Question

A cylinder of radius 12 cm. contains water to a depth of 20 cm. When a spherical iron ball is dropped in to the cylinder and thus the level of water is raised by 6.75 Find the
radius of the ball.​

Answer 1

Solution

Refer to attachment

Radius of the cylinder = 12 cm

Now, consider the raised portion

Radius of the raised portion cylinder $r_1$ = 12 cm

Height of the raised portion cylinder h = 6.75 cm

Let the radius of the iron ball be $r_2 \ cm$

Here,

Volume of the iron ball i.e sphere = Volume of the rasied portion cylinder

$\implies \dfrac{4}{3} \pi r_2 ^{3} = \pi r _1^{2} h$

Cancelling π on both sides

$\implies \dfrac{4}{3} r_2 ^{3} = r _1^{2} h$

Substituting the given values

$\implies \dfrac{4}{3} r_2 ^{3} = 12^{2} \times 6.75$

$\implies \dfrac{4}{3} r_2 ^{3} = 144 \times 6.75$

$\implies \dfrac{4}{3} r_2 ^{3} =972$

$\implies r_2 ^{3} =972 \times \dfrac{3}{4}$

$\implies r_2 ^{3} =972 \times \dfrac{3}{4}$

$\implies r_2 ^{3} =729$

$\implies r_2 ^{3} = {9}^{3}$

$\implies r_2 = {9}$

Hence, radius on the iron ball is 9 cm.