First slide

Fluid statics

Question

A U-tube contains two immiscible non-reacting liquids. P_A, P_B, P_C and P_D are pressures at points A, B, C and D respectively. Then

Moderate

A

$P_{A} = P_{B}$
B

$P_{C} > P_{D}$
C

$P_{A} < P_{B}$
D

$P_{A} > P_{B}$

Solution

If y be the height of a point in a liquid from the axis of the horizontal part of the tube and P be the pressure at that point, then $\frac{dP}{dy} = - ρg$ where $ρ$ is the density of liquid. Liquid 1 is lighter than liquid 2. So up to the interface of liquids in the left arm, $\frac{dP}{dy} = - ρ_{2} g$ is the same in both arms, hence $P_{C} = P_{D}$ . But above the interface of liquids $\frac{dP}{dy} = - ρ_{1} g$ in the left arm and $\frac{dp}{dy} = - ρ_{2} g$ in the right arm. Since $ρ_{1} < ρ_{2}$ , we can conclude $P_{A} > P_{B}$ .

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