Surface Areas and Volumes — Question 3

Given,

Diameter of the spherical ball (d) = 28 cm

Radius of the spherical ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{28}{2}$ cm = 14 cm.

By formula,

The amount of water displaced = volume of the sphere (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (14)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 2744 \\[1em] = \dfrac{88}{21} \times 2744 \\[1em] = \dfrac{241472}{21} \\[1em] = 11498\dfrac{2}{3} \text{ cm}^3.$

Hence, the amount of water displaced = $11498\dfrac{2}{3}$ cm³.