A) \[4T\sqrt{\frac{1}{15}}\]
B) \[2T\sqrt{\frac{1}{10}}\]
C) \[4T\sqrt{\frac{1}{14}}\]
D) \[2T\sqrt{\frac{1}{14}}\]
Correct Answer: A
Solution :
For a simple pendulum \[T=2\pi \sqrt{\frac{L}{{{g}_{eff}}}}\] situation 1 : when pendulum is in air \[\to {{g}_{eff}}=g\] situation 2 : when pendulum is in liquid \[\to {{g}_{eff}}=g\left( 1-\frac{{{\rho }_{liquid}}}{{{\rho }_{body}}} \right)=g\left( 1-\frac{1}{16} \right)=\frac{15g}{16}\] So,\[\frac{T'}{T}=\frac{2\pi \sqrt{\frac{L}{15g/16}}}{2\pi \sqrt{\frac{L}{g}}}\Rightarrow T'=\frac{4T}{\sqrt{15}}\] Option [a]
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