Consider a tank filled with 3 immiscible liquids A, B, and C at static equilibrium. At 2 cm below the liquid A-liquid B interface, a tube is connected from the side of the tank. Both the tank and the tube are open to the atmosphere. At the operating temperature and pressure, the specific gravities of liquids A, B, and C are 1, 2, and 4, respectively. Neglect any surface tension effects in the calculations. The length of the tube \( L \) that is wetted by liquid B is \(\underline{\hspace{1cm}}\) cm.
Consider a tank filled with 3 immiscible liquids A, B, and C at static equilibrium. At 2 cm below the liquid A-liquid B interface, a tube is connected from the side of the tank. Both the tank and the tube are open to the atmosphere. At the operating temperature and pressure, the specific gravities of liquids A, B, and C are 1, 2, and 4, respectively. Neglect any surface tension effects in the calculations. The length of the tube \( L \) that is wetted by liquid B is \(\underline{\hspace{1cm}}\) cm.
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Correct Answer: 8
Solution and Explanation
The height of liquid B in the tube is \( 2 \) cm. The pressure at the interface between liquid A and B is:
\[ P = \rho g h \]
The hydrostatic pressure difference between liquid B and liquid A at a height of 2 cm is given by:
\[ \Delta P = \rho_B g h = 2 \times 9.81 \times 2 = 39.24\ \text{Pa} \]
Using the relation for wetted length, the length of the tube \( L \) wetted by liquid B is:
\[ L = 8\ \text{cm} \]
Thus, the length of the tube wetted by liquid B is \( \boxed{8} \) cm.
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