Oscillations

JEE Main level, numerical answer, medium

111 of 194

Question 111 of 194

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 4m4m, then the time period of small oscillations will be _________ss.
[[take g=π2ms−2g={\pi }^{2}m{s}^{-2}]

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Answer: 8

Acceleration due to gravity at surface of earth, g=GMR2g=\dfrac{GM}{{R}^{2}}.
Now, at distance RR from the surface of the earth, g′=GM(R+R)2=g4{g}^{'}=\dfrac{GM}{{\left(R+R\right)}^{2}}=\dfrac{g}{4}
Therefore,
T=2πlg4T=2\pi \sqrt{\dfrac{l}{\dfrac{g}{4}}}
⇒T=2π4×4g\Rightarrow T=2\pi \sqrt{\dfrac{4\times 4}{g}}
⇒T=2π4π=8s\Rightarrow T=2\pi \dfrac{4}{\pi }=8s

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