A simple pendulum has a periodic time ' \(\mathrm{T}_1\) ' when it is on the surface of earth of radius ' \(R\) '. Its periodic time is ' \(T_2\) ' when it is taken to a height ' \(R\) ' above the earth's surface. The value of \(\frac{T_2}{T_1}\) is

\(\mathrm{T}_1=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)
At a height ' \(h=2 R\) ' from earth's surface,
\(\mathrm{T}_2=2 \pi \sqrt{\frac{l}{\mathrm{~g}_{\mathrm{h}}}} \)
\( \therefore \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{g}}{\mathrm{~g}_{\mathrm{h}}}}...\text{(i)}\)
\( \text {Now, } \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2} \)
\( \therefore \mathrm{~g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2} \)
\( \therefore \mathrm{~g}_{\mathrm{h}}=\frac{\mathrm{g}}{4} \ldots (\because \mathrm{R}+\mathrm{h}=2 \mathrm{R}) \) \(\ldots\text {(ii)}\)
\(\therefore \quad\) From equations (i) and (ii),
\(\frac{T_2}{T_1}=\sqrt{\frac{g}{g / 4}}=2\)