Answer
Substitute, $0.0186 \mathrm{ft}^{3}$ for $\Delta V$, $2 \mathrm{ft}$ for $d$
$$
\Delta l=\frac{0.0186}{\frac{\pi}{4} \times 2^{2}}
$$
$$
=0.00592 \mathrm{ft}
$$
Therefore, this small change in depth would not be noticeable.
Note: A slightly different value for ?/ will be obtained if specific weight of water is used rather than density. Since there is some uncertainty in the values and solution is sensitive to this uncertainity.
Work Step by Step
Since, mass must remain constant as the temperature changes, calculate the volume of water at $90^{\circ} \mathrm{F}$ by using the following equation:
$m_{40^{\circ} \mathrm{F}}=m_{90^{\circ} \mathrm{F}}$
$V_{40^{\circ} \mathrm{F}} \times \rho_{40^{\circ} \mathrm{F}}=V_{90^{\circ} \mathrm{F}} \times \rho_{90^{\circ} \mathrm{F}} \ldots \ldots$ (1)
Here, mass of fluid is $m$, volume of fluid is $V$ and density of fluid is $\rho$.
From physical properties of water table, obtain the density values at $40^{\circ} \mathrm{F}$ and $90^{\circ} \mathrm{F}$
$\rho_{H_{2} O @ 40^{\circ} \mathrm{F}}=1.94 \frac{\text { slugs }}{\mathrm{ft}^{3}}$
$\rho_{H_{2} O @ 90^{\circ} \mathrm{F}}=1.931 \frac{\text { slugs }}{\mathrm{ft}^{3}}$
Substitute, $1.94 \frac{\text { slugs }}{\mathrm{ft}^{3}}$ for $\rho_{H_{2} O @ 40^{\circ} \mathrm{F}}, \quad 1.931 \frac{\text { slugs }}{\mathrm{ft}^{3}}$ for $\rho_{H_{2} O @ 90^{\circ} \mathrm{F}}, 4 \mathrm{ft}^{3}$ for $V_{40^{\circ} \mathrm{F}}$ in $(1)$
$$
\begin{aligned}
V_{90^{\circ} \mathrm{F}} &=\frac{4 \times 1.94}{1.931} \\
&=4.0186 \mathrm{ft}^{3}
\end{aligned}
$$
Calculate the increase in volume by using the following equation:
$$
\Delta V=V_{90^{\circ} \mathrm{F}}-V_{40^{\circ} \mathrm{F}}
$$
Substitute, $4.0186 \mathrm{ft}^{3}$ for $V_{90^{\circ} \mathrm{F}}, 4 \mathrm{ft}^{3}$ for $V_{40^{\circ} \mathrm{F}}$
$$
\begin{aligned}
\Delta V &=4.0186-4 \\
&=0.0186 \mathrm{ft}^{3}
\end{aligned}
$$
Calculate the change in water depth by using by using following equation:
$$
\begin{aligned}
\Delta l &=\frac{\Delta V}{\text { area of the tank }} \\
&=\frac{\Delta V}{\frac{\pi}{4} d^{2}}
\end{aligned}
$$
Substitute, $0.0186 \mathrm{ft}^{3}$ for $\Delta V$, $2 \mathrm{ft}$ for $d$
$$
\Delta l=\frac{0.0186}{\frac{\pi}{4} \times 2^{2}}
$$
$$
=0.00592 \mathrm{ft}
$$
Therefore, this small change in depth would not be noticeable.
Note: A slightly different value for ?/ will be obtained if specific weight of water is used rather than density. Since there is some uncertainty in the values and solution is sensitive to this uncertainity.
