Surface Areas and Volumes — Question 4

Given,

Diameter of the spherical ball (d) = 0.21 m

Radius of the spherical ball (r) = Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{0.21}{2}$ m = 0.105 m

By formula,

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

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (0.105)^3 \\[1em] = \dfrac{88}{21} \times 0.001157625 \\[1em] = \dfrac{88}{21} \times 0.001157625 \\[1em] = \dfrac{0.101871}{21} \\[1em] = 0.004851\text{ m}^3.$

Hence, the amount of water displaced = 0.004851 m³.