Question

Two wires of the same material and same length but diameters in the ratio $1:2$ are stretched by the same force. The potential energy per unit volume of the two wires will be in the ratio:

Answer 1

Step 1: Given Data

The ratio of diameters of the two materials: $1:2$

$\frac{d_1}{d_2}=\frac{1}{2}$

It is already given in the question that the two wires are made of the same materials, therefore the strain on both the materials will be the same.

Step 2: Formula Used

$\mathrm{Strain}=\frac{\mathrm{Stress}}{Y}$

$Stress=\frac{F}{A}$, $A=\frac{\pi d^2}{4}$

$\mathrm{U}=\frac{1}{2}\mathrm{Stress}\times \mathrm{Strain}$

$U$is the potential energy

$F$ = Force applied
$A$ = Area on the surface

where $Y$ is Young's modulus

Stress is inversely proportional to the square of the diameter of the wire:

$\mathrm{Stress}\propto \frac{1}{{\mathrm{d}}^2}$

Step 3: Relationship amongst Potential energy, Stress, and Young's Modulus:

$$\begin{gathered} \mathrm{U}=\frac{1}{2}\mathrm{Stress}\times \mathrm{Strain} \\ \mathrm{U}=\frac{1}{2}\mathrm{Stress}\times \frac{\mathrm{Stress}}{Y} \\ \mathrm{U}=\frac{1}{2}\times \frac{{\left(\mathrm{Stress}\right)}^2}{Y}=\frac{F^2}{2Y{A}^2} \\ U=\frac{8F^2}{{\pi }^2{d}^{4}Y} \end{gathered}$$

{$F$ and $Y$ are the same for both wires}

Since Stress is inversely proportional to the square of the diameter of the wire, the potential energy will be inversely proportional to the diameter raised to its fourth power:

$\mathrm{U}\propto \frac{1}{{\mathrm{d}}^4}$

Step 4: Relationship between Potential energy and the two diameters:

$\frac{{\mathrm{U}}_1}{{\mathrm{U}}_2}={\left(\frac{{\mathrm{d}}_2^2}{{\mathrm{d}}_1^2}\right)}^2=\frac{2^4}{1^4}=16:1$

Hence, the potential energy per unit volume of the two wires will be in the ratio $16:1$.