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

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Solution

The correct option is D
2

If acceleration due to gravity is g at the surface of earth then at height R it value becomes $g^{'} = g (\frac{R}{R + h})^{2} = \frac{g}{4}$
$T_{1} = 2 π \sqrt{\frac{l}{g}} a n d T_{2} = 2 π \sqrt{l} \frac{g}{4} ∴ \frac{T_{2}}{T_{1}} = 2$

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