A) 400 radian
B) 200 radian
C) 100 radian
D) 250 radian
Correct Answer: A
Solution :
\[I=3\,kg\,{{m}^{2}},t=6\,Nm,t=20\,\sec \] From \[t=I\,\alpha \] Angular acceleration \[\,\alpha =\frac{t}{I}=\frac{6}{3}\] \[=2\,rad/{{\sec }^{2}}\] \[\therefore \]Angular displacement \[={{\omega }_{0}}t+\frac{1}{2}\alpha \,{{t}^{2}}\] \[=0\times 20+\frac{1}{2}(2){{(20)}^{2}}\] \[=400\,radian\]You need to login to perform this action.
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