A) 25 N
B) 50 N
C) 78.5 N
D) 157 N
Correct Answer: D
Solution :
[d] \[Here\,\alpha =2revolutions/{{s}^{2}}=4\pi \,rad/{{s}^{2}}\](given) \[{{I}_{cylinder}}=\frac{1}{2}M{{R}^{2}}=\frac{1}{2}(50){{(0.5)}^{2}}\] =\[\frac{25}{4}Kg-{{m}^{2}}\] As \[\tau =I\alpha \,so\,TR=I\,\alpha \] \[\Rightarrow T=\frac{I\alpha }{R}=\frac{\left( \frac{25}{4} \right)(4\pi )}{(0.5)}N=50\pi N=157N\]
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