Question

A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball ?

Answer 1

Let $r$ cm be the radius of iron ball.

Level of water raised by $9$ cm

Then,

volume of iron ball = level of water raised

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

$\frac{4}{3}\pi r^3=\pi \times (16{)}^2\times 9$

$r^3=64\times 9\times 3$

$r=4\times 3=12$ $cm$