A) \[T\]
B) \[\left( \frac{9}{10} \right)T\]
C) \[\sqrt{\frac{10}{9}}T\]
D) \[\sqrt{\frac{9}{10}}T\]
Correct Answer: C
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, \[T=2\pi \sqrt{\frac{L}{g}}\] \[T\alpha \sqrt{\frac{L}{g}}\] \[\Rightarrow \] \[T'=\sqrt{\frac{10}{9}}\,T\]You need to login to perform this action.
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