From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Find the volume of the remaining solid.

Given,

Diameter of solid wooden cylinder (D) = 6 cm

Radius of solid wooden cylinder (R) = $\dfrac{6}{2}$ = 3 cm

Height of solid wooden cylinder (H) = 28 cm

Diameter of cone (d) = 6 cm

Radius of cone (r) = $\dfrac{6}{2}$ = 3 cm

Height of the cone (h) = 10.5 cm

Volume of cylinder = πR²H

$= \dfrac{22}{7} \times 3^2 \times 28 \\[1em] = 22 \times 9 \times 4 \\[1em] = 792 \text{ cm}^3.$

Volume of single cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 3^2 \times 10.5 \\[1em] = \dfrac{22}{7} \times 9 \times 3.5 \\[1em] = \dfrac{693}{7} \\[1em] = 99 \text{ cm}^3.$

Volume of two conical cavities = 2 × 99 = 198 cm³

Volume of remaining solid = Volume of cylinder - Volume of 2 conical cavities

= 792 - 198

= 594 cm³.

Hence, volume of the remaining solid = 594 cm³.