A) 0.05 mm
B) 0.10 mm
C) 0.15 mm
D) 0.20 mm
E) 0.25 mm
Correct Answer: A
Solution :
\[l=\frac{FL}{AY}\]\ \[l\propto \frac{1}{A}\] (F,L and Y are constant] \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{4}{8}=\frac{1}{2}\]\[\Rightarrow {{l}_{2}}=\frac{{{l}_{1}}}{2}=\frac{0.1}{2}=0.05mm\]
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