Question

The length of a seconds pendulum at a height $h =2 R$ from earth surface will be :
(Given : $R =$ Radius of earth and acceleration due to gravity at the surface of earth $\left.g =\pi^2 ms ^{-2}\right)$

• A. $\frac{2}{9} m$
• B. $\frac{4}{9} m$
• C. $\frac{8}{9} m$
• D. $\frac{1}{9} m$

Answer 1

Answer: (D)

$ T =2 \pi \sqrt{\frac{ L }{ g }}, g ^{\prime}=\frac{ GM }{9 R ^2}=\frac{ g }{9}=\frac{\pi^2}{9} $
$ 2=2 \pi \sqrt{\frac{ L }{\pi^2} \times 9} $
$ \Rightarrow 1=\pi \sqrt{ L } \times \frac{3}{\pi} \Rightarrow L =\frac{1}{9} m $