Question

A uniform wire $(Y=2\times {10}^{11}N/m^2)$ is subjected to a longitudinal tensile stress of $5\times {10}^7$ $N/m^2$. If the overall volume change in the wire is $0.02\%$, the fractional decrease in its radius is $5k\times {10}^{-6}$. Find the value of k.

Answer 1

Strain $=\frac{\Delta l}{l}=\frac{stress}{Y}=\frac{5\times 1{-}^7}{2\times {10}^{11}}=2.5\times {10}^{-4}$
Volume $V=Al$

$\Delta V=A\Delta l+l\Delta A$

$\frac{\Delta V}{V}=\frac{\Delta l}{l}=\frac{\Delta A}{A}$

$\Rightarrow 2\times {10}^{-4}=2.5\times {10}^{-4}+\frac{\Delta A}{A}$
$\Rightarrow \frac{\Delta A}{A}=0.5\times {10}^{-4}$
$\therefore \frac{\Delta A}{A}=\frac{2\Delta r}{r}$

$\left|\frac{\Delta r}{r}\right|=\frac{1}{2}\left|\frac{\Delta A}{A}\right|=\frac{1}{2}\times 0.5\times {10}^{-4}=0.25\times {10}^{-4}$
Hence $k=5$.