Question

Three liquids having densities ${\rho }_1,{\rho }_2$ and ${\rho }_3$ are filled in a U-tube. Length of each liquid column is equal to I, ${\rho }_1>{\rho }_2>{\rho }_3$ and liquids remain at rest (relative to the cube) in the position shown in the figure. It is possible that:

• 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

Answer 1

Answer: (D)

None of the above

Since both the vertical limbs of U-tube are open to atmosphere, at surface of liquids, the pressure is equal to the atmospheric pressure. The liquids have different 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 accelerating down with acceleration g. In that case, liquids experience weightlessness. Hence, at every point of the tube, the pressure will become equal to atmospheric pressure.
But if the tube is not accelerating 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 greater than that at the bottom of the right vertical limb. It means on horizontal limb there will be a 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 so that the inertia force which is in the left direction balances the rightward force.
Obviously, only option (d) is correct.