Question

A $4.0\text{m}$ long copper wire of cross-sectional area $1.2{\text{cm}}^2$ is stretched by a force of $4.8\times {10}^3\text{N}$. The stress developed in the wire will be

• A. $4.0\times {10}^7{\text{N/mm}}^2$
• B. $4.0\times {10}^7{\text{kN/m}}^2$
• C. $4.0\times {10}^7{\text{N/m}}^2$
• D. None

Answer 1

The correct option is C $4.0\times {10}^7{\text{N/m}}^2$

The stress developed in the wire is given by

$\text{Stress}=\frac{F}{A}$

$=\frac{4.8\times {10}^3\text{N}}{1.2\times {10}^{-4}{\text{m}}^2}$

$\text{Stress}=4.0\times {10}^7{\text{N/m}}^2$

Hence, option $\left(\text{C}\right)$ is correct.