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Step-by-Step Solution
Step 1: Identify the given quantities and convert units to SI (if necessary)
• Cross-sectional area of each cylindrical vessel, $A = 16 \text{ cm}^2 = 16 \times 10^{-4} \text{ m}^2 = 0.0016 \text{ m}^2$.
• Initial heights of water are 100 cm ($1 \text{ m}$) and 150 cm ($1.5 \text{ m}$).
• Density of water, $\rho = 10^3 \text{ kg/m}^3$.
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$.
Step 2: Calculate the total volume of water
Since each vessel has cross-section $A$ and heights of $H_1 = 1 \text{ m}$ and $H_2 = 1.5 \text{ m}$, the volumes of water are:
$V_1 = A \times H_1$ and $V_2 = A \times H_2.$
Therefore,
$V_1 = 0.0016 \times 1 = 0.0016 \text{ m}^3,$
$V_2 = 0.0016 \times 1.5 = 0.0024 \text{ m}^3.$
Total volume of water,
$V_{\text{total}} = V_1 + V_2 = 0.0016 + 0.0024 = 0.004 \text{ m}^3.$
Step 3: Find the final common height of water in each vessel
When interconnected, both vessels will have the same final water height $H$. Because each vessel still has the same cross-section $A$, and there are two vessels, the total cross-sectional area combined is $2A$ if we think of them together. However, an easier approach is:
The total volume $V_{\text{total}}$ is distributed equally in two vessels that each have cross-section $A$, so each vessel will contain $V_{\text{total}}/2$ volume. The height in each vessel is the same, call it $H$:
$A \times H + A \times H = V_{\text{total}} \quad\Longrightarrow\quad 2A \times H = V_{\text{total}},$
so
$H = \frac{V_{\text{total}}}{2A} = \frac{0.004}{2 \times 0.0016} = 1.25 \text{ m}.$
Step 4: Calculate the initial potential energy ($E_{\text{in}}$)
Each column of water has its center of mass at half its height. Hence, for the first vessel:
• Mass: $m_1 = \rho \, A \, H_1 = 10^3 \times 0.0016 \times 1 = 1.6 \text{ kg}.$
• Center of mass at height $H_1/2 = 0.5 \text{ m}.$
• Potential energy: $U_1 = m_1 g \frac{H_1}{2}.$
Similarly, for the second vessel:
• Mass: $m_2 = \rho \, A \, H_2 = 10^3 \times 0.0016 \times 1.5 = 2.4 \text{ kg}.$
• Center of mass at height $H_2/2 = 0.75 \text{ m}.$
• Potential energy: $U_2 = m_2 g \frac{H_2}{2}.$
The total initial potential energy is thus:
$E_{\text{in}} = U_1 + U_2 \;=\; \rho \, g \, \frac{A}{2} (H_1^2 + H_2^2).$
Substituting $H_1=1 \text{ m}$ and $H_2=1.5 \text{ m}$:
$E_{\text{in}} = (10^3) \,(10)\,\frac{(0.0016)}{2}\,\Bigl(1^2 + (1.5)^2\Bigr) = 1000 \times 10 \times 0.0008 \times (1 + 2.25) = 8 \times 3.25 = 26 \text{ J}.$
In actual numeric detail, we will only need the difference between initial and final energies, but this shows the method.
Step 5: Calculate the final potential energy ($E_{\text{fin}}$)
After the water levels become equal, each vessel has the same height $H = 1.25 \text{ m}$. The mass in each is $m = \rho \, A \, H,$ and the center of mass for each column is at $H/2.$ The final potential energy in one vessel is:
$U_{\text{one\_vessel}} = \left(\rho A H\right)\, g \,\bigl(\tfrac{H}{2}\bigr)
= \rho \, g \, A \, \frac{H^2}{2}.$
With two vessels, the total final potential energy is:
$E_{\text{fin}} = 2 \times \rho g A \frac{H^2}{2} = \rho g A \, H^2.$
Substitute $H=1.25 \text{ m}$:
$E_{\text{fin}} = 1000 \times 10 \times 0.0016 \times (1.25)^2 = 16 \times 1.5625 = 25 \text{ J}.$
Step 6: Determine the work done by gravity
The work done by gravity is the decrease in gravitational potential energy:
$W = E_{\text{in}} - E_{\text{fin}}.$
From the simplified symbolic form:
$E_{\text{in}} = \rho g \frac{A}{2} (H_1^2 + H_2^2), \quad
E_{\text{fin}} = \rho g A (H^2).$
Substituting numbers (or directly using the difference):
$(H_1^2 + H_2^2) = 1^2 + 1.5^2 = 1 + 2.25 = 3.25,$
$H^2 = (1.25)^2 = 1.5625.$
Hence:
$E_{\text{in}} = \rho g \frac{A}{2} \times 3.25, \quad
E_{\text{fin}} = \rho g A \times 1.5625.$
Factor out $\rho g A$:
$W = \rho g A \left[\frac{3.25}{2} - 1.5625\right]
= \rho g A \bigl[1.625 - 1.5625\bigr]
= \rho g A \times 0.0625.$
And since $\rho g A = 1000 \times 10 \times 0.0016 = 16$:
$W = 16 \times 0.0625 = 1 \text{ J}.$
Final Answer
The work done by the force of gravity during the process is $1 \text{ J}$.
