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A steel rod has a radius of $10 \mathrm{~mm}$ and a length of $1 \mathrm{~m}$. A $80 \mathrm{kN}$ force stretches it along its length. If the Young's modulus of the rod is $2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$, then the change in length is
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The correct answer is:
$\frac{4}{\pi} \mathrm{mm}$
We know that, Young's modulus is given by
$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / a}{\Delta l / l}$ $\ldots$ (i)
or $\quad \Delta l=\frac{F}{a} \times \frac{l}{Y}$ $\ldots$ (ii)
Given, $\quad F=80 \mathrm{kN}=80 \times 10^3 \mathrm{~N}$
$l=1 \mathrm{~m}$
$a=\pi r^2=\pi \times\left(10 \times 10^{-3} \mathrm{~m}\right)^2=\pi \times 10^{-4} \mathrm{~m}^2$
$Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$\Rightarrow \quad \Delta l=\frac{80 \times 10^3 \times 1}{\pi \times 10^{-4} \times 2 \times 10^{11}} \mathrm{~m}$
$=\frac{8 \times 10^8}{\pi \times 2 \times 10^{11}} \mathrm{~m}=\frac{4}{\pi} \times 10^{-3} \mathrm{~m}$
$\Rightarrow \quad \Delta l=\frac{4}{\pi} \mathrm{mm}$
$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / a}{\Delta l / l}$ $\ldots$ (i)
or $\quad \Delta l=\frac{F}{a} \times \frac{l}{Y}$ $\ldots$ (ii)
Given, $\quad F=80 \mathrm{kN}=80 \times 10^3 \mathrm{~N}$
$l=1 \mathrm{~m}$
$a=\pi r^2=\pi \times\left(10 \times 10^{-3} \mathrm{~m}\right)^2=\pi \times 10^{-4} \mathrm{~m}^2$
$Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$\Rightarrow \quad \Delta l=\frac{80 \times 10^3 \times 1}{\pi \times 10^{-4} \times 2 \times 10^{11}} \mathrm{~m}$
$=\frac{8 \times 10^8}{\pi \times 2 \times 10^{11}} \mathrm{~m}=\frac{4}{\pi} \times 10^{-3} \mathrm{~m}$
$\Rightarrow \quad \Delta l=\frac{4}{\pi} \mathrm{mm}$
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