A) 0.02 Nm
B) 0.20 Nm
C) 0.10 Nm
D) 0.01 Nm
Correct Answer: C
Solution :
\[c.\] \[\therefore \]\[\alpha =\frac{10}{2}=5\,\text{rad/}{{\text{s}}^{\text{2}}}.\] \[I=M{{R}^{2}}=\frac{1}{2}{{(0.2)}^{2}}=0.02\,kg\,{{m}^{2}}\] \[\therefore \]\[\tau =I\alpha =5\times 0.02=0.10\,Nm.\] Hence, the correction option is (b).You need to login to perform this action.
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