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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
  • Hard
  • Numerical

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is \(4 \mathrm{~m}\), then the time period of small oscillations will be _______ \(s\). [take \(\mathrm{g}=\pi^2 \mathrm{~ms}^{-2}\)]

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Answer & Solution

Correct Answer

(D) \(8\)

Step-by-step Solution

Detailed explanation

Acceleration due to gravity \(\mathrm{g}^{\prime}=\frac{\mathrm{g}}{4}\)

\(T=2 \pi \sqrt{\frac{4 \ell}{g}}\)

\(T=2 \pi \sqrt{\frac{4 \times 4}{g}}\)

 \(T=2 \pi \frac{4}{\pi}=8 s\)

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