A) 100
B) 150
C) 250
D) 1000
Correct Answer: B
Solution :
From equation of rotational motion \[\omega ={{\omega }_{0}}+\alpha t\] Here,\[\omega =40\pi \,rad/s,\text{ }{{\omega }_{0}}=20\pi \text{ }rad/s,\text{ }t=10\text{ }s\] \[\therefore \] \[40\pi =20\pi +\alpha +10\] \[\Rightarrow \] \[\alpha =\frac{20\pi }{10}=2\pi \,rad/{{s}^{2}}\] Again, \[\theta ={{\omega }_{0}}t+\frac{1}{2}\alpha {{t}^{2}}\] \[=20\pi \times 10+\frac{1}{2}\times 2\pi \times {{(10)}^{2}}\] \[=200\pi \times 10+\frac{1}{2}\times 2\pi \times {{(10)}^{2}}\] \[=200\pi +100\pi \] \[=300\pi \] \[\therefore \]Number of revolutions \[n=\frac{\theta }{2\pi }=\frac{300\pi }{2\pi }=150\]
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