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=1.32\,mm $