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# A material has Poisson’s ratio 0.3. If a uniform rod of it suffers a longitudinal strain of $3 \times {10^{ - 3}}$, what will be the percentage increase in volume?

Last updated date: 13th Jun 2024
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Hint: In this question we have to find the value of increase in the volume of the material. The Poisson’s ratio and longitudinal strain of the material are given. To find the increase in volume we will use the formula for Poisson’s ratio,
${\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{\Delta R/R}}{{\Delta L/L}}} \right)$

Complete step by step solution:
Given,
$\Rightarrow \dfrac{{\Delta L}}{L} = 3 \times {10^{ - 3}}$
$\Rightarrow {\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{\Delta R/R}}{{\Delta L/L}}} \right)$
$\Rightarrow 0.3 = - \left( {\dfrac{{\Delta R/R}}{{3 \times {{10}^{ - 3}}}}} \right)$
$\Rightarrow \dfrac{{\Delta R}}{R} = - 0.3 \times 3 \times {10^{ - 3}}$
$\Rightarrow \dfrac{{\Delta R}}{R} = - 0.9 \times {10^{ - 3}}$
Volume of rod is $V = \pi {R^2}L$
To find an increase in volume we will convert it in this form.
$\Rightarrow \dfrac{{\Delta V}}{V} = 2\dfrac{{\Delta R}}{R} + \dfrac{{\Delta L}}{L}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \left( { - 2 \times 0.9 \times {{10}^{ - 3}} + 3 \times {{10}^{ - 3}}} \right)$
$\Rightarrow \dfrac{{\Delta V}}{V} = 1.2 \times {10^{ - 3}}$
Now, we will find the percentage increase in volume
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 1.2 \times {10^{ - 3}} \times 100$
$\Rightarrow \dfrac{{\Delta V}}{V} \times 100 = 0.12\%$
Hence, from the above calculation we have found the value of percentage increase in volume and it comes out to be, $\dfrac{{\Delta V}}{V} \times 100 = 0.12\%$.

$strain = \dfrac{{extension}}{{length}}$
If the extension in length is $\Delta L$and the total length of the body is $L$, then strain is given by following formula;
$strain = \dfrac{{\Delta L}}{L}$
Longitudinal strain $\dfrac{{\Delta L}}{L}$
Shearing strain $\dfrac{{\Delta L}}{L}$
Volumetric strain $\dfrac{{\Delta V}}{V}$
Note: Poisson’s ratio is a measure of deformation of a material in different directions perpendicular to the direction of the force applied. In other words, a Poisson’s ratio is the ratio of the transverse strain to the longitudinal strain. It is represented by $\sigma$. Since, it is a ratio so it does not have any dimension. It is a scalar quantity.
${\text{Poisson's ratio(}}\sigma {\text{)}} = - \left( {\dfrac{{Transverse{\text{ strain}}}}{{Longitudinal{\text{ strain}}}}} \right)$