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Q.
$T$ is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:
At surface of earth time period
$T=2 \pi \sqrt{\frac{\ell}{ g }}$
At height $h = R$
$ g ^{\prime}=\frac{ g }{\left(1+\frac{ h }{ R }\right)^2}=\frac{ g }{4} $
$\therefore xT =2 \pi \sqrt{\frac{\ell}{( g / 4)}} $
$\Rightarrow xT =2 \times 2 \pi \sqrt{\frac{\ell}{ g }}$
$ \Rightarrow xT =2 T \Rightarrow x =2$