A solid metallic sphere has a radius of r. The base radius of a solid metallic cone of height h is r. Both are melted together and recast into a solid right circular cone with a radius of base r.The height of the resulting cone is then $4\mathrm{r}+\mathrm{h}$

We would be able to see the newly formed object's surface.

only the curved surfaces of the two hemispheres and the curved surface of the cylinder

As a result, the total surface area of the new solid is the sum of the curved surfaces areas of each of the individual parts

Volume of resulting cone = Volume of sphere + Volume of cone which are melted.

Volume of a sphere of radius $r=\frac{4}{3}\pi r^3$

Volume of a cone $=\frac{1}{3}\pi {\mathrm{r}}^2 \mathrm{h}$, where $\mathrm{r}$ is the radius of the base of the cone and h is the height.

Let the height of the resulting cone be $\mathrm{H}$.

Hence, $\frac{1}{3}\pi {\mathrm{r}}^2\mathrm{H}=\frac{4}{3}\pi {\mathrm{r}}^3+\frac{1}{3}\pi {\mathrm{r}}^2 \mathrm{h}$

$\Rightarrow \mathrm{H}=4\mathrm{r}+\mathrm{h}$

Hence, the height of the resulting cone is $4\mathrm{r}+\mathrm{h}$.

Therefore, the answer is True.