A) 18
B) 12
C) 36
D) 48
Correct Answer: B
Solution :
By using equation\[{{\omega }^{2}}=\omega _{0}^{2}-2\alpha \theta \] \[{{\left( \frac{{{\omega }_{0}}}{2} \right)}^{2}}=\omega _{0}^{2}-2\alpha (2\pi n)\]Þ\[\alpha =\frac{3}{4}\frac{\omega _{0}^{2}}{4\pi \times 36}\], (n = 36) ..(i) Now let fan completes total n¢ revolution from the starting to come to rest \[0=\omega _{0}^{2}-2\alpha (2\pi {n}')\] Þ \[{n}'=\frac{\omega _{0}^{2}}{4\alpha \pi }\] substituting the value of a from equation (i) \[{n}'=\frac{\omega _{0}^{2}}{4\pi }\frac{4\times 4\pi \times 36}{3\omega _{0}^{2}}=48\]revolution Number of rotation = 48 ? 36 = 12
You need to login to perform this action.
You will be redirected in
3 sec
