If a solid sphere of radius r is melted and recast into the shape of a solid cone of height r then the radius of the base of the cone is :

1. r

2. 2r

3. 3r

4. 4r

Given,

Radius of sphere = r cm

Height of cone, h = r cm

Let radius of cone be R cm

Since, sphere is melted and recasted into cone.

∴ Volume of sphere = Volume of cone

$\Rightarrow \dfrac{4}{3} π\text{r}^3 = \dfrac{1}{3} π\text{R}^2\text{h} \\[1em] \Rightarrow \dfrac{4}{3} \text{r}^3 = \dfrac{1}{3} \text{R}^2 \times \text{r} \\[1em] \Rightarrow 3 \times \dfrac{4}{3} \times \text{r}^3 = \text{R}^2 \times \text{r} \\[1em] \Rightarrow 4 \times \dfrac{\text{r}^3}{\text{r}} = \text{R}^2 \\[1em] \Rightarrow 4 \times \text{r}^2 = \text{R}^2 \\[1em] \Rightarrow \text{R} = \sqrt{4 \times \text{r}^2} \\[1em] \Rightarrow \text{R} = 2\text{r}$

Hence, option 2 is the correct option.