A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is

1. 12 cm

2. 14 cm

3. 15 cm

4. 18 cm

Internal radius (r) = $\dfrac{4}{2}$ = 2 cm,

External radius (R) = $\dfrac{8}{2}$ = 4 cm.

Given, spherical shell is recasted into cone.

Volume of cone = Volume of spherical shell.

Radius of cone (r₁) = $\dfrac{8}{2}$ = 4 cm.

Let height of cone be h.

$\dfrac{1}{3}πr_1^2h = \dfrac{4}{3}π(R^3 - r^3)$

Multiplying both sides by $\dfrac{3}{π}$.

$\Rightarrow r$1^2h = 4(R^3 - r^3) \\[1em] \Rightarrow h = \dfrac{4(R^3 - r^3)}{r1^2} \\[1em] = \dfrac{4 \times (4^3 - 2^3)}{4^2} \\[1em] = \dfrac{4 \times (64 - 8)}{16} \\[1em] = \dfrac{4 \times 56}{16} \\[1em] = \dfrac{224}{16} \\[1em] = 14 \text{ cm}

Hence, Option 2 is the correct option.