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 bottoms. Find the volume of water left in the cylinder, if the radius of the cylinder is $60cm$ and its height is $180cm$.

Answer 1

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

Volume of cylinder:

Radius, $r=60cm$

Height, $h=180cm$

Volume of outer cylinder = $\pi r^{2}h=\frac{22}{7}\times {60}^2\times 180=\frac{14256000}{7}$

Volume of solid$=$ Volume of cone $+$ Volume of hemisphere

Volume of cone:

Radius, $r=60cm$

Height,$h=120cm$

Volume of cone = $\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\times \frac{22}{7}\times {60}^2\times 120$

$=\frac{3168000}{7}c{m}^3$

Volume of hemisphere:
radius, $r=60cm$
Volume = $\frac{2}{3}\pi r^3$

$=\frac{2}{3}\times \frac{22}{7}\times {60}^3$

$=\frac{3168000}{7}$

Volume of solid $=\frac{3168000}{7}+\frac{3168000}{7}=\frac{6336000}{7}$

Now, Volume of water left in cylinder $=$ Volume of cylinder - Volume of solid

$=\frac{14256000}{7}-\frac{6336000}{7}$

$=\frac{7920000}{7}c{m}^3$
$=1131428.57c{m}^3$
$=1131428.57\times \frac{1}{100}\times \frac{1}{100}\times \frac{1}{100}{m}^3$ [Since, $1cm=\frac{1}{100}m$]
$=1.131m^3$