Question

A 4 m long copper wire of cross-sectional area 1.2 cm2 is stretched by a force of $4.8\times {10}^{3}N,$ Young's modulus for copper is $y=1.2\times {10}^{11}N/m^2$ the increase in length of the wire is

• A. 1.32mm
• B. 0.8mm
• C. 0.48 mm
• D. 5.36 mm

Answer 1

Answer: (A)

$Stress=\frac{Force}{Area}$ $=\frac{4.8\times {10}^3}{1.2\times {10}^{-4}}=4\times {10}^{7}N/m^2$ Young's modulus, $Y=\frac{Stress}{Strain}$ $\Rightarrow$ $Strain=\frac{4\times {10}^7}{1.2\times {10}^{11}}=3.3\times {10}^{-4}$ $\therefore$ Increase in length = longitudinal strain x initial length $=(3.3\times {10}^{-4})\times 4$ $=13.2\times {10}^{-4}m=1.32mm$