A cylindrical tub of radius 12 cm contains water upto a depth of 20 cm. A spherical iron ball is dropped into the tub and is fully immersed in it. Thus, the level of water is raised by 6.75 cm. Find the radius of the ball.

Let the radius of the sphere be r cm.

Radius of cylinder, R = 12 cm

SInce, a spherical iron ball is dropped into the tub and is fully immersed in it.

Height of water raised by 6.75 cm.

∴ h = 6.75 cm.

Volume of water rise in cylinder = Volume of sphere

$\Rightarrow π\text{R}^2\text{h} = \dfrac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \dfrac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \dfrac{3}{4} \times \text{R}^2 \times \text{h} \\[1em] \Rightarrow \text{r}^3 = \dfrac{3}{4} \times 12^2 \times 6.75 \\[1em] \Rightarrow \text{r}^3 = \dfrac{3}{4} \times 144 \times 6.75 \\[1em] \Rightarrow \text{r}^3 = \dfrac{2916}{4} \\[1em] \Rightarrow \text{r}^3 = 729 \\[1em] \Rightarrow \text{r} = \sqrt[3]{729} \\[1em] \Rightarrow \text{r} = 9 \text{ cm.}$

Hence, the radius of the ball is 9 cm.