Question

A uniform cylindrical wire is subjected to a longitudinal tensile stress of $5\times {10}^{7}N/m^2$. Young's modulus of the material of the wire is $2\times {10}^{11}N/m^2$. The volume change in the wire is $0.02\%$. The fractional change in the radius is

• A. $0.25\times {10}^{-4}$
• B. $0.5\times {10}^{-4}$
• C. $0.1\times {10}^{-4}$
• D. $1.5\times {10}^{-4}$

Answer 1

Answer: (A)

$0.25\times {10}^{-4}$

$Y=\frac{Stress}{\Delta L/L}$
$\frac{\Delta L}{L}=\frac{stress}{Y}=\frac{5\times {10}^7}{2\times {10}^{11}}=2.5\times {10}^{-4}$

Now $V=\pi r^{2}L$
$\frac{\Delta V}{V}=\frac{\pi \Delta \left(r^{2}L\right)}{\pi r^{2}L}$$=\frac{r^2\Delta L+L\times 2r\Delta r}{r^{2}L}$

$\frac{\Delta L}{L}+2\frac{\Delta r}{r}$
$2\frac{\Delta r}{r}=\frac{\Delta V}{V}-\frac{\Delta L}{L}$$=\frac{0.02}{100}-2.5\times {10}^{-4}$
$\frac{\Delta r}{r}=1\times {10}^{-4}-\frac{2.5}{2}\times {10}^{-4}=-0.25\times {10}^{-4}$