Question

A material has Poisson's ratio $0.2$. If a uniform rod of it suffers longitudinal strain \[4 \times {10^{ - 3}}\], calculate the percentage change in its volume.
A) $0.15\% $
B) $0.02\% $
C) $0.24\% $
D) $0.48\% $

Answer 1

Hint:You can easily approach the solution to this question by recalling what Poisson’s ratio actually means. Poisson’s ratio actually accounts for the width or radius variance due to the longitudinal strain and vice-versa. It can be mathematically represented as:
$\sigma = \, - \dfrac{{\Delta r/r}}{{\Delta l/l}}$

Complete step by step answer:
We will be solving the question exactly like we told we could in the hint section of the solution to the question. Firstly, we will find the strain or change in the radius of the rod since its length has changed due to external force applied. Once we find what’s the new radius in terms of original radius, we can easily compare the volume of before and after the force or strain applied on the rod.

Let us define Poisson’s ratio first. Poisson’s ratio accounts for the change in width or radius of an object because of the change in length brought to by an external force applied, and vice-versa. Using Poisson’s ratio, we can easily find the new radius since we have been given both, the Poisson’s ratio and the longitudinal strain of the rod. Which makes our job easier.

Let’s first see what the question has given to us:
Longitudinal strain $\left( {\Delta l/l} \right) = 4 \times {10^{ - 3}}$
Poisson’s ratio $\left( \sigma \right) = 0.2$
Now, we already know that:
$\sigma = \, - \dfrac{{\Delta r/r}}{{\Delta l/l}}$
Putting in the values that are given to us, we get:
$0.2 = \, - \dfrac{{\Delta r/r}}{{4 \times {{10}^{ - 3}}}}$
After substituting and solving, we get:
$\Delta r/r = - 0.8 \times {10^{ - 3}}$
The negative sign depicts that the radius has shrunk since the length has increased, this is very easy to observe.
Now that we have found the change in radius, we can easily compare the volume change:
Volume change =$\dfrac{{\Delta V}}{V}\, \times 100$
Using system error and analysis, we can reach at:
$\dfrac{{\Delta V}}{V} \times 100\, = \,\left( {2\dfrac{{\Delta r}}{r} + \dfrac{{\Delta l}}{l}} \right) \times 100$
Putting in the values that are given in the question and the one that we found out, we get:
Volume change = $\left( {2\left( { - 0.8 \times {{10}^{ - 3}}} \right) + 4.0 \times {{10}^{ - 3}}} \right) \times 100$
Solving this, we get:
Volume change $ = \,24 \times {10^{ - 3}} \times 100 = 0.24\% $

Hence, the correct option is option C

Note:Main mistake that students make while solving the question is that they do not consider the negative sign in the ratio of radius change. Other than that, many students do not use the error and analysis method in the final step and reach a wrong answer.