A) 8 rad/sec
B) 4 rad/sec
C) 2 rad/sec
D) 1 rad/sec
Correct Answer: B
Solution :
Centripetal force = breaking force Þ\[m{{\omega }^{2}}r=\]breaking stress ´ cross sectional area Þ\[m{{\omega }^{2}}r=p\times A\] Þ \[\omega =\sqrt{\frac{p\times A}{mr}}=\sqrt{\frac{4.8\times {{10}^{7}}\times {{10}^{-6}}}{10\times 0.3}}\] \ \[\omega =4\,rad/\sec \]
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