Question

A cylindrical tub of radius $12cm$ contains water upto the height of $20cm$. A spherical iron ball of radius $9cm$ is dropped into the tub and thus water level is raised by $hcm$. What is the value of $h$.

Answer 1

$\begin{array}{c}Volumeofcylinder+Volumeofsphere=Volumeofhighcylinder\\ \Rightarrow \pi \times 12\times 12\times 20+\frac{4}{3}\times \pi \times 9^3=\pi \times 12\times 12\left(20+h\right)\\ \Rightarrow \pi \left[144\times 20+4\times 81\times 3\right]=\pi \times 144\left(20+h\right)\\ \Rightarrow 144\times 20+121\times 81=144\times 20+144h\\ \Rightarrow \frac{12\times 81}{144}=h\\ \Rightarrow h=\frac{27}{4}=6.75cm\end{array}$