Question

A steel rod is having a radius of $10\text{mm}$ and length $1.0\text{m}$. A force of $100\text{kN}$ is used to stretch it along its length. If the Young's modulus of steel is $2\times {10}^{11}{\text{Nm}}^{-2}$, then the strain in the rod will be

• A. $0.002$
• B. $0.003$
• C. $0.008$
• D. $0.0016$

Answer 1

The correct option is D $0.0016$

Given :
Applied force $\left(F\right)=100\text{kN}$ or ${10}^5\text{N}$
Young's modulus of elasticity $\left(Y\right)=2\times {10}^{11}{\text{Nm}}^{-2}$
Length of the steel rod $\left(L\right)=1.0\text{m}$
Area of the steel rod $\left(A\right)=\pi \times r^2=\pi \times (10\times {10}^{-3}{)}^2=\pi \times {10}^{-4}{\text{m}}^2$

We know that, $Y=\frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}$
$\Rightarrow \text{Longitudinal Strain}\left(\frac{\Delta L}{L}\right)=\frac{F}{A\times Y}$
From the given data,
$\frac{\Delta L}{L}=\frac{{10}^5}{(\pi \times {10}^{-4})\times (2\times {10}^{11})}=\frac{5\times {10}^{-3}}{\pi }$
$\therefore \frac{\Delta L}{L}=0.0016$
Thus, option (d) is the correct answer.