Question

A material has poisson's ratio $0.3$. If a uniform rod of it suffers a longitudinal strain of $25\times {10}^{-3}$, then the percentage increase in its volume is

• A. $1\%$
• B. $2\%$
• C. $3\%$
• D. $4\%$

Answer 1

Answer: (C)

$3\%$

$\begin{array}{cc}\mu & =\frac{1}{2}-\frac{Y}{6B}\\ \Rightarrow & \frac{3}{10}=\frac{1}{2}-\frac{Y}{6B}\\ \Rightarrow & \frac{Y}{6B}=\frac{1}{2}-\frac{3}{10}=\frac{1}{5}\\ \Rightarrow & \frac{Y}{6B}=\frac{1}{5}\\ \Rightarrow & \frac{Y}{B}=\frac{6}{5}\end{array}$

$\begin{array}{cc}\Rightarrow \frac{\left(\Delta V/V\right)}{\left(\Delta l/l\right)}& =\frac{6}{5}\\ \Rightarrow \left(\Delta v/v\right)& =\frac{25\times {10}^{-3}\times 6}{5}=30\times {10}^{-3}\\ \Rightarrow \frac{\Delta V}{V}\%& =30\times {10}^{-3}\times 100=3\%\\ Ans:& \left(C\right)\end{array}$