Question

A solid consisting of a right circular cone of height $120cm$ and radius $60cm$ standing on a hemisphere of radius $60cm$ 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, (in $m^3$) if the radius of the cylinder is $60cm$ and its height is $180cm$?

Answer 1

In the given question,

Volume of water in the cylinder $=$ Volume of cylinder $-$ Volume of Solid

Given:

Radius of the base of the cone $r_1=60cm$

Height of the cone $h_1=120cm=2r1$

Radius of the hemisphere $r_2=60cm=r_1$

Height of the cylinder $h_2=180cm=3r_1$

Radius of the base of the cylinder $r_3=60cm=r_1$

Now,

Volume of the solid $=\frac{2}{3}\pi (r_1{)}^3+\frac{1}{3}\pi (r_1{)}^2\left(2r_1\right)$

$=\frac{2}{3}\pi (60{)}^3+\frac{1}{3}\pi (60{)}^2(2\times 60)$

$=\frac{1}{3}\pi \left(60{)}^3\right[2+2]$

$=\frac{4}{3}\pi (60{)}^3$

Again,

Volume of the cylinder $=\pi \left(r_1{)}^2\right(3r_1)$

$=3\pi (60{)}^3$

Therefore,

Required volume of water $=\pi (60{)}^3[3-\frac{4}{3}]$

$=\frac{5}{3}\pi (60{)}^3$

$=1130973.35c{m}^3$

$=1.131m^3$ (approx.)