A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone, which exactly coincides with hemisphere, is 7 cm and its height is 8 cm. The solid is placed in a cylindrical vessel of internal radius 7 cm and height 10 cm. How much water, in cm³, will be required to fill the vessel completely?

Given,

Diameter of cone = 7 cm

Radius of cone (r) = $\dfrac{7}{2}$ = 3.5 cm

Height of cone (h) = 8 cm

Radius of hemisphere = radius of cone = r = 3.5 cm

Radius of cylindrical vessel (R) = 7 cm

Height of cylindrical vessel (H) = 10 cm.

Volume of solid = Volume of cone + Volume of hemisphere

$= \dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3 = πr^2\Big(\dfrac{1}{3}h + \dfrac{2}{3}r) \\[1em] = \dfrac{22}{7} \times (3.5)^2 \times \Big(\dfrac{1}{3} \times 8 + \dfrac{2}{3} \times 3.5\Big) \\[1em] = 22 \times 0.5 \times 3.5 \times \Big(\dfrac{8}{3} + \dfrac{7}{3}\Big) \\[1em] = 38.5 \times \dfrac{15}{3} \\[1em] = 38.5 \times 5 \\[1em] = 192.5 \text{ cm}^3.$

By formula,

Volume of cylindrical vessel = πR²H

$= \dfrac{22}{7} \times 7^2 \times 10 \\[1em] = 22 \times 7 \times 10 \\[1em] = 1540 \text{ cm}^3.$

Volume of water required to fill the vessel completely = Volume of cylindrical vessel - Volume of solid

= 1540 - 192.5

= 1347.5 cm³.

Hence, volume of water required to fill the vessel completely = 1347.5 cm³.