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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2023
  • Medium
  • Numerical

\(T\) is the time period of simple pendulum on the earth's surface. Its time period becomes \(x T\) when taken to a height \(R\) (equal to earth's radius) above the earth's surface. Then, the value of \(x\) will be:

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Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

At surface of earth time period

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

At height \(h = R\)

\(g ^{\prime}=\frac{ g }{\left(1+\frac{ h }{ R }\right)^2}=\frac{ g }{4}\)

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

\(\Rightarrow xT =2 \times 2 \pi \sqrt{\frac{\ell}{ g }}\)

\(\Rightarrow xT =2 T \Rightarrow x =2\)

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