A) \[\frac{2\pi }{13}N-m\]
B) \[\frac{\pi }{14}N-m\]
C) \[\frac{\pi }{15}N-m\]
D) \[\frac{\pi }{20}N-m\]
Correct Answer: C
Solution :
Given, \[l=2\,kg-{{m}^{2}},{{\omega }_{0}}=\frac{60}{60}\times 2\pi \,rad/s\] \[\omega =0\]and \[t=60\,s\] The torque required to stop the wheels rotation is \[\tau =l\alpha =l\left( \frac{{{\omega }_{0}}-\omega }{t} \right)\] \[\tau =\frac{2\times 2\times 60}{60\times 60}=\frac{\pi }{15}\,N-m\]
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