The length of a seconds pendulum at a height h = 2R from earth surface will be:
(Given : R = Radius of earth and acceleration due to gravity at the surface of earth $g = π^{2} m / s^{- 2}$ )

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$\frac{4}{9} m$

$\frac{1}{9} m$

$\frac{2}{9} m$

$\frac{8}{9} m$

answer is D.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{L}{g}}, g^{'} = \frac{GM}{9 R^{2}} = \frac{g}{9} = \frac{π^{2}}{9} \\ 2 = 2 π \sqrt{\frac{L}{π^{2}} \times 9} \\ \Rightarrow 1 = π \sqrt{L} \times \frac{3}{π} \Rightarrow L = \frac{1}{9} m \end{array}$

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