T is the time period of a simple pendulum on | vedclass

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,where $g$ is the acceleration due to gravity at the surface.At a

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$2$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,where $g$ is the acceleration due to gravity at the surface.At a height $h = R$ above the surface,the acceleration due to gravity $g'$ is given by $g' = \frac{g}{(1 + h/R)^2}$.Substituting $h = R$,we get $g' = \frac{g}{(1 + R/R)^2} = \frac{g}{(2)^2} = \frac{g}{4}$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{\ell}{g'}} = 2\pi \sqrt{\frac{\ell}{g/4}} = 2 	imes 2\pi \sqrt{\frac{\ell}{g}} = 2T$.Given that the new time period is $xT$,we have $xT = 2T$,which implies $x = 2$.

$T$ is the time period of a simple pendulum 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. Then,the value of $x$ will be:

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