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Three liquids having densities `rho_(1),rho_(2), rho_(3)` are filled in a `U`-tube. Length of each liquid column is equal to `I`, `rho_(1)gtrho_(2)gtrho_(3)` and liquids remain at rest (relative to the tube) in the position shown in the figure. It is possible that
image
A. `U`-tube is accelerating leftwards
B. `U`-tube is accelerating upwards with acceleration `g`
C. `U-`tube is moving with a constant velocity
D. none of the above

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Correct Answer - D
Since both the verticla limbs of `U`-tube are open to atmosphere, at surface of liquids, the pressure is equal to the atmospheric pressure. Since heights of the liquids in the two vertical limbs are equal and the liquids have diferent densities, pressure exerted by them will be different from each other. THe liquids can remain in static equilibrium relative to the tube only, when the system is acelerating down with acceleration `g`. In that case liquids experiece weightlessness. Hence, at every point of the tube, the pressure will become equal to atmospheric pressure. But if the tube is not acelerating down under gravity, then the heavier liquid will exert more pressure at the bottom. Hence, at the bottom of the left vertical limb, pressure will be grater than that at the bottom of the left vertical limb, pressure will be greater than that at the bottom of the right vertical limb. It means on horizontal limb there will be resultant horizontal force which will be towards the right as shown in the figure.
Therefore the liquid in the horizontal limb will have a rightward acceleration. image
In fact, this whole system must have a horizontally rightward acceleration. Obviously, only option `d` is correct.

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