Variation Of Pressure

Pressure is a fundamental concept in physics that influences countless aspects of our daily lives. From the air we breathe to the food we cook, pressure variations play a crucial role in both natural and engineered systems. In weather forecasting, atmospheric pressure changes can predict storms and calm days alike. In the kitchen, pressure cookers use elevated pressure to cook food faster, highlighting how manipulating pressure can enhance efficiency. Moreover, divers experience the effects of pressure changes as they descend into deeper waters, which can affect their health and equipment. In this article, we will exploring the variation of pressure, we gain insights into how these fluctuations impact everything from weather patterns to culinary techniques, demonstrating the profound influence of pressure in both the natural world and human-made technologies.

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This Story also Contains

1. Variation of Pressure
2. Gauge Pressure
3. Variation of Pressure Along Horizontally
4. Solved Examples Based on Variation of Pressure
5. Summary

Variation of Pressure

Pressure, a vital physical parameter, exhibits significant variation across different environments and conditions. Its effects are evident in diverse areas ranging from meteorology to engineering. For instance, changes in atmospheric pressure drive weather patterns, influencing everything from wind speeds to precipitation.

Variation of Pressure with Depth

The variation of pressure with depth is a fundamental concept in fluid mechanics and geophysics, describing how pressure increases as one moves deeper into a fluid or solid medium. In Earth's oceans, for instance, pressure rises significantly with depth due to the weight of the overlying water. This increase in pressure can be dramatic, reaching several hundred atmospheres at the deepest ocean trenches.

Have a look at the below figure

Here $P_0=$ Atmospheric pressure at the upper surface
And $\mathrm{h}=$ depth below the upper surface
$\rho=$ density of liquid
$\mathrm{P}=$ Hydrostatic pressure for a point at depth h below the upper surface
Then P is given by $P=P_0+\rho g h$

This means Pressure increases with depth linearly.

Hydrostatic pressure $=$ Absolute Pressure $=P=P_0+\rho g h$

Absolute Pressure is always positive, It can never be zero.

From equation $P=P_0+\rho g h$

We can say that

Hydrostatic pressure depends on

$\mathrm{h}=$ depth of the point below the surface
$\rho=$ nature of liquid
$\mathrm{g}=$ acceleration due to gravity

Hydrostatic pressure does not depend on

amount of liquid

the shape of the container

From this, we can say that for the below figure where the liquid is filled in vessels of

different shapes to the same height,

the pressure at the base in each vessel will be the same, though

the volume or weight of the liquid in different vessels will be different.

I.e In the above figure $P_A=P_B=P_C$

Gauge Pressure

Gauge pressure refers to the pressure of a system relative to the atmospheric pressure surrounding it. Unlike absolute pressure, which measures the total pressure exerted by a fluid including atmospheric pressure, gauge pressure focuses on the pressure difference between the system and the ambient environment.

So Gauge Pressure is given as $P-P_0=$ gauge pressure
In the equation
$
P=P_0+\rho g h
$

The term $\rho g h$ is known as pressure due to the liquid column of height $h$
We can rewrite the above equation as $\rho g h=P-P_0$
Or we can say that Gauge Pressure $=\rho g h=P-P_0$

It may be positive, negative or zero

Variation of Pressure Along Horizontally

The pressure is uniform on a horizontal line.

The below figure

In a horizontal line or in a horizontal plane in stationary liquid

$P_A=P_B=P_C$

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Solved Examples Based on Variation of Pressure

Example 1: The water is flowing in a pipe as shown in the figure. What is the value of the h difference in pressure head from the data given

1) 15 m

2) 10 m

3) 12 m

4) 16 m

Solution:

$\begin{aligned} & h=\frac{V_2^2-V_1^2}{2 g} \\ & h=\frac{(20)^2-(10)^2}{2 \times 10} \\ & h=15 \mathrm{~m}\end{aligned}$

Hence, the answer is the option (1).

