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 \(\mathrm{g}=\pi^2 \mathrm{~ms}^{-2}\)]
Pick an option to check your answer.
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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