Question

A solid is in the form of right circular cone mounted on a hemisphere.The radius of the hemisphere is 3.5 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 cylindrical tub is 5 cm and its height is 10.5 cm,find the volume of water left in the cylindrical tub.

Answer 1

Question:

A solid is in the form of right circular cone mounted on a hemisphere.The radius of the hemisphere is 3.5 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 cylindrical tub is 5 cm and its height is 10.5 cm,find the volume of water left in the cylindrical tub.

Solution:

Given, radius of the hemisphere,

$r = 3.5 cm$

Now,

Since the solid is in the form of a right circular cone mounted on a hemisphere, 

then, Radius of base of the cone = Radius of the hemisphere

→  radius of the base of the cone $= r = 3.5cm$

Height of the cone, $h = 4cm$

So,

Volume of the solid = volume of the cone + volume of the hemisphere

→ $\red{volume \: of \: the \: solid = \frac{1}{31} {πr}^{2} h+ \frac{2}{3} {πr}^{3}}$

→ $\red{volume \: of \: the \: solid = \frac{1}{3} {πr}^{2} (h+2r)}$

→ $\red{volume \: of \: the \: solid = \frac{1}{3} \times \frac{22}{7} \times (4+7)}$

$= {141.16cm}^{3}$

Now, radius of the base of the cylindrical vessel,

$r_{1} = 5cm$

Height of the cylindrical vessel,

$h_{1} = 10.5cm$

∴ Volume of the water in the cylindrical vessel $= \red{ {πr_{1}}^{2} h_{1} = \frac{22}{7} \times 25 \times 10.5 = {825cm}^{3} }$

Now, when the solid is completely submerged in the cylindrical vessel full of water,

then volume of the water displaced by the solid = volume of solid

Hence, volume of the water left in the vessel = volume of the water in the vessel - volume of solid

→ ${(825-141.16)cm}^{3} = {683.84cm}^{3}$