A) 0.2 Newton
B) 2.0 Newton
C) 0.02 Newton
D) 20.0 Newton
Correct Answer: A
Solution :
\[F=\frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}\] Here, \[m=0.10\,\,kg,\,\,r=0.5\,\,metre\] and \[\omega =\frac{2\pi n}{t}=\frac{2\times 3.14\times 10}{31.4}=2\,\,rad/\sec \] \[\therefore \]\[F~=0.10\times 0.5\times {{(2)}^{2}}=0.2\]Newton
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