Chapter 8, Problem 2TS

A $6\text{-kg}$ piece of metal displaces $1\text{L}$ of water when submerged. Show that its density is $6000\text{kg}/{\text{m}}^{\text{3}}$. How does this compare with the density of water?

To determine

The density of water is $6\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ and the comparison of the density of metal piece with water.

Answer to Problem 2TS

The density of water is $6\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ and the density of metal piece is $6000$ times denser than water.

Explanation of Solution

Given info: The mass of the metal piece is $6\text{kg}$, and the volume of displaced water is $1\text{L}$.

Write the expression for density.

$d=\frac{m}{V}$

Here,

$m$ is the mass of the metal piece

$V$ is the volume of displaced water

Substitute $6\text{kg}$ for $m$ and $1\text{L}$ for $V$ in the above equation to get $d$.

$\begin{array}{l}d=\frac{6\text{kg}}{1\text{L}\times \frac{1{\text{m}}^3}{1000\text{L}}}\\ d=6\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Since, the density of water is $6\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, so it is $6000$ times denser than water.

Conclusion:

Therefore, the density is $6\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ and the density of metal piece is $6000$ times denser than water.