begin{aligned}
& \frac{f_d}{f}=\frac{\sqrt{3}}{2} \\
& f_d=\frac{\sqrt{3}}{2} f=\frac{\sqrt{3} n}{2} \quad \ldots(\because f=n)
\end{aligned}\)