1. Time period of a simple pendulum

   For small oscillations, $T = 2\pi \sqrt{\frac{l}{g}}$ where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

2. Gravity at height $h$ above Earth

   At a height $h$ above the Earth's surface, $g_h = g\left(\frac{R}{R+h}\right)^2$ where $R$ is the Earth's radius.

3. Given height

   Here, $h = R$.

   So, $g_h = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{R}{2R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$

4. New time period at height $R$

   Let the new time period be $T'$.

   $T' = 2\pi \sqrt{\frac{l}{g_h}} = 2\pi \sqrt{\frac{l}{g/4}}$

   $T' = 2\pi \sqrt{\frac{4l}{g}} = 2\,\left(2\pi \sqrt{\frac{l}{g}}\right) = 2T$

5. Compare with given form

   Since the new time period becomes $xT$, $xT = 2T \implies x = 2$

6. Option check

   • A: $4$ ❌
   • B: $\frac{1}{2}$ ❌
   • C: $2$ ✅
   • D: $\frac{1}{4}$ ❌

Therefore, the correct answer is Option C.