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# A U-tube contains two immiscible non-reacting liquids. PA, PB, PC and PD are pressures at points A, B, C and D respectively. Then

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a
PA=PB
b
PC>PD
c
PA
d
PA>PB

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detailed solution

Correct option is D

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 dPdy=−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, dPdy=−ρ2g is the same in both arms, hence PC=PD. But above the interface of liquids dPdy=−ρ1g in the left arm and dpdy=−ρ2g in the right arm. Since ρ1<ρ2, we can conclude PA>PB.

Similar Questions

Two immiscible liquids P and Q of different densities are  contained in a wide U-tube as shown in Fig. 12.54. The heights of the two liquids above the horizontal line XX' which cuts the boundary between the liquids are Hp and HQ respectively. The U-tube is transported to a planet where the acceleration of free fall is $\frac{2}{3}$ that on the earth, where the 3 liquids do not evaporate and where the heights of liquid  (measured relative to XX' ) are hp and hQ respectively.
Which of the given statements is correct?