A simple pendulum has a time period T₁ when on the earth's surface, and T₂ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T₂/T₁ is

The acceleration due to gravity at a height h above the surface of the earth is given by 

                    $g_2=g_1{\left(\frac{R}{R+h}\right)}^2$

where g₁ is the value at the surface of the earth. Now

                    $T_2=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{{\mathrm{g}}_2}} \mathrm{and} {\mathrm{T}}_1=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{{\mathrm{g}}_1}}$

$$\begin{gathered} \therefore \frac{T_2}{T_1}=\sqrt{\frac{g_1}{g_2}}=\frac{R+h}{R} \\ =\frac{R+R}{R}=2 \left(\because h=R \right) \end{gathered}$$

Hence the correct choice is (d).