Question

A cylinderical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm.
What is the radius of the ball?

Answer 1

Radius of the cylinderical tub (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a spherical ball in it, the water raised by 6.75 cm
$\therefore Volumeofball=\pi r^{2}h=\pi \times 12\times 12\times 6.75c{m}^3=6.75\times 144\pi c{m}^3$
$\therefore$ Radius of the ball $=3\sqrt{\frac{Volume\times 3}{4\pi }}=3\sqrt{36\times 3\times 6.75}=3\sqrt{729}=9cm$