T is the time period of simple pendulum on the earth's surface. Its...

Time period of a simple pendulum at Earth's surface is given by

$T = 2\pi\sqrt{\dfrac{l}{g}}$.

When distance from centre of Earth, i.e., at $h = R$, acceleration due to gravity is $g ' = \dfrac{g}{{\left(1 + \dfrac{h}{R}\right)}^{2}} = \dfrac{g}{{\left(2\right)}^{2}} = \dfrac{g}{4}$

Now, time period is $xT = 2\pi\sqrt{\dfrac{l}{\left(\dfrac{g}{4}\right)}}$

$⇒ xT = 2 \times 2\pi\sqrt{\dfrac{l}{g}}$

$⇒ xT = 2T ⇒ x = 2$

Thus, $x = 2$.