Chapter 1, Problem 85A

To determine

To Calculate:The density of the metal.

Answer to Problem 85A

  $8.0{\text{g/cm}}^3$

Explanation of Solution

Given:

Volume of water is $V_w=10.0{\text{cm}}^{\text{3}}$

Mass of the sample metal, $m=17.6\text{g}$

Total volume when metal is placed in cylinder, $V=12.20{\text{cm}}^{\text{3}}$

Formula Used:

Density of an object can be obtained by:

  $\rho =\frac{m}{V}$

Where, $\rho$ is the density, m is the mass and V is the volume.

Calculations:

When metal is placed in cylinder containing water, the rise in level of water is equivalent to volume of metal.

  $\begin{array}{c}{V}_m=V-V_w\\ =12.20{\text{cm}}^3-10{\text{cm}}^3\\ =2.20{\text{cm}}^{\text{3}}\end{array}$

Density of the rectangular bar in ${\text{g/cm}}^{\text{3}}$ .

  $\begin{array}{c}\rho =\frac{17.6\text{g}}{\text{2}\text{.20}{\text{cm}}^3}\\ =8.0{\text{g/cm}}^3\end{array}$

Conclusion:

Thus, the density of the metal is $8.0{\text{g/cm}}^3$ .