Two separate wires $\mathrm{A}$ and $\mathrm{B}$ are stretched by $2 \mathrm{~mm}$ and $4 \mathrm{~mm}$ respectively, when they are subjected to a force of $2 \mathrm{~N}$. 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 $a / b$ can be expressed as $1 / \mathrm{x}$ where $\mathrm{x}$ is
$(32)$
For $\mathrm{A} \frac{\mathrm{E}}{\pi \mathrm{r}^{2}}=\mathrm{y} \frac{2 \mathrm{~mm}}{\mathrm{a}} \ldots$ (1)
For $\mathrm{B} \frac{\mathrm{E}}{\pi \cdot 16 \mathrm{r}^{2}}=\mathrm{y} \frac{4 \mathrm{~mm}}{\mathrm{~b}} \ldots(2)$
$\therefore(1) /(2)$
$16=\frac{2 b}{4 a}$
$\frac{a}{b}=\frac{1}{32}$
$\therefore$ Answer $=32$
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