JEE Mains · Physics · STD 11 - 7. gravitation
A simple pendulum doing small oscillations at a place \(\mathrm{R}\) height above earth surface has time period of \(T_1=4 \mathrm{~s}\). \(T_2\) would be it's time period if it is brought to a point which is at a height \(2 R\) from earth surface. Choose the correct relation. \([R=\) radius of Earth]:
- A \(\mathrm{T}_1=\mathrm{T}_2\)
- B \(2 \mathrm{~T}_1=3 \mathrm{~T}_2\)
- C \(3 \mathrm{~T}_1=2 \mathrm{~T}_2\)
- D \(2 \mathrm{~T}_1=\mathrm{T}_2\)
Answer & Solution
Correct Answer
(C) \(3 \mathrm{~T}_1=2 \mathrm{~T}_2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_1=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(2 \mathrm{R})^2}\) \(\mathrm{~T}_2=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(3 \mathrm{R})^2}\) \(\therefore \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{2}{3}\)
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