A) \[40\]
B) \[30\]
C) \[20\]
D) \[10\]
Correct Answer: B
Solution :
From equation of rotational motion, we have \[\omega ={{\omega }_{0}}+\alpha t\]and\[\omega =2\pi n\] First case: \[{{\omega }_{0}}=0,\,\,\omega =2\pi n=2\pi \times 10,\,\,t=3\,\,s\] \[\therefore \] \[2\pi \times 10=\alpha \times 3\] \[\Rightarrow \] \[\alpha =\frac{20\pi }{3}\] Second case: \[\omega ={{\omega }_{0}}+\alpha t\] Putting \[{{\omega }_{0}}=2\pi n\] \[=20\pi ,\,\,\alpha =\frac{20\pi }{3},\,\,t=6\,\,s\] \[\therefore \] \[\omega =20\pi +\frac{20\pi }{3}\times 6\] \[\Rightarrow \] \[2\pi n'=60\pi \] \[\Rightarrow \] \[n'=30\]
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