Chapter 18, Problem 42A

a.

To determine

Volume of the cube.

Answer to Problem 42A

  $V=64{\text{cm}}^3$

Explanation of Solution

Given:

Side of cube, $a=4\text{cm}$

Mass of cube, $m=672\text{g}$

Formula used:

Volume of the cube is,

  $V=a^3$

Where $a$ is the side of cube.

Calculation:

Putting the values in above formula,

Volume of the cube is

  $\begin{array}{l}V={\left(4\text{cm}\right)}^3\\ \Rightarrow V=64{\text{cm}}^3\end{array}$

Conclusion:

Therefore, the volume of the cube is $V=64{\text{cm}}^3$ .

b.

To determine

Total surface area of cube.

Answer to Problem 42A

  $S=24{\text{cm}}^2$

Explanation of Solution

Given:

Side of cube, $a=4\text{cm}$

Mass of cube, $m=672\text{g}$

Formula used:

The total surface area of the cube is

  $S=6a^2$

Where $a$ is the side of cube.

Calculation:

On substituting the value in above formula,

Total surface area of the cube,

  $\begin{array}{l}S=6{\left(2\text{cm}\right)}^2\\ \Rightarrow S=24{\text{cm}}^2\end{array}$

Conclusion:

Therefore, the total surface area of the cube is $S=24{\text{cm}}^2$ .

c.

To determine

The density of the cube.

Answer to Problem 42A

  $\rho =10.5{\text{g/cm}}^3$

Explanation of Solution

Given:

Side of cube, $a=4\text{cm}$

Mass of cube, $m=672\text{g}$

Formula used:

The density is given by,

  $\rho =\frac{m}{V}$

Where $m$ is mass and $V$ is volume.

Calculation:

From previous sub-part,

  $V=64{\text{cm}}^3$

On substituting the values in above formula, the density is

  $\begin{array}{l}\rho =\frac{672}{64}{\text{g/cm}}^3\\ \Rightarrow \rho =10.5{\text{g/cm}}^3\end{array}$

Conclusion:

Therefore, the density of the cube is $\rho =10.5{\text{g/cm}}^3$ .