Question

A steel rod has a radius of $10 \,mm$ and a length of $1 \,m$. A $80 \,kN$ 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. $\frac{2}{\pi} m m$
• B. $\frac{4}{\pi} m m$
• C. $\frac{3}{\pi} m m$
• D. $1 \,mm$

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} \ldots (ii) $
Given, $F=80 kN =80 \times 10^{3} N$
$l=1 m $
$a=\pi r^{2}=\pi \times\left(10 \times 10^{-3} m \right)^{2}$
$=\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 $
$\Rightarrow \Delta l=\frac{4}{\pi} mm$