Using a scale, a piece of alloy has a measured mass of 86 g in air and 73 g when immersed in water. Find its volume and its density.
The apparent change in measured mass is due to the buoyant force of the water. Figure 13-7 shows the situation when the object is in water. From the figure, $F_{B}+F_{T}=m g$, so
$F_{B}=(0.086)(9.81) \mathrm{N}-(0.073)(9.81) \mathrm{N}=(0.013)(9.81) \mathrm{N}$
But $F_{B}$ must be equal to the weight of the displaced water.
$$
\begin{aligned}
F_{B} &=\text { Weight of water }=\text { (Mass of water) }(g) \\
&=\text { (Volume of water)(Density of water })(g)
\end{aligned}
$$
or $(0.013)(9.81) \mathrm{N}=V\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)$
from which $V=1.3 \times 10^{-5} \mathrm{~m}^{3}$. This is also the volume of the piece of alloy. Therefore.
$$
\rho \text { of alloy }=\frac{\text { Mass }}{\text { Volume }}=\frac{0.086 \mathrm{~kg}}{1.3 \times 10^{-5} \mathrm{~m}^{3}}=6.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$