MHT CET · Physics · Gravitation
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)
- A \(\sqrt{\frac{2}{3}} \mathrm{n}\)
- B \(\sqrt{\frac{3}{2}} \mathrm{n}\)
- C \(\sqrt{\frac{1}{3}} n\)
- D \(\sqrt{\frac{1}{2}} n\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{2}{3}} \mathrm{n}\)
Step-by-step Solution
Detailed explanation
\(g' = g \left(1 - \frac{d}{R}\right) = g \left(1 - \frac{R/3}{R}\right) = g \frac{2}{3}\) \(n = \frac{1}{2\pi} \sqrt{\frac{g}{L}}\)
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