Time period of simple pendulum on earth's surface is ' \(\mathrm{T}\) '. Its time period becomes ' \(\mathrm{xT}\) ' when taken to a height \(\mathrm{R}\) (equal to earth's radius) above the earth's surface. Then the value of ' \(x\) ' will be

\(
\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}
\)
At a height 'h' from earth's surface,
\(
\begin{array}{ll}
& \mathrm{xT}=2 \pi \sqrt{\frac{l}{g_{\mathrm{h}}}} \\
\therefore \quad & \mathrm{x}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}} \quad \ldots . .(\mathrm{i}) \\
& \text { Now, } \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2} \\
\therefore \quad & \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2} \\
\therefore \quad & \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{4} \\
\therefore \quad & \text { From equations (i) and (ii), } \\
& \mathrm{x}=\sqrt{\frac{\mathrm{g}}{\mathrm{g} / 4}}=\sqrt{4}=2
\end{array}
\)
\(\therefore \quad\) From equations (i) and (ii),
\(
x=\sqrt{\frac{g}{g / 4}}=\sqrt{4}=2
\)