Chapter 19, Problem 24A

An 8.6-kg piece of metal displaces 1 liter of water when submerged. Calculate its density.

To determine

Density of metal.

Answer to Problem 24A

Density $\rho =8600{\text{kg/m}}^3$

Explanation of Solution

Given:

Mass of metal, $m=8.6\text{kg}$

Volume of water displaced, $V=1\text{liter}$

Formula Used:

Density is given as,

  $\rho =\frac{m}{V}$

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

Calculation:

As, $1{\text{m}}^3=1000\text{liter}$

So, $1\text{liter}={10}^{-3}{\text{m}}^3$

Now, the density of metal is

  $\begin{array}{l}\rho =\frac{m}{V}=\frac{8.6}{{10}^{-3}}{\text{kg/m}}^3\\ \Rightarrow \rho =8600{\text{kg/m}}^3\end{array}$

Conclusion:

Therefore, the density of metal is $\rho =8600{\text{kg/m}}^3$ .