TS ICET 24 May 2018 Shift 1 Paper

Let the radius of the sphere be r.
In the melting and casting, the volume of the cylinder equals the volume of the sphere
$\mathit{\pi }{\mathit{r}}_{\mathit{c}\mathit{y}\mathit{l}}^2{\mathit{h}}_{\mathit{c}\mathit{y}\mathit{l}}=\frac{4}{3}\mathit{\pi }{\mathit{r}}_{\mathit{s}\mathit{p}\mathit{h}}^3$
$\Rightarrow {\left(0.75\right)}^2\times 1=\left(\frac{4}{3}\right)\times {\mathit{r}}^3$
$\Rightarrow {\mathit{r}}^3={0.75}^3$
$\Rightarrow \mathit{r}=0.75$
The required answer, diameter of the sphere $=2\mathit{r}=2\times 0.75=1.5$ meters