The time taken by a simple pendulum for one o | vedclass

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.On the surface of the earth, $g = \frac{GM}{R^2}$, so $T = 2\pi \sq

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$2$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.On the surface of the earth, $g = \frac{GM}{R^2}$, so $T = 2\pi \sqrt{\frac{L}{GM/R^2}} = 2\pi \sqrt{\frac{LR^2}{GM}}$.At a height $h = R$ above the surface, the effective acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} ight)^2$.Substituting $h = R$, we get $g' = g \left( \frac{R}{R+R} ight)^2 = g \left( \frac{R}{2R} ight)^2 = \frac{g}{4}$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2 	imes 2\pi \sqrt{\frac{L}{g}} = 2T$.Comparing $T' = 2T$ with $T' = xT$, we get $x = 2$.

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

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