A) \[1\,rad{{\sec }^{-1}}\]
B) \[2\,rad{{\sec }^{-1}}\]
C) \[4\,rad{{\sec }^{-1}}\]
D) \[8\,rad{{\sec }^{-1}}\]
Correct Answer: C
Solution :
[c] Given that F/A = \[4.8\times {{10}^{7}}N{{m}^{-2}}\] \[\therefore \,\,F=4.8\times {{10}^{7}}\times A\] or \[\frac{m{{v}^{2}}}{r}=4.8\times {{10}^{7}}\times {{10}^{-6}}=48\] or \[\frac{m{{r}^{2}}{{\omega }^{2}}}{r}=48\,\,or\,\,{{\omega }^{2}}=\frac{48}{mr}\] \[\omega =\sqrt{\left( \frac{48}{10\times 0.3} \right)=}\sqrt{16}=4rad/\sec \]You need to login to perform this action.
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