If $$R$$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $$g = \pi^2$$ m s$$^{-2}$$, then the length of the second's pendulum at a height $$h = 2R$$ from the surface of earth will be:

Find the length of a seconds pendulum at height $$h = 2R$$ where $$g = \pi^2$$ m/s$$^2$$.

At height $$h=2R$$ above Earth’s surface the acceleration due to gravity becomes $$g' = g\left(\frac{R}{R+h}\right)^2 = \pi^2 \left(\frac{R}{R+2R}\right)^2 = \pi^2 \times \frac{1}{9} = \frac{\pi^2}{9}$$ m/s$$^2$$.

A seconds pendulum has a time period $$T = 2$$ s and its period satisfies $$T = 2\pi\sqrt{\frac{l}{g'}}$$. Substituting $$T=2$$ and $$g'=\pi^2/9$$ gives $$2 = 2\pi\sqrt{\frac{l}{\pi^2/9}}$$. Inside the square root one finds $$\sqrt{\frac{l}{\pi^2/9}} = \sqrt{\frac{9l}{\pi^2}} = \frac{3\sqrt{l}}{\pi}$$. Hence $$2 = 2\pi\times\frac{3\sqrt{l}}{\pi} = 6\sqrt{l}$$, which leads to $$6\sqrt{l} = 2$$, so $$\sqrt{l} = \frac{2}{6} = \frac{1}{3}$$ and therefore $$l = \frac{1}{9}$$ m.

The correct answer is Option B: $$\frac{1}{9}$$ m.