A) \[{{60}^{o}}\]
B) \[{{120}^{o}}\]
C) \[{{150}^{o}}\]
D) \[{{\tan }^{-1}}\left( -\frac{1}{2} \right)\]
Correct Answer: B
Solution :
Apparent acceleration due to gravity because of force of buoyancy \[g'=g\left( 1-\frac{{{P}_{l}}}{{{P}_{s}}} \right){{P}_{l}}=\frac{{{P}_{s}}}{10}\] \[\Rightarrow \] \[g'=g\left( 1-\frac{1}{10} \right)=\frac{9}{10}g\] As, we know that for simple pendulum, \[T=2\pi \sqrt{\frac{L}{g}}\] \[T\alpha \sqrt{\frac{L}{g}}\] \[\Rightarrow \] \[T'=\sqrt{\frac{10}{9}}\,T\]
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