A) \[\text{1}.\text{6}\times \text{1}{{0}^{\text{5}}}\text{ N}-\text{m}\]
B) \[\text{1}.\text{6}\times \text{1}{{0}^{-\text{4}}}\text{ N}-\text{m}\]
C) \[2.5\times \text{1}{{0}^{-3}}\text{ N}-\text{m}\]
D) \[2.5\times \text{1}{{0}^{-2}}\text{ N}-\text{m}\]
Correct Answer: D
Solution :
From equation of rotational motion, we have \[{{\omega }^{2}}=\omega _{0}^{2}=2\alpha \theta \] ... (i) where \[\omega \] is angular velocity, \[\alpha \] the angular acceleration and \[\theta \] the angular displacement. Also, \[\omega =2\pi f\] ... (ii) Where \[f\] is number of revolutions. From Eqs. (i) and (ii), we get For \[\omega =0\] (sphere stop), \[\theta =2\pi \times 2\pi =4{{\pi }^{2}}\], \[\therefore \] n = 300 rev/min \[=\frac{300}{60}=5\] rev/s \[\therefore \] \[0={{(2\pi \times 5)}^{2}}-2\alpha \,\,{{(2\pi )}^{2}}\] \[\Rightarrow \] \[\alpha =\frac{100{{\pi }^{2}}}{8{{\pi }^{2}}}=12.5\] rad/s Torque applied \[\tau =I\alpha \]where \[I\] is moment of inertia, a the angular acceleration. For a sphere, \[I=\frac{2}{5}m{{R}^{2}}\] \[\therefore \] \[\tau =\frac{2}{5}m{{R}^{2}}\alpha \] \[\Rightarrow \] \[\tau =\frac{2}{5}\times 2\times {{(5\times {{10}^{-2}})}^{2}}\times 12.5\] \[\Rightarrow \] \[\tau =2.5\times {{10}^{-2}}N-m\].
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