Question

A solid consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder, given that the radius of the cylinder is 3 cm, and its height is 6 cm. The radius of the hemisphere and cone is 2 cm, and the height of the cone is 4 cm. $(Take\pi =\frac{22}{7})$.

Answer 1

Volume of water left in cylinder $=$ Volume of cylinder $-$ volume of cone $-$ volume of hemisphere

We know that, volume of cylinder $=\pi r^{2}h$

Given, $r=3cm$ and $h=6cm$

$\therefore V_{cyl}=\frac{22}{7}\times 3\times 3\times 6[\because \pi =\frac{22}{7}]$

$=169.71c{m}^3$

Also, volume of cone $=\frac{1}{3}\pi r^{2}h$

Given, $r=2cm$ and $h=4cm$

$\therefore V_{cone}=\frac{1}{3}\times \frac{22}{7}\times 2\times 2\times 4[\because \pi =\frac{22}{7}]$

$=16.76c{m}^3$

Now, volume of hemisphere $=\frac{4}{3}\pi r^3$

Given, $r=2cm$

$\therefore V_{hemisphere}=\frac{4}{3}\times \frac{22}{7}\times 2\times 2\times 2[\because \pi =\frac{22}{7}]$

$=33.52c{m}^3$

Hence, volume of water $\left(V\right)=V_{cyl}-V_{cone}-V_{hemisphere}$

$V=169.71-16.76-33.52$

$V=119.43c{m}^3$

Hence, the volume of water left in the cylinder is $119.43c{m}^3$.