Question

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

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

Answer 1

Answer: (A)

$1.32mm$

Young's modulus, $Y=\frac{stess}{strain}=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

So, increase in length $\Delta l=\frac{Fl}{AY}=\frac{(4.8\times {10}^3)\left(4\right)}{(1.2\times {10}^{-4})(1.2\times {10}^{11})}=1.33\times {10}^{-3}m=1.33mm$