Question

A wire of length $1\text{m}$ is stretched by a force of $10\text{N}$. The area of cross-section of the wire is $2\times {10}^{-6}{\text{m}}^2$ and Young's modulus $\left(Y\right)$ is $2\times {10}^{11}{\text{N/m}}^2$. Increase in length of the wire will be -

• A. $2.5\times {10}^{-5}\text{cm}$
• B. $2.5\times {10}^{-5}\text{mm}$
• C. $2.5\times {10}^{-5}\text{m}$
• D. None of these

Answer 1

The correct option is C $2.5\times {10}^{-5}\text{m}$

As we know,

$\text{Young's modulus}\left(Y\right)=\frac{\text{Stress}}{\text{Strain}}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$

$\text{Stress}=\frac{F}{A}=\frac{10}{(2\times {10}^{-6})}=5\times {10}^6{\text{N/m}}^2$

From $\text{Y}=\frac{\text{Stress}}{\text{Strain}}$

$2\times {10}^{11}=\frac{5\times {10}^6}{\frac{\Delta L}{L}}$

$\Rightarrow \frac{\Delta L}{L}=2.5\times {10}^{-5}$

$\Rightarrow \Delta L=1\times 2.5\times {10}^{-5}$

$\Rightarrow \Delta L=2.5\times {10}^{-5}\text{m}$

Hence, option $\left(C\right)$ is the correct answer.

Why this Question? Key point: Youngs modulus of elasticity is given by, $\text{Young's modulus}\left(Y\right)=\frac{\text{Stress}}{\text{Strain}}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$ $\Rightarrow Y=\frac{FL}{A\Delta L}$