Question

The Poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in cross-sectional area by 4%. The percentage increase in length is-
A. 1%
B. 2%
C. 2.5%
D. 4%

Answer 1

Hint: The density of the material will always remain constant. Split volume as product of area and length. Take logs and solve.

Complete step-by-step answer:
Step1: As density will remain constant so change in volume will become zero.
$\rho =m\times V$
Where $\rho$= density (constant), m= mass and V= volume
$V=A\times l$= constant
A= cross-sectional area $l$= original length
Taking log both sides,
$$\begin{gathered} \ln V=\ln \left(A\times l\right) \\ \ln V=\ln A+\ln l \end{gathered}$$
Now differentiating the above equation,
${\displaystyle \frac{dV}{V}}={\displaystyle \frac{dA}{A}}+{\displaystyle \frac{dl}{l}}$
$\Rightarrow {\displaystyle \frac{dA}{A}}+{\displaystyle \frac{dl}{l}}=0$

Step2: Now From given data substitute the values in the above equation to calculate the % increase in length.
Therefore, ${\displaystyle \frac{dl}{l}}=-{\displaystyle \frac{dA}{A}}$
Given that ${\displaystyle \frac{dA}{A}}=-4\mathrm{\%}$ (Negative sign indicates decrease in area)
$$\begin{gathered} {\displaystyle \frac{dl}{l}}=-\left(-4\right) \\ {\displaystyle \frac{dl}{l}}=4\mathrm{\%} \end{gathered}$$
So the length will increase by 4%. Hence option (D) is the correct option.

Note: Keep in mind that on the application of external forces on the body only dimensions of the body changes but the volume will always remain the same until there is change in density of the material.