Surface Areas and Volumes — Question 5

Given,

The diameter of the metallic ball (d) = 4.2 cm

Radius of the metallic ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{4.2}{2}$ cm = 2.1 cm

The density of the metal = 8.9 g per cm³

We know that,

Density = $\dfrac{\text{Mass}}{\text{Volume}}$

Mass = Density × Volume ...........(1)

By formula,

Volume of the metallic ball (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (2.1)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 9.261 \\[1em] = \dfrac{88}{21} \times 9.261 \\[1em] = \dfrac{814.968}{21} \\[1em] = \text{38.808 cm}^3$

Substituting value in equation (1), we get :

Mass of the metallic ball = 8.9 × 38.808

= 345.39 g

Hence, the mass of the ball is 345.39 g.