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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
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  • MCQ

The length of a seconds pendulum at a height \(h=2 R\) from earth surface will be _______. (Given: \(R =\) Radius of earth and acceleration due to gravity at the surface of earth \(g =\pi^{2}\,m / s ^{-2}\) )

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Correct Answer

(D) \(\frac{1}{9}\,m\)

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Detailed explanation

\(T =2 \pi \sqrt{\frac{ L }{ g }}, \quad g ^{\prime}=\frac{ GM }{9 R ^{2}}=\frac{ g }{9}=\frac{\pi^{2}}{9}\)

\(2=2 \pi \sqrt{\frac{ L }{\pi^{2}} \times 9}\)

\(1=\pi \sqrt{ L } \times \frac{3}{\pi} \Rightarrow L =\frac{1}{9}\,m\)

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