Question

A steel rod has a radius of $10mm$ and a length of $1m$. A $80kN$ force stretches it along its length. If the Young's modulus of the rod is $2\times 10^{11}N/m^2$, then the change in length is

• A. $\frac{2}{\pi }mm$
• B. $\frac{4}{\pi }mm$
• C. $\frac{3}{\pi }mm$
• D. $1mm$

Answer 1

Answer: (B)

We know that, Young's modulus is given by
$Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{F/a}{\Delta l/l}..(i)$
or $\Delta l=\frac{F}{a}\times \frac{l}{Y}\dots (ii)$
Given, $F=80kN=80\times 10^{3}N$
$l=1m$
$a=\pi r^2=\pi \times {\left(10\times 10^{-3}m\right)}^2$
$=\pi \times 10^{-4}{m}^2$
$Y=2\times 10^{11}N/m^2$
$\Rightarrow \Delta l=\frac{80\times 10^3\times 1}{\pi \times 10^{-4}\times 2\times 10^{11}}m$
$=\frac{8\times 10^8}{\pi \times 2\times 10^{11}}m=\frac{4}{\pi }\times 10^{-3}m$
$\Rightarrow \Delta l=\frac{4}{\pi }mm$