question_answer

    A sphere of mass 100 g is attached to an inextensible string of length 1.3m whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 50 cm. The time period of the revolution is

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,                 \[T=\frac{2\pi }{\omega }\]                 \[=2\pi \sqrt{\frac{r}{g\tan \theta }}\]                 \[=2\times 3.14\sqrt{\frac{50}{980\times \frac{50}{130}}}\]                 \[=6.28\sqrt{\frac{50\times 130}{980\times 50}}\simeq 2.3\,s\]