A spherical shell of lead whose external and internal diameters are 24 cm and 18 cm, is melted and recast into a right circular cylinder 37 cm high. Find the diameter of the base of the cylinder.

Given,

Height of the solid right circular cylinder, h = 37 cm

Internal radius of metallic spherical shell, r = $\dfrac{\text{diameter}}{2} = \dfrac{18}{2}$ = 9 cm

External radius of metallic spherical shell, R = $\dfrac{\text{diameter}}{2} = \dfrac{24}{2}$ = 12 cm

Let the radius of cylinder be a cm.

As, metallic spherical shell is recasted into right circular cylinder.

∴ Volume of spherical shell = Volume of cylinder

$\Rightarrow \dfrac{4}{3} π(\text{R}^3 - \text{r}^3) = π\text{a}^2\text{h} \\[1em] \text{Divide by π on both sides, we get:} \Rightarrow \dfrac{4}{3} (\text{R}^3 - \text{r}^3) = \text{a}^2\text{h} \\[1em] \Rightarrow \dfrac{4}{3} \times (12^3 - 9^3) = \text{a}^2 \times 37 \\[1em] \Rightarrow \dfrac{4}{3} \times (1728 - 729) = \text{a}^2 \times 37 \\[1em] \Rightarrow \dfrac{4}{3} \times 999 = \text{a}^2 \times 37 \\[1em] \Rightarrow \text{a}^2 = \dfrac{4 \times 999}{3 \times 37} \\[1em] \Rightarrow \text{a}^2 = \dfrac{3996}{111} \\[1em] \Rightarrow \text{a}^2 = 36 \\[1em] \Rightarrow \text{a} = \sqrt{36} \\[1em] \Rightarrow \text{a} = 6 \text{ cm.}$

Diameter = 2a = 2 × 6 = 12 cm.

Hence, diameter of the base of the cylinder is 12 cm.