$\begin{matrix} A simple pendulum has a time period T_{1} when on the earth's surface and T_{2} when taken to height R above the earth's surface, where R is the radius of the earth. The value of \frac{T_{2}}{T_{1}} is \end{matrix}$

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\sqrt{2}

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Solution

The correct option is A 2
$\begin{matrix} T_{1} = 2 π \sqrt{\frac{l}{g}} g^{'} = \frac{G M}{{(R + R)}^{2}} = \frac{G M}{4 R^{2}} = \frac{g}{4} T_{2} = 2 π \sqrt{\frac{4 l}{g}} = 2 T_{1} \Rightarrow \frac{T_{2}}{T_{1}} = 2 \end{matrix}$
Hence,
option $(A)$ is correct answer.

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