From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume and total surface area of the remaining solid.

Given,

Height of the cylinder (H) = 16 cm

Radius of the base of cylinder (R) = 12 cm

Height of the cone (h) = 8 cm

Radius of the base of cone (r) = 6 cm

(i) Volume of remaining part (V) = Volume of cylinder - Volume of cone

$V = πR^2H - \dfrac{1}{3}πr^2h \\[1em] = \dfrac{22}{7} \times 12^2 \times 16 - \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 8 \\[1em] = \dfrac{22}{7} \times [(12^2 \times 16) - (\dfrac{1}{3} \times 6^2 \times 8)] \\[1em] = \dfrac{22}{7} \times \Big(2304 - \dfrac{288}{3}\Big) \\[1em] = \dfrac{22}{7} \times \dfrac{6912 - 288}{3} \\[1em] = \dfrac{22}{7} \times \dfrac{6624}{3} \\[1em] = \dfrac{22}{7} \times 2208 \\[1em] = \dfrac{48576}{7} \\[1em] = 6939.43 \text{ cm}^3.$

Hence, volume of remaining part = 6939.43 cm³.

(ii) By formula,

⇒ l² = r² + h²

⇒ l² = 6² + 8²

⇒ l² = 36 + 64

⇒ l² = 100

⇒ l² = 10²

⇒ l = 10 cm.

Thus,

Total surface area of remaining solid (T) = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder + Area of circular ring on upper side of cylinder

(T) = 2πRH + πrl + πR² + π(R² - r²)

$= π(2RH + rl + R^2 + R^2 - r^2) \\[1em] = π(2RH + rl + 2R^2 - r^2) \\[1em] = \dfrac{22}{7} \times (2 \times 12 \times 16 + 6 \times 10 + 2 \times 12^2 - 6^2) \\[1em] = \dfrac{22}{7} \times (384 + 60 + 288 - 36) \\[1em] = \dfrac{22}{7} \times 696 \\[1em] = \dfrac{15312}{7} \\[1em] = 2187.43 \text{ cm}^2.$

Hence, total surface area of remaining solid = 2187.43 cm².