CTET 2 Math and Science 2011 Paper

Let h and r be the original height and radius of iron rod
Since, iron-rod is in cylindrical form
∴ initial volume $=\mathit{\pi }{\mathit{r}}^2\mathit{h}$
After recasting
Volume $=\mathit{\pi }{\mathit{r}}^2\mathit{h}$ where h' is the new height
$=\mathit{\pi }{\left(\frac{\mathit{r}}{4}\right)}^2\mathit{h}\prime$
Since, its volume remains constant
$\therefore \mathit{\pi }{\mathit{r}}^2\mathit{h}=\mathit{\pi }{\mathit{r}}^2\mathit{h}\prime$
$\mathit{\pi }{\mathit{r}}^2\mathit{h}=\mathit{\pi }{\left(\frac{\mathit{r}}{4}\right)}^2\mathit{h}\prime$
$\mathit{h}=\frac{{\mathit{h}}^{\mathit{I}}}{16}\Rightarrow \mathit{h}\prime =16\mathit{h}$
Hence, the new length will become 16 times of the original