JEE Main Class 11 Syllabus 2025 PDF for Paper 1 and 2

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.

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Variation of Pressure in an accelerated fluid

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