Q104 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics

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]

  • A
    $\frac{\mathrm{n}}{3}$
  • Correct answer
    B
    $\frac{\mathrm{n}}{\sqrt{2}}$
  • C
    $2 \mathrm{n}$
  • D
    $\frac{\mathrm{n}}{2}$

Solution

Frequency of a simple pendulum is given by $$ \begin{aligned} & \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \\ & \therefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \\ & \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2} \\ & \therefore \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}} \end{aligned} $$

Hence, the correct answer is option: 'B'.

Relevant Questions