A) 143
B) 272
C) 314
D) 722
Correct Answer: C
Solution :
From equation of rotational motion \[{{\omega }^{2}}=\omega _{0}^{2}-2\alpha \theta \] \[\Rightarrow \] \[0=4{{\pi }^{2}}{{n}^{2}}-2\alpha \theta \] \[\theta =\frac{4{{\pi }^{2}}{{\left( \frac{1200}{60} \right)}^{2}}}{2\times 4}=200\,\,{{\pi }^{2}}rad\] \[\therefore \] \[2\pi n=200{{\pi }^{2}}\] So, the number of revolution \[n=100\pi =314\]
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