Question

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

Answer 1

Given the radius of the cylinder, r = 12 cm
It is also given that a spherical iron ball is dropped into the cylinder and the water level raised by 6.75 cm
Hence volume of water displaced = volume of the iron ball
Height of the raised water level, h = 6.75 m
Volume of water displaced = $\pi$$r^2$h
= π $\times$ 12 $\times$ 12 $\times$ 6.75 $c{m}^3$
⇒ Volume of iron ball = π $\times$ 12$\times$ 12 $\times$6.75 cm³ → (1)
But, volume of iron ball = $\frac{4}{3}$$\pi$$r^3$ ---(2)
From (1) and (2) we get

$\frac{4}{3}$$\pi$$r^3$ ​​​​​​​ = $\pi$ $\times$ 12 $\times$ 12 $\times$ 6.75​​​​​​​

$r^3$ = $\pi$ $\times$ 12 $\times$ 12 $\times$ 6.75 $\times$ $\frac{3}{4}$​​​​​​​

$r^3$​​​​​​​ = 729

$r^3$​​​​​​​ = $3^3$

r = 9

Thus the radius of the iron ball is 9 cm