Question

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

• A. $U\text{tube}$ is accelerating leftwards
• B. $U\text{tube}$ is accelerating upwards with acceleration $g$
• C. $U\text{tube}$ is moving with a constant velocity
• D. None of the above

Answer 1

The correct option is D None of the above

Since both the vertical limbs of $U\text{tube}$ are open to atmosphere, at the surface of liquids, the pressure is equal to the atmospheric pressure.

$\left(1\right)$ Since the heights of the liquids in the two vertical limbs are equal and the liquids have different densities, pressure exerted by them at the bottom will be different from each other.
Thus, 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.

$\left(2\right)$ 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 a horizontal limb, there will be a resultant horizontal force, which will be towards the right.
Therefore, the liquid in the horizontal limb will have a rightward acceleration.
In fact, this whole system must have a horizontally right-ward acceleration.

Therefore, option $\left(\text{A}\right)$, $\left(\text{B}\right)$ and $\left(\text{C}\right)$ is wrong.
Hence, option $\left(\text{D}\right)$ is correct.