A) 5.0s
B) 4.5s
C) 3.2s
D) 2.3s
Correct Answer: D
Solution :
In equilibrium, \[T\cos \theta =mg\] ...(i) and \[T\sin \theta =\frac{m{{v}^{2}}}{T}\] or \[T\sin \theta =mr{{\omega }^{2}}\] ...(ii) From Eqs. (i) and (ii), \[\frac{r{{\omega }^{2}}}{g}=\tan \theta \] \[\omega =\sqrt{\frac{g\tan \theta }{r}}\] \[\therefore \]Time period, \[T=\frac{2\pi }{\omega }\] \[=2\pi \sqrt{\frac{r}{g\tan \theta }}\] \[=2\times 3.14\sqrt{\frac{50}{980\times \frac{50}{130}}}\] \[=6.28\sqrt{\frac{50\times 130}{980\times 50}}\simeq 2.3\,s\]You need to login to perform this action.
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