Question

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

Answer 1

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball of radius 9cm into the tub, the height of the raised water is hcm. Therefore, the volume of the raised water is

$V=\pi \times {12}^2\times hc{m}^3$

The volume of the spherical ball is

$V_1=\frac{4}{3}\pi \times (9{)}^{3}c{m}^3$

Since, the volume of the raised water is same as the volume of the spherical ball, we have

$V_1=V$

$\frac{4}{3}\pi \times 9^3=\pi \times {12}^2\times h$

$h=\frac{4\times 9^3}{3\times {12}^2}$

$h=\frac{27}{4}=6.75cm$

Therefore, the height of the raised water is
H = ​​​​6.75 cm