Question

Copper of fixed volume $V$ is drawn into wire of length $l$. When this wire is subjected to a constant force $F$, the extension produced in the wire is $\Delta l$. Which of the following graphs will be a straight line ?

• A. $\Delta l$ versus $\frac{1}{l}$
• B. $\Delta l$ versus $l^2$
• C. $\Delta l$ versus $\frac{1}{l^2}$
• D. $\Delta l$ versus $l$

Answer 1

The correct option is B $\Delta l$ versus $l^2$

The wire of volume $V$ will be having a cylindrical shape with $A$ as cross-sectional area and $l$ as the length.
$\therefore V=A\times l...\left(i\right)$
$F$ is the applied external force
$\Rightarrow$ From Hooke's law:
$\frac{F}{A}=Y\frac{\Delta l}{l}.....\left(ii\right)$
where, $\frac{F}{A}=\text{Stress},\frac{\Delta l}{l}=\text{Strain}Y=\text{Young's modulus}$

From Eq. $\left(i\right)$ and $\left(ii\right)$:
$F=\frac{VY\Delta l}{l^2}$
$\Rightarrow \Delta l=\frac{F{l}^2}{VY}...\left(iii\right)$
Eq. $\left(iii\right)$ is representing a straighline of form $y=mx$, where $y\to \Delta l,x\to l^2$ and slope is given by,
$m=\frac{F}{V}=\text{constant}$

$\therefore$ Graph of $\Delta l$ versus $l^2$ will be a straight line passing through origin, as shown in figure below.