Answer
From the physical properties of water tables, at $20^{\circ} \mathrm{C}$ obtain
The Density of water, $\rho_{H_{2} o}=998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}$
The specific weight of water $\gamma_{H_{2} o}=9789 \frac{\mathrm{N}}{\mathrm{m}^{3}}$
The Specific gravity of water $S G=0.9982$
For water at $20^{\circ} \mathrm{C}, \quad \gamma_{H_{2} \mathrm{O}}=9789 \frac{\mathrm{N}}{\mathrm{m}^{3}} ; \rho_{H_{2} \mathrm{O}}=998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}} ; S G=0.9982$
A comparison of these values for water with those for the pop shows that the specific weight, density and specific gravity of the pop are slightly lower than the corresponding values for water.
Work Step by Step
Calculate the Total weight of can including fluid:
$$
W=m g
$$
Here, mass of fluid is $m$; acceleration due to gravity is $g$
Substitute, $0.369 \mathrm{~kg}$ for $m, 9.81 \frac{\mathrm{m}}{\mathrm{sec}^{2}}$ for $g$
$$
\begin{aligned}
W &=(0.369 \mathrm{~kg})\left(9.81 \frac{\mathrm{m}}{\mathrm{sec}^{2}}\right) \\
&=3.6189 \mathrm{~kg} \cdot \frac{\mathrm{m}}{\mathrm{sec}^{2}} \\
&=3.62 \mathrm{~N}
\end{aligned}
$$
Calculate the weight of fluid:
Weight of fluid $W_{f}=($ Total weight of can including fluid) $-$ (weight of can)
$$
\begin{array}{l}
=(3.62)-(0.153) \\
=3.467 \mathrm{~N}
\end{array}
$$
Calculate the specific weight:
$$
\begin{array}{l}
\gamma=\frac{\text { weight of fluid }}{\text { volume of fluid }} \\
=\frac{W_{f}}{V_{f}} \ldots \ldots(1)
\end{array}
$$
Convert the units of Volume of fluid from milli Liters to $\mathrm{m}^{3}$ :
$$
\begin{aligned}
V &=\left(355 \times 10^{-3} \mathrm{~L}\right)\left(10^{-3} \frac{\mathrm{m}^{3}}{\mathrm{~L}}\right) \\
&=355 \times 10^{-6} \mathrm{~m}^{3}
\end{aligned}
$$
Substitute, $3.467 \mathrm{~N}$ for $W_{f}, \quad 355 \times 10^{-6} \mathrm{~m}^{3}$ for $V$ in $(1)$
$$
\begin{aligned}
\gamma &=\frac{3.467}{355 \times 10^{-6}} \\
&=9766 \frac{\mathrm{N}}{\mathrm{m}^{3}}
\end{aligned}
$$
Calculate the Density of fluid:
$$
\rho=\frac{\gamma}{g}
$$
Substitute, $9766 \frac{\mathrm{N}}{\mathrm{m}^{3}}$ for $\gamma, 9.81 \frac{\mathrm{m}}{\mathrm{sec}^{2}}$ for $g$
$$
\begin{aligned}
\rho &=\frac{9766.19 \frac{\mathrm{N}}{\mathrm{m}^{3}}}{9.81 \frac{\mathrm{m}}{\mathrm{sec}^{2}}} \\
&=995.53 \frac{\mathrm{kg}}{\mathrm{m}^{3}}
\end{aligned}
$$
Calculate the Specific gravity,
$$
S G=\frac{\rho}{\rho_{W}}
$$
Density of water is $\rho_{w}=1000 \mathrm{~kg} / \mathrm{m}^{3}$
Substitute, $1000 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho_{W}, 995.53 \frac{\mathrm{kg}}{\mathrm{m}^{3}}$ for $\rho$
$$
\begin{aligned}
S G &=\frac{995.53}{1000} \\
&=0.99553
\end{aligned}
$$
Therefore, the Density, The specific weight of water and The Specific gravity of the fluid is
$$
995.53 \frac{\mathrm{kg}}{\mathrm{m}^{3}}, \sqrt{9766 \frac{\mathrm{N}}{\mathrm{m}^{3}}} \text { and } 0.99553
$$
From the physical properties of water tables, at $20^{\circ} \mathrm{C}$ obtain
The Density of water, $\rho_{H_{2} o}=998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}$
The specific weight of water $\gamma_{H_{2} o}=9789 \frac{\mathrm{N}}{\mathrm{m}^{3}}$
The Specific gravity of water $S G=0.9982$
For water at $20^{\circ} \mathrm{C}, \quad \gamma_{H_{2} \mathrm{O}}=9789 \frac{\mathrm{N}}{\mathrm{m}^{3}} ; \rho_{H_{2} \mathrm{O}}=998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}} ; S G=0.9982$
A comparison of these values for water with those for the pop shows that the specific weight, density and specific gravity of the pop are slightly lower than the corresponding values for water.
