A) 1 cm
B) 2 cm
C) 4cm
D) 8cm
Correct Answer: B
Solution :
We know\[F=mr{{\omega }^{2}}\] \[\Rightarrow \] \[r{{\omega }^{2}}=cons\tan t\] \[\Rightarrow \] \[{{\omega }^{2}}\propto \frac{1}{r}\] \[\therefore \] \[{{\left( \frac{{{\omega }_{2}}}{{{\omega }_{1}}} \right)}^{2}}=\frac{{{r}_{1}}}{{{r}_{2}}}\] \[\Rightarrow \] \[\frac{4\omega _{1}^{2}}{\omega _{1}^{2}}=\frac{8}{{{r}_{2}}}\] \[\therefore \] \[{{r}_{2}}=2\,cm\]
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