The time taken by simple pendulum for one oscillation is T on earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. T

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The time taken by simple pendulum for one oscillation is T on earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. The value of x is

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The time period of a simple pendulum on the surface of the earth is given by $T = 2 π \frac{L}{g}$ .

The acceleration due to gravity at a height $h$ above the earth's surface is $g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}$ .

Given $h = R$ , we have $g^{'} = \frac{g}{( 1 + \frac{R}{R} ) ^{2}} = \frac{g}{4}$ .

The new time period $T^{'}$ at height $R$ is $T^{'} = 2 π \frac{L}{g ^{'}} = 2 π \frac{L}{\frac{g}{4}} = 2 \times 2 π \frac{L}{g} = 2 T$ .

Since $T^{'} = x T$ , we get $x = 2$ .

Answer: $2$

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