Question

A cylindrical tub of radius $12cm$ contains water to a depth of $20cm$. A spherical form ball of radius $9cm$ is dropped into the tub and thus the level of water is raised by $hcm$. What is the value of $h$?

Answer 1

Since rise in water level is solely caused by ball,

$\pi \times \left({12}^2\right)\times h=\frac{4}{3}\pi \times \left(9^3\right)$

$\Rightarrow h=6.75cm$