A pendulum is oscillating with frequency ' $n$ ' on the surface of earth. If it is taken to a depth $\frac{{ }^{\prime} \mathrm{R}^{\prime}}{3}$ below the surface of earth, new frequency of oscillation is ( $R=$ radius of earth)

The frequency of oscillation of a simple pendulum is given by
$n = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$, where $g$ is gravitational acceleration and $L$ is the pendulum length. At depth $d = \frac{R}{3}$ below Earth's surface, the reduced gravitational acceleration is
$g' = g\left(1 - \frac{d}{R}\right) = g\left(1 - \frac{1}{3}\right) = \frac{2}{3}g$.

The new frequency becomes
$n' = \frac{1}{2\pi}\sqrt{\frac{g'}{L}} = \frac{1}{2\pi}\sqrt{\frac{2g}{3L}} = \sqrt{\frac{2}{3}} \cdot \frac{1}{2\pi}\sqrt{\frac{g}{L}}$.

Since $\frac{1}{2\pi}\sqrt{\frac{g}{L}} = n$, it follows that
$n' = \sqrt{\frac{2}{3}}n$.

The result corresponds to option A.