From a circular cylinder of diameter 10 cm and height 12 cm, a conical cavity of the same base radius and of the same height is hollowed out. Find the volume and the whole surface of the remaining solid. Leave the answer in π

Given,

Height of the cylinder (H) = 12 cm

Radius of the base of the cylinder (R) = $\dfrac{\text{diameter}}{2} = \dfrac{10}{2}$ = 5 cm

Height of the cone (h) = 12 cm

Radius of the cone (r) = 5 cm

Volume of the remaining part = Volume of cylinder - Volume of cone

= πR²H - $\dfrac{1}{3}$ πr²h

$= π \times 5^2 \times 12 - π \times \dfrac{1}{3} \times 5^2 \times 12 \\[1em] = π(25 \times 12 - 25 \times 4) \\[1em] = π(300 - 100) \\[1em] = 200 π \text{ cm}^3$

∴ The volume of the remaining solid is 200 π cm³.

By formula,

l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm

Total surface area of remaining solid = Curved surface area of cylinder + curved surface area of cone + base area of cylinder

= 2πRH + πrl + πR²

= π(2RH + rl + R²)

= π(2 × 5 × 12 + 5 × 13 + 5²)

= π(120 + 65 + 25)

= 210 π cm².

Hence, the volume of the remaining solid is 200 π cm³ and total surface area of remaining solid is 210 π cm².