A solid, in the form of a right circular cone, is mounted on a hemisphere of the same radius. The radius of the hemisphere is 2.1 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the tub is 5 cm and its height is 9.8 cm, find the volume of water left in the tub.

Given,

For the solid formed :

Radius of cone = radius of hemisphere = r = 2.1 cm

Height of cone (h) = 4 cm

For cylindrical vessel :

Radius (R) = 5 cm

Height (H) = 9.8 cm

Volume of water left in tub = Volume of cylindrical vessel - Volume of solid

= Volume of cylindrical vessel - (Volume of cone + Volume of hemisphere)

$= πR^2H - \Big(\dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3\Big) \\[1em] = \dfrac{22}{7}\times 5^2 \times 9.8 - \Big(\dfrac{1}{3} \times \dfrac{22}{7} \times (2.1)^2 \times 4 + \dfrac{2}{3} \times \dfrac{22}{7} \times (2.1)^3\Big) \\[1em] = 22 \times 25 \times 1.4 - [22 \times 0.1 \times 2.1 \times 4 + 44 \times 0.1 \times (2.1)^2] \\[1em] = 770 - [18.48 + 19.404] \\[1em] = 770 - 37.884 \\[1em] = 732.116 \text{ cm}^3.$

Hence, volume of water left = 732.116 cm³.