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Variation Of Pressure In An Accelerated Fluid MCQ - Practice Questions with Answers

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

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  • 16 Questions around this concept.

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The minimum horizontal acceleration of the container so that the pressure at point A of the container becomes atmospheric is (the tank is of sufficient height) 

 A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of the vessel is 5 cm and its rotational speed is 2  rotations per second, then the difference in the heights between the centre and the sides, in cm, will be:

 

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 body having volume V and density $\rho$ is attached to the bottom of a container as shown. Density of the liquid is $\mathrm{d}(>\rho)$. Container has a constant upward acceleration a. Tension in the string is

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 $=P_1 A$
The force on the bottom face of the element $=P_2 A$
If $a$ is the acceleration of the liquid then
We can write

$$
P_2 A-\left(h A \rho g+P_1 A\right)=m a
$$


Where $m$ is the mass of the element of the liquid and which $i$

$$
m=\rho g A
$$


Where $\rho=$ density of liquid
So using this we get

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

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

          I.  constant  downward acceleration (a< g)

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

             And Pressure at point A is given as 

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

              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 a Horizontal  direction

If a liquid in the tank that 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 I and cross-section area $A$
So the force on the left face of the cylinder is $F_1=P_1 A$
While force on the right face of the cylinder is $F_2=P_2 A$
And we can also write

$$
P_1=\rho g y_1 \text { and } P_2=\rho g y_2
$$


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

$$
m=\rho g A
$$


Where $\rho=$ density of liquid
So using Newton's second law for the element

$$
\begin{aligned}
& \quad F_1-F_2=m a \\
& \text { or } P_1 A-P_2 A=m a \\
& \text { or }\left(\rho g y_1-\rho g y_2\right) A=A l \rho a \\
& \text { or } \frac{y_1-y_2}{l}=\frac{a}{g}=\tan \theta
\end{aligned}
$$


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.

The Pressure gradient in the x-direction is given as

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

Where the -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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