Question

The ratio of diameters of two wires of same material is $n:1$. The length of each wire is $4\text{m}$. On applying the same load, the increase in the length of the thin wire will be?

• A. $n$ times
• B. $2n$ times
• C. $2n+1$ times
• D. $n^2$ times

Answer 1

The correct option is D

$n^2$ times

Explanation for the correct option:

Sep 1: Given

Given, ratio of diameter of the the two wires is $n:1$ and the wires are made of same material.
Let $l_1$ and $l_2$ be the lengths of the two wire.
We have,
$l_1=l_2=4\text{m}$

Step 2: Formulas used

We know that Young's modulus is given by,
$Y=\frac{F·l}{A·∆l}$
where $Y$ is Young's modulus, $F$ is the force applied and $A$ is the area, $l$ is the original length of the material and $∆l$ is the change in length after application of load.

Step 3: Solution

We know that the wires are of the same material so their Young's modulus is also same.

Let $∆l_1$ and $∆l_2$ be the change in length of the wires and $A_1$ and $A_2$ be their respective areas.
We know that wires are circular. So,
$$\begin{gathered} A_1=\mathrm{\pi }{\left(\frac{d_1}{2}\right)}^2 \\ {\mathrm{A}}_2=\mathrm{\pi }{\left(\frac{d_2}{2}\right)}^2 \end{gathered}$$

We have,
$\frac{F·l_1}{A_1·∆l_1}=\frac{F·l_2}{A_2·∆l_2}$
Since the applied load is same,
$\begin{array}{cc}\Rightarrow & \frac{1}{\mathrm{\pi }{\left({\displaystyle \frac{d_1}{2}}\right)}^2·∆l_1}=\frac{1}{\mathrm{\pi }{\left({\displaystyle \frac{d_2}{2}}\right)}^2·∆l_2}\\ \Rightarrow & {\left(d_1\right)}^2∆l_1={\left(d_2\right)}^2∆l_2\\ \Rightarrow & \frac{∆l_2}{∆l_1}={\left(\frac{d_1}{d_2}\right)}^2\\ \Rightarrow & \end{array}$

We know,
$\frac{d_1}{d_2}=\frac{n}{1}$

So,
$\begin{array}{cc}\Rightarrow & \frac{∆l_2}{∆l_1}=\frac{n^2}{1}\\ \Rightarrow & ∆l_2=n^2·∆l_1\end{array}$

Therefore, the increase in the length of the thin wire is $n^2$ times.

Hence, option D is correct.