Question

Two separate wires A and B are stretched by $2mm$ and $4mm$ respectively, when they are subjected to a force of $2N$. Assume that both the wires are made up of the same material and the radius of wire B is $4$ times that of the radius of wire A. The length of the wires A and B are in the ratio of $a:b$, Then $\frac{a}{b}$ can be expressed as $\frac{1}{x}$. What is the value of where $x$?

Answer 1

Step 1: Given

Change in length of wire A: $∆L_A=2mm$

Change in length of wire A: $∆L_B=4mm$

Radius of wire A: $r_A=r$

Radius of wire B: $r_B=4r$

Ratio of length of wire: $\frac{L_A}{L_B}=\frac{a}{b}=\frac{1}{x}$

Step 2: Formula Used

Young's Modulus of Elasticity: $\gamma =\frac{Stress}{Strain}=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆L}{L}}}$, where $F$ is force acting on wire, $A$ is area of cross-section, $∆L$ is change in length of wire and $L$ is the length of wire.

Step 3: Find the value of $x$

Find an expression of the length of wire A using the formula for Young's modulus of elasticity. Substitute $A=\pi r^2$, for the area of cross-section of the wire.

$$\begin{gathered} \gamma =\frac{{\displaystyle \frac{F_A}{A_A}}}{{\displaystyle \frac{∆L_A}{L_A}}} \\ \Rightarrow \gamma =\frac{F_A{L}_A}{A_A∆L_A} \\ \Rightarrow L_A=\frac{\gamma \left(\pi {r_A}^2\right)∆L}{F} \\ \Rightarrow L_A=\frac{\gamma \left(\pi r^2\right)\times 2}{F} \end{gathered}$$

Find an expression of length of wire B using the formula for Young's modulus of elasticity. Substitute $A=\pi r^2$, for the area of cross-section of wire.

$$\begin{gathered} \gamma =\frac{{\displaystyle \frac{F_B}{A_B}}}{{\displaystyle \frac{∆L_B}{L_B}}} \\ \Rightarrow \gamma =\frac{F_B{L}_B}{A_B∆L_B} \\ \Rightarrow L_B=\frac{\gamma \left(\pi {r_B}^2\right)∆L}{F} \\ \Rightarrow L_B=\frac{\gamma \left(\pi {\left(4r\right)}^2\right)\times 4}{F} \\ \Rightarrow L_B=\frac{\gamma \left(16\pi r^2\right)\times 4}{F} \end{gathered}$$

Divide the length of wire A by that of B to find their ratio

$\begin{array}{rcl}\frac{L_A}{L_B}& =& \frac{\frac{\gamma \left(\pi r^2\right)\times 2}{F}}{\frac{\gamma \left(16\pi r^2\right)\times 4}{F}}\\ & =& \frac{2}{64}\\ & =& \frac{1}{32}\end{array}$

Hence, the value is $x=32$