Question

When at rest, a liquid stands at the same level in the tubes as shown in the figure. But as indicated, a height difference h occurs when the system is given an acceleration a towards the right. Then h is equal to:

A. $\dfrac{{aL}}{{2g}}$
B. $\dfrac{{gL}}{{2a}}$
C. $\dfrac{{gL}}{a}$
D. $\dfrac{{aL}}{g}$

Answer 1

Hint: It is known fact that the liquid has some mass. Now due to that mass there will be weight and pressure acting at every point of the liquid. If the liquid is in static condition then the pressure at points of same height level will be the same for the liquid of same density and different pressures can’t exist for one point. By using this property we will solve this question
Formula used:
$p = \rho {a_{eff}}{x_{eff}}$

Complete answer:
The basic property of any liquid in static condition is it will align its surface perpendicular to the net force. Let us take the example of a liquid in the beaker. The force acting will be its weight and it will be acting vertically downward. Now the surface will be horizontal i.e perpendicular to the net force.
If the surface is not perpendicular then the liquid will be having acceleration along the direction of force and liquid will no longer be in equilibrium.

If we consider the point which is exactly ‘h’ distance down to the top surface of the left limb, the vertical pressure due to gravity will be $p = \rho {a_{eff}}{x_{eff}}$
Along the vertical effective acceleration is acceleration due to gravity i.e ‘g’ and effective ‘x’ is ‘h’
$p = \rho {a_{eff}}{x_{eff}}$
$ \Rightarrow p = \rho gh$
the horizontal pressure due to pseudo acceleration ‘a’ will be $p = \rho {a_{eff}}{x_{eff}}$
Along the horizontal effective acceleration is acceleration due to pseudo force i.e ‘a’ and effective ‘x’ is ‘L’
$p = \rho {a_{eff}}{x_{eff}}$
$ \Rightarrow p = \rho aL$
Since two different pressures can’t exist for the same point of the liquid, equate the horizontal pressure to the vertical pressure. Then we will get
$\eqalign{
  & \rho gh = \rho aL \cr
  & \Rightarrow h = \dfrac{{aL}}{g} \cr} $

So, the correct answer is “Option D”.

Note:
The resultant force will be the resultant of weight and the pseudo force and the surface of the liquid will be perpendicular to the resultant force vector. We can solve the same problem by assuming liquid is in a beaker and moves with the same acceleration and then also we would get the same answer.