Example 2: There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d₁ and d₂ are filled in the tube. Each liquid subtends 90⁰ angles at the centre. The radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\frac{d_1}{d_2}$ is :

1) $\frac{1+\sin \alpha}{1-\sin \alpha}$
2) $\frac{1+\cos \alpha}{1-\cos \alpha}$
3) $\frac{1+\tan \alpha}{1-\tan \alpha}$
4) $\frac{1+\sin \alpha}{1-\cos \alpha}$

Solution:

Pressure at interface A must be the same from both sides in equilibrium.

So

$\begin{aligned} & \quad(R \cos \alpha+R \sin \alpha) d_2 g=(R \cos \alpha-R \sin \alpha) d_1 g \\ & \frac{d_1}{d_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}\end{aligned}$

Hence, the answer is the option (3).

Example 3: An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by an additional 46 cm. What will be the length (cm) of the air column above the mercury in the tube now?

(Atmospheric pressure =76 cm of Hg)

1) 16

2) 22

3) 38

4) 6

Solution:

Absolute Pressure

$
P=P_0+\rho g h
$
wherein
$P \rightarrow$ hydrostatics Pressure
$P_0 \rightarrow$ atmospheric Pressure

let mercury rise by distance X

For air trapped in tube T = constant

$
\begin{aligned}
& \Rightarrow P_1 V_1=P_2 V_2 \Rightarrow P_1=P_{\text {atm }}=\rho g(76) \\
& V_1=A \pi 8 \mathrm{~cm} \\
& P_2=(76-x) \rho g \\
& V_2=A \cdot(54-x) \\
& \Rightarrow \rho g(76) \times(8 A)=\rho g(76-x)(A(54-x)) \\
& \Rightarrow x^2-130 x-3496=0 \\
& \Rightarrow x=38 \mathrm{~cm} \\
&
\end{aligned}
$

So the length of the air column is $(54-\mathrm{x})=54-38=16 \mathrm{~cm}$

Hence, the answer is the option (1).

Example 4: The thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $p_1$ and $p_2\left(p_1>p_2\right)$, fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is

1) $\theta=\tan ^{-1} \pi\left(\frac{p_1}{p_2}\right)$
2) $\theta=\tan ^{-1} \frac{\pi}{2}\left(\frac{p_2}{p_1}\right)$
3) $\theta=\tan ^{-1}\left[\left(\frac{p_1-p_2}{p_1+p_2}\right)\right]$
4) $\theta=\tan ^{-1} \frac{\pi}{2}\left[\left(\frac{p_1+p_2}{p_1-p_2}\right)\right]$

Solution:

Equating pressure at point A

$\begin{aligned} & p_1 g R(\cos \theta-\sin \theta)=p_2 g R(\sin \theta+\cos \theta) \\ & \frac{p_1}{p_2}=\frac{\sin \theta+\cos \theta}{\cos \theta-\sin \theta}=\frac{\tan \theta+1}{1-\tan \theta} \\ & p_1-p_1 \tan \theta=p_2+p_2 \tan \theta \\ & \Rightarrow\left(p_1+p_2\right) \tan \theta=p_1-p_2 \\ & \Rightarrow \tan \theta=\frac{p_1-p_2}{p_1++p_2} \Rightarrow \theta=\tan ^{-1}\left(\frac{p_1-p_2}{p_1+p_2}\right)\end{aligned}$

Hence, the answer is the option (3).

Summary

Pressure variations are critical in various applications, from understanding atmospheric and hydrostatic pressures to practical uses in engineering and daily life. Hydrostatic pressure increases linearly with depth, influenced by the fluid's density and gravity, and is independent of the container's shape or the liquid's volume. Gauge pressure measures the pressure relative to atmospheric pressure and can vary based on the fluid column height. Examples illustrate how to apply these principles in different scenarios, such as pressure differences in pipes, interactions of immiscible liquids, and changes in trapped air columns, highlighting the importance of understanding pressure in both theoretical and practical contexts.