A) 10rad/s
B) 20rad/s
C) 40rad/s
D) 5rad/s
Correct Answer: B
Solution :
As the string pulls the wheel,. then torque is given by \[\tau =F\times r=I\alpha =I\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}\] \[\Rightarrow \] \[{{\omega }_{2}}-{{\omega }_{1}}=\frac{F\times r\times t}{I}\] \[=\frac{10\times 0.2\times 4}{0.4}\] \[\therefore \] \[\Delta \omega =20\,rad/s\]
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