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 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 of these

Answer 1

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

Given, Force $F=4.8\times {10}^3\text{N}$
Area of cross-section $A=1.2\times {10}^{-4}{\text{m}}^2$

As we know,
Stress $=\frac{F}{A}$
$=\frac{4.8\times {10}^3\text{N}}{1.2\times {10}^{-4}{\text{m}}^2}=4.0\times {10}^7{\text{N/m}}^2$