Chapter 10, Problem 28P

To determine

To Find: The true mass of the aluminum piece

Answer to Problem 28P

The mass of the aluminum is $2.0009\mathrm{kg}$ .

Explanation of Solution

Given info:

Density of air, ${\rho }_0=1.225{\text{kg/m}}^{\text{3}}$

Density of aluminum, ${\rho }_{Al}=2.7\times {10}^3{\text{kg/m}}^{\text{3}}$

Apparent mass of Al, $m_{Al}=2.0000\mathrm{kg}$

Formula used:

Weight in vacuum −air buoyancy = $mg$

Calculation:

Weight in vacuum,

  $\begin{array}{l}W=mg\\ W=\rho Vg\end{array}$

Air buoyancy acting on aluminum,

  $B={\rho }_{0}Vg$

Volume is the same because the volume of the aluminum displaces an equal amount of volume air. So,

Apparent weight is given as,

  $\begin{array}{l}\rho Vg-{\rho }_{0}Vg=mg\\ \rho V-{\rho }_{0}V=m\\ V\left(\rho -{\rho }_0\right)=m\\ V=\frac{2.00}{\left(\rho -{\rho }_0\right)}\left(\because m=2.000\text{kg}\right)\end{array}$

But $m=\rho V$

So, $\therefore m=\frac{2.00\rho }{\left(\rho -{\rho }_0\right)}=\frac{2\times \left(2.7\times 10^3\right)}{\left(2.7\times {10}^3-1.225\right)}=2.0009\mathrm{kg}$

Conclusion:

The actual mass of the aluminum is $2.0009\mathrm{kg}$ .