A) 0.5 N-m
B) 0.9 N-m
C) 1.2N-m
D) 2.3 N-m
Correct Answer: B
Solution :
Given, \[{{\omega }_{1}}=10\,\,\text{rad}{{\text{s}}^{-1}},\] \[{{\omega }_{2}}=0,\] \[t=10\,s\] \[\therefore \] \[\alpha =\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}=\frac{0-10}{10}\] \[=-1\,\,rad\,{{s}^{-2}}\] (Negative sign shows retardation) Now, moment of inertia \[I=m{{r}^{2}},\] \[=10\times {{(0.3)}^{2}}=0.9\,kg\text{-}{{m}^{2}}\] \[\therefore \]Torque, \[\tau =I\,\alpha =0.9\times 1=0.9\,N\text{-}m\]You need to login to perform this action.
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