Solution
Analysis:
The tube connects the two vessels. Initially, the tube ends are at a depth of $18 \text{ cm}$ below the surface.
Since water density ($\rho_w = 1.0$) is greater than oil density ($\rho_o = 0.8$), the pressure at the left opening ($18g$) is higher than at the right opening ($14.4g$). When the valve opens, water flows from the left vessel, through the tube, into the right vessel.
Assumptions:
Since the vessels are identical, if water level drops by $x$ in the left vessel, a layer of water of height $x$ is formed at the bottom of the right vessel (as water is denser than oil). The oil layer (height $27 \text{ cm}$) floats on top of this new water layer.
Final State Calculation:
The connection points (tube openings) are fixed relative to the vessel bottom.
Initial level = $27 \text{ cm}$. Tube openings are at depth $18 \text{ cm}$, meaning they are at height $h_{tube} = 27 – 18 = 9 \text{ cm}$ from the bottom.
For equilibrium, pressures at the tube openings (height $9 \text{ cm}$) must be equal in both vessels.
Left Vessel (Water):
Final height $H_L = 27 – x$.
Pressure at $9 \text{ cm}$ height:
$$ P_L = \rho_w g (H_L – 9) = 1.0 \cdot g \cdot (27 – x – 9) = g(18 – x) $$
Right Vessel (Oil + Water):
Water layer at bottom has height $x$. Oil layer above has height $27$.
Assume $x < 9 \text{ cm}$ (the water layer is below the tube opening). The tube opening is inside the oil layer.
Depth of tube opening from free surface:
Total height $H_R = x + 27$. Height of opening = $9$.
Depth = $H_R - 9 = x + 18$.
Pressure at opening:
$$ P_R = \rho_o g (\text{Depth}) = 0.8 \cdot g \cdot (x + 18) $$
Equating Pressures:
$$ 1.0(18 – x) = 0.8(18 + x) $$
$$ 18 – x = 14.4 + 0.8x $$
$$ 3.6 = 1.8x \implies x = 2 \text{ cm} $$
Final Heights:
Water Vessel: $27 – x = 27 – 2 = \mathbf{25 \text{ cm}}$
Oil Vessel: $27 + x = 27 + 2 = \mathbf{29 \text{ cm}}$