NIRF Ranking 2024: List of Top Engineering Colleges in India

Variation Of Pressure In An Accelerated Fluid - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

• 7 Questions around this concept.

Solve by difficulty

When a liquid is subjected to horizontal acceleration:

A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration towards right. Pressure is-

(i) maximum at, and (ii) minimum at-

An object of uniform density is allowed to float in water kept in a beaker. The object has triangular cross-section as shown in the figure. If the water pressure measured at three point A, B and C below the object are PA, PB and PC respectively. Then-

A cylindrical tank contains water upto a height H. If the tank is accelerated with acceleration a, then pressure at the point A is P1. If the tank is accelerated downwards with acceleration a the pressure at A is P2 then-

Concepts Covered - 1

Variation of Pressure in an accelerated fluid

Case I-  When Acceleration in the vertical direction

1. When the liquid container is moving with constant acceleration in an upward direction

Consider a cylindrical element of height h and Area A as shown in the below figure.

The force on the top face of the element =$P_1A$

The force on the bottom face of the element = $P_2A$

If a is the acceleration of the liquid then

We can write

$P_2A-(hA\rho g+P_1A)=ma$

Where m is the mass of the element of the liquid and which is given by

$m=\rho hA$

Where $\rho$=density of liquid

So using this we get

$P_2-P_1=\rho (g+a)h=\rho g_{eff}h$

1. When the liquid container is moving with constant acceleration in a downward direction

I.  constant  downward acceleration (a< g)

So $g_{eff}$  for the below figure is given by $g_{eff}=(g-a)$

And Pressure at point A is given as

$P=\rho (g-a)h=\rho g_{eff}h$

II.  constant  downward acceleration (a=g)

The pressure became zero everywhere when a=g

III.   constant  downward acceleration (a> g)

In this case, the fluid occupies the upper part of the container as shown in the figure.

Case II-  When Acceleration in Horizontal  direction

If a liquid in the tank is moving on a horizontal surface with some constant acceleration a

Then the free surface of the liquid takes the shape as shown by the dotted line in the figure.

Now consider a cylindrical element of length l and cross-section area A

So the force on the left face of the cylinder is $F_1=P_1A$

While  force on the right face of the cylinder is $F_2=P_2A$

And we can also write

$P_1=\rho gy_1 \ and P_2=\rho gy_2$

And the mass of the element of the liquid and which is given by

$m=\rho lA$

Where $\rho$=density of liquid

So using Newton's second law for the element

$F_1-F_2=ma\\ or \ P_1A-P_2A=ma \\ or \ ( \rho gy_1-\rho gy_2)A=Al\rho a\\ or \ \frac{y_1-y_2}{l}=\frac{a}{g}=tan\theta$

So we can say that

The free surface of the liquid makes an angle $\theta$ with horizontal

Or the free surface of the liquid orient itself perpendicular to the direction of net effective gravity.

So for the below figure, we can say that

Pressure will vary in the horizontal direction.

And the Pressure gradient in the x-direction is given as

$\frac{d p }{d x}=-\rho a_x$

Where -ve sign indicates pressure increases in a direction opposite to the direction of acceleration.

Study it with Videos

Variation of Pressure in an accelerated fluid

"Stay in the loop. Receive exam news, study resources, and expert advice!"