JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2010

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


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