Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.0000 kg when weighed in air.

The difference between the true and apparent mass of a body is the air mass displaced by the aluminum. The difference is due to the effect of air buoyancy.

The apparent mass of the aluminum in the air is m_apparent= 4.0000 kg.

As discussed above,

\begin{aligned}{m_{{\rm{actual}}}}-{m_{{\rm{apparent}}}}={m_{{\rm{air}}}}\\{m_{{\rm{actual}}}}-{m_{{\rm{apparent}}}}={\rho_{{\rm{air}}}}\cdot{V_{{\rm{Al}}}}\\{m_{{\rm{actual}}}}-{m_{{\rm{apparent}}}}={\rho_{{\rm{air}}}}\frac{{{m_{{\rm{actual}}}}}}{{{\rho_{{\rm{Al}}}}}}\end{aligned}

Here, m_actual is the true/actual mass of the aluminum, ρ_air is the density of air, ρ_AI is the density of aluminum, and V_AI is the aluminum piece’s volume.

\begin{aligned}{m_{{\rm{actual}}}}-{\rho_{{\rm{air}}}}\frac{{{m_{{\rm{actual}}}}}}{{{\rho_{{\rm{Al}}}}}}={m_{{\rm{apparent}}}}\\{m_{{\rm{actual}}}}\left({1-\frac{{{\rho_{{\rm{air}}}}}}{{{\rho_{{\rm{Al}}}}}}}\right)={m_{{\rm{apparent}}}}\\{m_{{\rm{actual}}}}=\frac{{{m_{{\rm{apparent}}}}}}{{\left({1-\frac{{{\rho_{{\rm{air}}}}}}{{{\rho_{{\rm{Al}}}}}}}\right)}}\end{aligned}

Now, considering the standard values of ρ_air and ρ_AI we have,

ρ_air =1.29 kg/m³

ρ_AI = 2.70 x 10 kg/m³

Substitute these values in the above formula.

\begin{aligned}{m_{{\rm{actual}}}}=\frac{{4.0000\;{\rm{kg}}}}{{1-\frac{{1.29\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}}}{{2.70\times{{10}^3}\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}}}}}\\{m_{{\rm{actual}}}}=4.0019\;{\rm{kg}}\end{aligned}

Hence the true mass of aluminum is 4.0019 kg.