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: \[{{\omega }_{0}}=0,\omega =2\pi n=2\pi \times 10,t=3s\] \[\therefore \] \[\alpha =\frac{20\pi }{3}\] Second case: \[\omega ={{\omega }_{0}}+\alpha t\] Putting \[{{\omega }_{0}}=2\pi \,n\] \[=20\pi ,\alpha =\frac{20\pi }{3},t=6s\] \[\therefore \] \[\omega =20\pi +\frac{20\pi }{3}\times 6\] \[\Rightarrow \] \[2\pi n'=60\pi \] \[\Rightarrow \] \[n'=30\]
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