Two separate wires $$A$$ and $$B$$ are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of 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 $$\frac{1}{x}$$, where $$x$$ is ________.

Young’s modulus $$Y$$ for a wire is defined as
$$Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F/A}{\Delta L/L} = \dfrac{F L}{A \,\Delta L}\;$$ where   $$F$$ = stretching force,   $$A$$ = cross-sectional area,   $$L$$ = original length,   $$\Delta L$$ = extension.

Both wires $$A$$ and $$B$$ are made of the same material, so their Young’s modulus is equal:
$$Y_A = Y_B.$$

Let the radius of wire $$A$$ be $$r$$, so its area is $$A_A = \pi r^{2}.$$

The radius of wire $$B$$ is 4 times that of $$A$$, so $$A_B = \pi (4r)^{2} = 16 \pi r^{2}.$$

Substituting into the formula for $$Y$$ gives for each wire

Wire $$A$$ : $$Y = \dfrac{F L_A}{A_A\,\Delta L_A} = \dfrac{F L_A}{\pi r^{2}\,(2\text{ mm})}$$

Wire $$B$$ : $$Y = \dfrac{F L_B}{A_B\,\Delta L_B} = \dfrac{F L_B}{16\pi r^{2}\,(4\text{ mm})}$$

Set the two expressions equal because $$Y_A = Y_B$$:

$$\dfrac{F L_A}{\pi r^{2}\,(2)} = \dfrac{F L_B}{16\pi r^{2}\,(4)}$$

Cancel the common factors $$F,\;\pi,\;r^{2}$$ from both sides:

$$\dfrac{L_A}{2} = \dfrac{L_B}{16 \times 4}$$

Simplify:

$$\dfrac{L_A}{2} = \dfrac{L_B}{64} \;\;\Longrightarrow\;\; \frac{L_A}{L_B} = \frac{2}{64} = \frac{1}{32}$$

Hence the ratio of their lengths is $$L_A : L_B = 1 : 32,$$ so $$\dfrac{a}{b} = \dfrac{1}{32} = \dfrac{1}{x}.$$

Therefore $$x = 32.$$

Answer : 32