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JEE Mains · Physics · STD 11 - 7. gravitation
\(T\) 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:
- A \(4\)
- B \(2\)
- C \(\frac{1}{2}\)
- D \(\frac{1}{4}\)
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 \(h = R\) \(g ^{\prime}=\frac{ g }{\left(1+\frac{ h }{ R }\right)^2}=\frac{ g }{4}\) \(xT =2 \pi \sqrt{\frac{\ell}{( g / 4)}}\) \(\Rightarrow xT =2 \times 2 \pi \sqrt{\frac{\ell}{ g }}\)…
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