A) 80
B) 100
C) 120
D) 150
Correct Answer: D
Solution :
As \[{{\omega }_{2}}={{\omega }_{1}}+\alpha t\] \[\therefore \] \[40\pi =20\pi +\alpha \times 10\] or \[\alpha =2\pi \text{ }rad/{{s}^{2}}\] From \[\omega _{2}^{2}-\omega _{1}^{2}=2\alpha \theta \] \[{{(40\pi )}^{2}}-{{(20\pi )}^{2}}=2\times 2\pi \theta \] \[\therefore \] \[\theta =\frac{1200{{\pi }^{2}}}{4\pi }=300\pi \] Number of rotations completed \[=\frac{\theta }{2\pi }\] \[=\frac{300\pi }{2\pi }\] \[=150\]
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