A) 24
B) 36
C) 18
D) 12
Correct Answer: D
Solution :
Angular motion \[={{\omega }^{2}}=\omega _{0}^{2}-2\alpha \theta \] (here, (\[\omega =0\]) \[\therefore \,\,0=\,{{\left( \frac{{{\omega }_{0}}}{2} \right)}^{2}}-2\times \,\frac{\omega _{0}^{2}.\,\theta }{16\times 12\pi }\] Or \[\theta =24\,\pi \,or\,\,\theta =12\times 2\pi \] But \[2\pi =1\] cycle, So, \[\theta =12\] cycle
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