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

Taking π = $\dfrac{22}{7}$, find the volume of remaining solid.

Given,

Diameter of solid wooden cylinder (D) = 6 cm

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

Height of solid wooden cylinder (H) = 28 cm

Diameter of cone (d) = 6 cm

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

Height of cone (h) = 10.5 cm

Volume of cylinder (V) = πr²h

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

Volume of single cone (v) = $\dfrac{1}{3}$ πR²H

$v = \dfrac{1}{3} \times \dfrac{22}{7} \times 3^2 \times 10.5\\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 9 \times 10.5\\[1em] = 22 \times 3 \times 1.5\\[1em] = 99 \text{ cm}^2$

Volume of two conical cavities = 2 x 99 = 198 cm²

Volume of remaining solid = Volume of cylinder - Volume of 2 conical cavities = 792 - 198 = 594 cm².

Hence, volume of remaining solid = 594 cm².