A) 0.005 m
B) 0.01 m
C) 0.02 m
D) 0.002 m
Correct Answer: A
Solution :
\[l=\frac{FL}{\pi {{r}^{2}}Y}\]\ \[l\propto \frac{L}{{{r}^{2}}}\] (Y and F are constant) \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{L}_{2}}}{{{L}_{1}}}\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=(2)\times {{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{2}\] Þ \[{{l}_{2}}=\frac{{{l}_{1}}}{2}=\frac{0.01m}{2}=0.005m\]
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