A) 1.32 mm
B) 0.8 mm
C) 0.48 mm
D) 5.36 mm
E) 2.45 mm
Correct Answer: A
Solution :
\[Stress=\frac{Force}{Area}=\frac{4.8\times {{10}^{3}}}{1.2\times {{10}^{-4}}}=4\times {{10}^{7}}N/{{m}^{2}}\] Youngs modulus,\[Y=\frac{stress}{strain}\] \[\Rightarrow \] \[Strain=\frac{4\times {{10}^{7}}}{1.2\times {{10}^{-4}}}=3.3\times {{10}^{-4}}\] \[\therefore \]Increase in length = longitudinal strain\[\times \]initial length \[=(3.3\times {{10}^{-4}})\times 4\] \[=13.2\times {{10}^{-4}}m=1.32\,mm\]
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