Question

A solid consisting of a right 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, if the radius of the cylinder is $60cm$ and its height is $180cm$, the radius of the hemisphere is $60cm$ and height of the cone is $120cm$, assuming that the hemisphere and the cone have common base.

Answer 1

For the cylinder we have
Radius of the base$=r=60cm$
Height $=h=180cm$

Volume of water that the cylinder contains $=V=\pi r^{2}h=\{\pi \times {60}^2\times 180\}{cm}^3$

For conical part, we have
Radius of the base $=r=60cm$
Height $=h_1=120cm$

Volume of conical part $V_1=\frac{1}{3}\pi r^2{h}_1=\frac{1}{3}\pi \times {60}^2\times 120{cm}^3=\{\pi \times {60}^2\times 40\}{cm}^3$
For hemispherical part, we have
Radius $=r=60cm$

Volume of the hemisphere $V_2=\{\frac{2}{3}\pi \times {60}^3\}{cm}^3$

$\Rightarrow V_2=\{2\pi \times {60}^2\times 20\}{cm}^3=(40\pi \times {60}^2){cm}^3$

Hence volume of the water left out in the cylinder

$\Rightarrow V=V_1-V_2=\{\pi \times {60}^2\times 180-\pi \times {60}^2\times 40-40\pi \times {60}^2\}{cm}^3=\pi \times {60}^2\times \{180-40-40\}{cm}^3$

$=\frac{22}{7}\times 3600\times 100{cm}^3$

$=\frac{22\times 36}{700}{m}^3$

$=1.1314m^3$