Question

A copper wire 4m long has a diameter of 1mm, if load of 10kg wt is attached at the other end. What extension is produced, if Poisson’s ratio is 0.26? How much lateral compression is produced in it? $(Y_{cu}=12.5\times 10^{10} N/m^2)$

Answer 1

Hint: First, find out the longitudinal strain in the wire due to the load by using the formula of Young’s modulus and then by using the formula of Poisson’s ratio, we can find out the lateral strain and then the lateral expansion.

Formulae used:
Young’s modulus, $Y=\dfrac{Stress}{Strain}=\dfrac{FL}{A\Delta L}$
Poisson’s ratio, $\sigma =\dfrac{\text{Lateral strain}}{\text{Longitudinal strain}}$

Complete step by step solution:
It has been that the length (l) and diameter (d) of the copper wire is 4m and 1mm i.e. 0.001m respectively.
Load on the other end of the copper wire, F = 10kg wt= 98N. (taking $g=9.8ms^{-2}$)
According to the question, $Y_{cu}=12.5\times 10^{10} N/m^2$
And we know that, the formula for Young’s modulus is given by,
$Y=\dfrac{Fl}{A\Delta l}$, where F is the force acting, A is the area of cross-section, l is the length and $\Delta l$ is the change in length due to the application of force.
So, we can write $Y_{cu}=\dfrac{Fl}{A\Delta l}$
Rearranging the terms to get the value of extension, $\Delta l$
$\therefore \Delta l=\dfrac{Fl}{AY_{cu}}=\dfrac{98\times 4}{\dfrac{\pi (0.001)^2}{4}\times 12.5\times 10^{10}}=4\times 10^{-3}\; m$
Now, we know that Poisson’s ratio is the ratio of lateral strain and longitudinal strain and is given by,
$\sigma =\dfrac{\Delta d/d}{\Delta l/l}\implies \Delta d=\dfrac{\sigma \times \Delta l\times d}{l}=\dfrac{0.26\times 4\times 10^{-3}\times 10^{-3}}{4}$
$\therefore \Delta d=2.6\times 10^{-5}\; m$
Hence, the lateral expansion in the wire will be $2.6\times 10^{-5}\; m$. The negative sign is because the cross-section of the wire will decrease due to an increase in the length.

Note: The sign of the lateral or longitudinal change depends on the direction of application of the force. If the force acting is along the length, then the change in length will be positive and change in cross-section will be negative and vice versa.