Answer
Calculate the specific gravity of the whipped cream using the following relation:
$$
S G=\frac{\rho_{\text {whipped cream }}}{\rho_{\text {water } @ 4^{\circ} C}}
$$
Here, the density of water at $4^{\circ} \mathrm{C}$ is $\rho_{\text {water } \mathbb{2} 4^{\circ} \mathrm{C}}$.
Substitute $335 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho_{\text {whippedcream }}$ and $1000 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho_{\text {water } \mathbb{4} 4^{\circ} \mathrm{C}}$.
$$
\begin{aligned}
S G &=\frac{335 \mathrm{~kg} / \mathrm{m}^{3}}{1000 \mathrm{~kg} / \mathrm{m}^{3}} \\
&=0.335
\end{aligned}
$$
Therefore, the specific gravity $S G$ of the whipped cream is $0.335$.
Work Step by Step
Calculate the mass of the cream using the following relation:
$m_{\text {cream }}=\rho V_{\text {cup }}$
Here, the density of cream is $\rho$ and the volume of cup is $V_{\text {cup }}$.
Substitute $1005 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho$.
$$
m_{\text {crom }}=\left[1005 \times V_{\text {cup }}\right] \mathrm{kg}
$$
Since, the mass of the cream is equal to mass of the whipped cream.
$$
\begin{aligned}
m_{\text {whipped crom }} &=m_{\text {cream }} \\
&=\left[1005 \times V_{\text {cap }}\right] \mathrm{kg}
\end{aligned}
$$
Calculate the density of whipped cream using the following relation:
$\rho_{\text {whipped croam }}=\frac{m_{\text {whipped cream }}}{3 V_{\text {cup }}}$
Substitute $\left[1005 \times V_{\text {cup }}\right] \mathrm{kg}$ for $m_{\text {whippedcream }} .$
$$
\begin{aligned}
\rho_{\text {whippedeream }} &=\frac{\left[1005 \times V_{\text {cup }}\right]}{3 V_{\text {cup }}} \\
&=\frac{1005}{3} \\
&=335 \mathrm{~kg} / \mathrm{m}^{3}
\end{aligned}
$$
Calculate the specific gravity of the whipped cream using the following relation:
$$
S G=\frac{\rho_{\text {whipped cream }}}{\rho_{\text {water } @ 4^{\circ} C}}
$$
Here, the density of water at $4^{\circ} \mathrm{C}$ is $\rho_{\text {water } \mathbb{2} 4^{\circ} \mathrm{C}}$.
Substitute $335 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho_{\text {whippedcream }}$ and $1000 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho_{\text {water } \mathbb{4} 4^{\circ} \mathrm{C}}$.
$$
\begin{aligned}
S G &=\frac{335 \mathrm{~kg} / \mathrm{m}^{3}}{1000 \mathrm{~kg} / \mathrm{m}^{3}} \\
&=0.335
\end{aligned}
$$
Therefore, the specific gravity $S G$ of the whipped cream is $0.335$.
