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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 15 Questions around this concept.

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Find the depth at which an air bubble of radius 0.7 \mathrm{~mm} will remain in equilibrium in water. Given, surface tension of water \mathrm{=7.0 \times 10^{-2} \mathrm{Nm}^{-1}}. Take \mathrm{g=10 \mathrm{~ms}^{-2}.}

An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of its side it was found that one third of volume of water spilled out. The acceleration was

A U-tube has uniform cross section and has a lower portion in shape of a quarter circle of radius 0.1 \mathrm{~m}. A valve is provided at the right tube as shown. Water is filled such that the height of the liquid column in the left tube is 0.2 \mathrm{~m} and water is filled upto the closed valve in the right tube. The valve is suddenly opened. The pressure at the lowest point (A) immediately after opening the valve is

Figure shows a U-tube of uniform cross-sectional area A accelerated with acceleration ' a ' as shown.

If ' d ' is the separation between the limbs, then the difference in the levels of the liquid in the U-tube is.

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 gA   

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 which 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 gA   

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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