Question

A copper wire 3m long is stretched to increase its length by 0.3cm. Find the lateral strain produced in the wire, if the Poisson’s ratio for copper is 0.25
$\begin{align}
  & a)5\times {{10}^{-4}} \\
 & b)2.5\times {{10}^{-4}} \\
 & c)5\times {{10}^{-3}} \\
 & d)2.5\times {{10}^{-3}} \\
\end{align}$

Answer 1

Hint: When an object is subject to a longitudinal stress at time its length increases as well its diameter decreases. The decrease in the length of the diameter to the original length is defined as the lateral strain and the increase in the length of the wire with respect to its original length is defined as the longitudinal strain. The ratio of the lateral strain to that of the longitudinal strain is defined as the Poisson’s ratio. Hence from this definition we can determine the lateral strain.
Formula used:
If a rod of length (L) and diameter (D), is subjected to a longitudinal stress and as a resultant of this increases the length of the rod by $\Delta L$and decreases the diameter by $-\Delta D$, then the Poisson’s ratio for the rod is given by,
$\begin{align}
  & \sigma =\dfrac{\text{lateral strain}}{\text{Longitudinal strain }} \\
 & \sigma =\dfrac{-\Delta D/D}{\Delta L/L}....(1) \\
\end{align}$

Complete solution:
In the above question it is given to us that the length of the rod increases by 0.3cm and the length of the wire is given as 3m. Therefore the longitudinal strain in the wire is equal to,
$\begin{align}
  & \text{Longitudinal strain}=\dfrac{\Delta L}{L} \\
 & \text{Longitudinal strain}=\dfrac{0.3\times {{10}^{-2}}}{3}=0.1\times {{10}^{-2}} \\
\end{align}$
The Poisson’s ratio for the copper wire is given to us as 0.25. Therefore lateral strain from equation 1 is,
$\begin{align}
  & \sigma =\dfrac{\text{lateral strain}}{\text{Longitudinal strain }} \\
 & 0.25=\dfrac{\text{lateral strain}}{\text{0}\text{.1 }} \\
 & \text{lateral strain}=0.1\times {{10}^{-2}}\times 0.25 \\
 & \Rightarrow \text{lateral strain}=0.1\times {{10}^{-2}}\times 0.25 \\
 & \Rightarrow \text{lateral strain}=2.5\times {{10}^{-4}} \\
\end{align}$

Hence the correct answer of the above question is option b.

Note:
It is to be noted that the lateral strain as well as the longitudinal strain are the ratios of the same physical quantities. Therefore we can say that both of them are dimensionless. Since the Poisson's ratio is the ratio of the physical quantities which are dimensionless, therefore it is also dimensionless.