A) \[12.5\]
B) \[40\]
C) \[32.6\]
D) \[15.6\]
Correct Answer: A
Solution :
\[0={{\omega }_{0}}-\alpha t\] \[\alpha =\frac{{{\omega }_{0}}}{t}=\frac{(100\times 2\pi )/60}{15}=0.7\,\,rad/s\] Now, angle rotated before coming to rest \[\theta =\frac{\omega _{0}^{2}}{2\alpha }\] \[\theta =\frac{{{\left( \frac{100\times 2\pi }{60} \right)}^{2}}}{2\times 0.7}=78.33\,\,rad\] Numbers of rotations \[\eta =\frac{\theta }{2\pi }=12.5\]
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