Question

A cylindrical tub of radius $5cm$ and length $9.8cm$ is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemisphere is $3.5cm$ and height of the cone outside the hemisphere is $5cm$, find the volume of the water left in the tub (Take $\pi =22/7$)

Answer 1

Volume of water in tub$=\pi r^{2}h=\pi \times 5^2\times 9.8c{m}^3$

Volume of Hemishpere$=\frac{2}{3}\times \pi \times {3.5}^{3}c{m}^3$

Volume of cone$=\frac{1}{3}\pi \times {3.5}^2\times 5c{m}^3$

Volume of water left in the tub = Volume of water in tub-Volume of toy

Volume left = $\pi \times 25\times 9.8-(\frac{2}{3}\times \pi \times {3.5}^3+\frac{1}{3}\pi \times {3.5}^2\times 5)$

$V=\pi (245-49)=\frac{22}{7}\times 196=22\times 28=616c{m}^3$