Question.
The diameter of a metallic ball is $4.2 \mathrm{~cm}$. What is the mass of the ball, if the density of the metal is $8.9 \mathrm{~g}$ per $\mathrm{cm}^{3} ?\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
The diameter of a metallic ball is $4.2 \mathrm{~cm}$. What is the mass of the ball, if the density of the metal is $8.9 \mathrm{~g}$ per $\mathrm{cm}^{3} ?\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Radius $(r)$ of metallic ball $=\left(\frac{4.2}{2}\right) \mathrm{cm}=2.1 \mathrm{~cm}$
Volume of metallic ball $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}\right] \mathrm{cm}^{3}$
$=38.808 \mathrm{~cm}^{3}$
Density $=\frac{\text { Mass }}{\text { Volume }}$
Mass $=$ Density $\times$ Volume
$=(8.9 \times 38.808) \mathrm{g}$
$=345.3912 \mathrm{~g}$
Hence, the mass of the ball is $345.39 \mathrm{~g}$ (approximately).
Radius $(r)$ of metallic ball $=\left(\frac{4.2}{2}\right) \mathrm{cm}=2.1 \mathrm{~cm}$
Volume of metallic ball $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}\right] \mathrm{cm}^{3}$
$=38.808 \mathrm{~cm}^{3}$
Density $=\frac{\text { Mass }}{\text { Volume }}$
Mass $=$ Density $\times$ Volume
$=(8.9 \times 38.808) \mathrm{g}$
$=345.3912 \mathrm{~g}$
Hence, the mass of the ball is $345.39 \mathrm{~g}$ (approximately).