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A pendulum is oscillating with frequency ' $n$ ' on the surface of the earth. It is taken to a depth $\frac{R}{2}$ below the surface of earth. New frequency of oscillation at depth $\frac{\mathrm{R}}{2}$ is
[ $\mathrm{R}$ is the radius of earth]
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[ $\mathrm{R}$ is the radius of earth]
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Verified Answer
The correct answer is:
$\frac{\mathrm{n}}{\sqrt{2}}$
Frequency of a simple pendulum is given by
$$
\begin{aligned}
& \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \\
& \therefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \\
& \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2} \\
& \therefore \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \\
& \therefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \\
& \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2} \\
& \therefore \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}}
\end{aligned}
$$
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