A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Given,

Height of right circular cone (h) = 120 cm

Height of cylinder (H) = 180 cm

From figure,

Radius of cylinder = Radius of cone = Radius of hemisphere = r = 60 cm.

Volume of water left in cylinder = Volume of cylinder - Volume of solid

= Volume of cylinder - (Volume of cone + Volume of hemisphere)

$= πr^2H - \Big(\dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3\Big) \\[1em] = πr^2(H - \dfrac{h}{3} - \dfrac{2}{3}r) \\[1em] = \dfrac{22}{7} \times 60^2 \Big(180 - \dfrac{120}{3} - \dfrac{2}{3} \times 60\Big) \\[1em] = \dfrac{22}{7} \times 3600 \times (180 - 40 - 40) \\[1em] = \dfrac{22}{7} \times 3600 \times 100 \\[1em] = \dfrac{7920000}{7} \\[1em] = 1131428.57143 \text{ cm}^3 \\[1em] = 1131428.57143 \times 10^{-6} \text{ m}^3 \\[1em] = 1.131 \text{ m}^3.$

Hence, volume of water left in cylinder = 1.131 m³.