A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place.

Internal radius (r) = 3 cm

Height (h) = 21 cm

Thickness = External radius (R) - Internal radius

⇒ 0.5 = External radius - 3 cm

⇒ External radius = 0.5 + 3 = 3.5 cm

Volume of hollow cylinder = π(R² - r²)h,

Putting values we get,

∴ Volume of metal = π(3.5² - 3²) × 21

$= \dfrac{22}{7} \times (12.25 - 9) \times 21 \\[1em] = 22 \times 3.25 \times 3 \\[1em] = 214.5 \text{ cm}^3$

Given the tube is melted and cast into a right circular cone of height (H) 7 cm.

So, the volume of metal and volume of cone will be same.

∴ 214.5 = $\dfrac{1}{3}$ πr²H

$\Rightarrow 214.5 = \dfrac{1}{3} \times \dfrac{22}{7} \times \text{r}^2 \times 7 \\[1em] \Rightarrow 214.5 = \dfrac{1}{3} \times 22 \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \dfrac{214.5 \times 3}{22} \\[1em] \Rightarrow \text{r}^2 = 29.25 \\[1em] \Rightarrow \text{r} = \sqrt{29.25} \\[1em] \Rightarrow \text{r} = 5.4 \text{ cm.}$

Hence, radius of the cone is 5.4 cm.