Chapter 5, Problem 67A

Interpretation Introduction

Interpretation : Whether the spherical ball sink or float on the water should be checked according to the given information.

Concept Introduction :

The density of the substance is the mass per unit volume. It is given as,

  $\rho =\frac{m}{V}$

Where, $\rho$ is the density of the object

  $m$ is the mass of the object

  $V$ is the volume of the object

Answer to Problem 67A

The density of the ball is higher than the density of the water. So, it sinks into the water.

Explanation of Solution

Given information :

The radius of the ball is, $r=0.5\text{0 cm}$

Mass of the ball is, $m=\text{ 0}\text{.2 g}$

The density and volume can be calculated as follows:

1. $\rho =\frac{m}{V}$
2. $V_{ball}=\frac{4}{3}\pi r^3$

Firstly, the volume of the spherical ball ids calculated by substituting the values in the formula, such that,

  $\begin{array}{c}{V}_{\text{ball}}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\times 3.14\times {\left(0.50cm\right)}^3\\ =0.524c{m}^3\end{array}$

Now use the volume for calculating the density of the ball,

  $\begin{array}{c}\rho =\frac{m}{V}\\ =\frac{\text{0}\text{.2g}}{\text{0}{\text{.524cm}}^{\text{3}}}\\ \text{=3}{\text{.82g/cm}}^{\text{3}}\end{array}$

Conclusion

The density of the ball is $\text{3}{\text{.82 g/cm}}^{\text{3}}$ which is greater than the density of the water.