A) \[t={{t}_{0}}\]
B) \[t={{t}_{0}}/2\]
C) \[t=2{{t}_{0}}\]
D) \[t=4{{t}_{0}}\]
Correct Answer: C
Solution :
\[\because \,\,{{t}_{o}}=2\,\pi \sqrt{\frac{l}{g}}\] Effective weight of bob inside water, \[{W}'=mg-\text{thrust}=V\rho g-V{\rho }'g\] \[\Rightarrow V\rho {{g}_{eff}}=V(\rho -{\rho }')g,\] where, \[\rho \]= Density of bob \[\Rightarrow {{g}_{eff}}=\left( 1-\frac{{{\rho }'}}{\rho } \right)\,g\] and \[{\rho }'\]= Density of water \[\therefore t=2\,\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\,\pi \sqrt{\frac{l}{(1-{\rho }'/\rho )g}}\]\[\therefore \frac{t}{{{t}_{0}}}=\sqrt{\frac{1}{1-{\rho }'/\rho }}=\sqrt{\frac{1}{1-\frac{3}{4}}}\] \[\begin{align} & (\because {\rho }'={{10}^{3}}kg/{{m}^{3}} \\ & \,\,\,\,\,\rho =\frac{4}{3}\times {{10}^{3}}kg/{{m}^{3}} \\ \end{align}\] \[\Rightarrow t=2\,{{t}_{0}}\].
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