Question

The length of a wire is increased by $0.3\%$. Find the percentage increase in its volume, if Poisson’s ratio of wire material is$0.5$.
(A) $0.3\%$
(B) $0.9\%$
(C) $0.27\%$
(D) $Zero\,\%$

Answer 1

Hint
To solve this question, we need to obtain the change in the diameter of the wire by using the value of Poisson’s ratio. Then by using the formula of volume, we can find out the required percentage increase in the volume.
The formulae used to solve this question are:
$\Rightarrow \upsilon = - \dfrac{{{\text{Lateral Strain}}}}{{{\text{Longitudinal Strain}}}}$, where $\upsilon$ is the value of Poisson’s ratio.
$\Rightarrow V = \pi {r^2}l$, where $V$ is the volume of the wire of length $l$ and radius $r$

Complete step by step answer
We know that the Poisson’s ratio of a material is
$\Rightarrow \upsilon = - \dfrac{{{\text{Lateral Strain}}}}{{{\text{Longitudinal Strain}}}}$
We know that strain is the ratio of change in dimension to the original dimension.
So, $\upsilon = - \dfrac{{\Delta d/d}}{{\Delta l/l}}$ (1)
Now, according to the question, the increase in the length of the wire is $0.3\%$.
So, $\Delta l = 0.3l$ (2)
Also, the Poisson’s ratio is $\upsilon = 0.5$
Substituting these in (1), we have
$\Rightarrow 0.5 = - \dfrac{{\Delta d/d}}{{0.3l/l}}$
$\Rightarrow 0.5 = - \dfrac{{\Delta d/d}}{{0.3}}$
Multiplying both sides by$- 0.3$, we get
$\Rightarrow \dfrac{{\Delta d}}{d} = - 0.3 \times 0.5$
$\Rightarrow \dfrac{{\Delta d}}{d} = - 0.15$
Which gives
$\Rightarrow \Delta d = - 0.15d$
$\Rightarrow \Delta d = - \dfrac{{15}}{{100}}d$ (3)
So, the diameter of the wire is reduced by $15\%$
We know that the volume of a wire is given as
$\Rightarrow V = \pi {r^2}l$
Substituting$r = \dfrac{d}{2}$ , we get
$\Rightarrow V = \pi {\left( {\dfrac{d}{2}} \right)^2}l$
$\Rightarrow V = \dfrac{{\pi {d^2}l}}{4}$
Now, using the concept of relative errors, we have
$\Rightarrow \dfrac{{\Delta V}}{V} = 2\dfrac{{\Delta d}}{d} + \dfrac{{\Delta l}}{l}$
Substituting (2) and (3), we have
$\Rightarrow \dfrac{{\Delta V}}{V} = 2\left( {\dfrac{{ - 0.15d}}{d}} \right) + \dfrac{{0.3l}}{l}$
$\Rightarrow \dfrac{{\Delta V}}{V} = 2( - 0.15) + 0.3$
On solving, we get
$\Rightarrow \dfrac{{\Delta V}}{V} = - 0.3 + 0.3$
Finally, we have
$\Rightarrow \dfrac{{\Delta V}}{V} = 0$
Or, $\Delta V = 0$
So, the percentage increase in the volume of the wire is $0$
Hence, the correct answer is option (D), $Zero\,\%$.

Note
Do not forget the negative sign in the expression of the Poisson’s ratio. The negative sign indicates that with the increase in the longitudinal dimensions of a material, its lateral dimensions decreases, and vice-versa. Also, include the negative sign of change in dimension while calculating the change in volume.