EPFO 2012 Solved Paper

Let the thickness of the wire = $2\mathit{x}$ cm
radius of wire = $\frac{2\mathit{x}}{2}=\mathit{x}$ cm
According to the question,
Volume of iron rod = Volume of wire
$\mathit{\pi }\times {\left(\frac{1}{2}\right)}^2\times 8=\mathit{\pi }\times {\mathit{x}}^2\times 18\times 100$
{∵ volume of cylindrical shape = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$ and $1\mathit{m}=100\mathrm{cm}\}$
⇒ $\frac{1}{4}\times 8={\mathit{x}}^2\times 1800$
⇒ ${\mathit{x}}^2=\frac{2}{1800}$
⇒ ${\mathit{x}}^2=\frac{1}{900}$
$\Rightarrow \mathit{x}=\sqrt{\frac{1}{900}}$
⇒ $\mathit{x}=\sqrt{\frac{1}{30\times 30}}$
∴ $\mathit{x}=\frac{1}{30}\mathit{c}\mathit{m}$
Hence, thickness of the wire $=2\times \frac{1}{30}=\frac{1}{15}\mathit{c}\mathit{m}$