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If R is the radius of earth, then the length of seconds pendulum at a height h=2R above the surface of earth ,will be

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4/9 m

1/9m

2/9 m

8/9 m

answer is D.

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Detailed Solution

$T = 2 π \sqrt{\frac{l}{g}},$
But $\frac{g^{1}}{g} = {(\frac{R}{R + 2 R})}^{2} = \frac{1}{9}$
$\begin{array}{l} \Rightarrow 2 = 2 π \sqrt{\frac{l}{(\frac{g}{9})}} \\ l = \frac{1}{9} m \end{array}$

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