A) 2.2s
B) 1.1s
C) 0.5s
D) 3.1s
Correct Answer: B
Solution :
[b]: Let us depict the forces acting on the oscillating ball at an arbitrary angular position \[\theta \], relative to equilibrium position where \[{{F}_{B}}\]is the force of buoyancy. The equation of motion for ball \[-mgl\sin \theta +{{F}_{B}}l\sin \theta =m{{l}^{2}}\ddot{\theta }\] ...(i) Using \[m=\frac{4}{3}\pi {{r}^{3}}\sigma ,{{F}_{B}}=\frac{4}{3}\pi {{r}^{3}}\rho g\]and \[\theta \simeq \theta \]for small \[\theta \], in equation (i), we get\[\ddot{\theta }=-\frac{g}{l}\left( 1-\frac{\rho }{\sigma } \right)\theta \] Thus the time period of the ball \[T=2\pi \frac{1}{\sqrt{\frac{g}{l}\left( 1-\frac{\rho }{\sigma } \right)}}=2\pi \sqrt{\frac{l/g}{1-\frac{1}{3}}}\] \[=6.28\sqrt{\frac{0.2/9.8}{2/3}}=1.1s\]You need to login to perform this action.
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