A) 0.5 mm
B) 1.0 mm
C) 0.05 mm
D) 0.06 mm
Correct Answer: C
Solution :
If the material and length are same as well as the stretching weight, then F, L and Y are fixed in the relation \[\Delta L=\frac{FL}{AY}.\]Then, \[\Delta L\propto \frac{1}{A},A\Delta L=\]constant \[{{A}_{1}}\Delta {{L}_{1}}={{A}_{2}}\Delta {{L}_{2}}\] After substituting the values, we get\[\Delta {{L}_{2}}=0.05\,mm\]
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