Question

The $\sigma$ of a material is $\mathrm{0.20.}$ If a longitudinal strain of $4.0\times {10}^{-3}$ is caused in a rod, by what percentage will the volume change?

• A. $0.48\%$
• B. $0.32\%$
• C. $0.24\%$
• D. $0.50\%$

Answer 1

The correct option is C $0.24\%$

given, $\sigma =0.20$

$\begin{array}{c}\text{Longitudinal strain}=4\times {10}^{-3}\\ \frac{\Delta l}{l}=4\times {10}^{-3}\\ \sigma =-\frac{\Delta r/\gamma }{\Delta l/l}\Rightarrow \frac{\Delta r}{\gamma }=-\sigma \times \frac{\Delta l}{1}\\ =-\frac{1}{5}\times 4\times {10}^{-3}\\ \text{For rod,}V=\pi r^{2}l\end{array}$ Taking log on both sides: $\log v=\log \left(\pi r^{2}l\right)$ $\begin{array}{c}\text{Differentiating, we get :}\frac{1}{v}dV=\frac{2dr}{r}+\frac{dl}{l}\\ \therefore \frac{dV}{V}=-1.6\times {10}^{-3}+4\times {10}^{-3}\\ =2.4\times {10}^{-3}\end{array}$ change in volume is $0.24\%$