Question

A uniform metallic wire is elongated by $0.04m$ when subjected to a linear force $F$. The elongation, if its length and diameter is doubled and subjected to the same force will be ________$cm$.

Answer 1

Step 1. Given Data:

Change in length on applying force is, $∆L=0.04m$

Step 2. Solving using Young's modulus relation

Young's modulus,

$$\begin{gathered} Y=\frac{F/A}{∆L/L} \\ \Rightarrow Y=\frac{F/A}{0.04/L}\dots \left(1\right) \end{gathered}$$ [$A={\mathrm{\pi r}}^2$, $r$ is the radius]

Where, the $A$ is area and $L$ is the length of wire, $F$ is the force applied.

Step 3. Finding the new area

Now, when the length is doubled to$2L$ and the new diameter, $d\text{'}$ is doubled to$2d$, the radius changes to $r\text{'}$ and the area changes to $A\text{'}$.

By using the formula of area,

$A\text{'}=\mathrm{\pi r}{\text{'}}^2$ [The diameter is twice the radius.]

$$\begin{gathered} A\text{'}=\mathrm{\pi }{\left(\frac{\mathrm{d}\text{'}}{2}\right)}^2 \\ \mathrm{A}\text{'}=\mathrm{\pi }\frac{\mathrm{d}{\text{'}}^2}{4} \\ \mathrm{A}\text{'}=\mathrm{\pi }\left(\frac{{\left(2\mathrm{d}\right)}^2}{4}\right) \\ \mathrm{A}\text{'}={\mathrm{\pi d}}^2 \end{gathered}$$ $[d\text{'}=2d]$

$$\begin{gathered} A\text{'}=\mathrm{\pi }{\left(2\mathrm{r}\right)}^2 \\ \mathrm{A}\text{'}=4{\mathrm{\pi r}}^2 \\ \mathrm{A}\text{'}=4\mathrm{A} \end{gathered}$$ [$A={\mathrm{\pi r}}^2$]

Step 4. Finding the change in the new length, $∆L\text{'}$

Putting the value of the new area, in $\left(1\right)$ we get,

$$\begin{gathered} Y=\frac{F/4A}{∆L\text{'}/2L}[\because A\text{'}=4A,L=2L] \\ Y=\frac{F/2A}{∆L\text{'}/L}\dots \left(2\right) \end{gathered}$$

Since, the material is the same for both the lengths hence, $Y$ will be the same even if a change in dimension occurs.

From equation $\left(1\right)$ and equation $\left(2\right)$ we can write,

$$\begin{gathered} \Rightarrow \frac{F/2A}{∆L\text{'}/L}=\frac{F/A}{0.04/L} \\ \Rightarrow \frac{{\displaystyle \frac{1}{2}}}{∆L\text{'}}=\frac{1}{0.04} \\ \Rightarrow ∆L\text{'}=\frac{0.04}{2}m \end{gathered}$$

$\Rightarrow ∆L\text{'}=0.02m$

$\Rightarrow ∆L\text{'}=2cm$

Hence, the correct answer is $2cm$.