A) 4 mm
B) 16 mm
C) 1 mm
D) 0.25 mm
Correct Answer: C
Solution :
\[l=\frac{FL}{AY}\therefore l\propto \frac{F}{{{r}^{2}}}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{F}_{2}}}{{{F}_{1}}}{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=(4)\times {{\left( \frac{1}{2} \right)}^{2}}=1\]\\[{{l}_{2}}={{l}_{1}}=1mm\]
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