What is fluid pressure and how is it different from solid pressure?

Fluid pressure is the force exerted by a fluid (liquid or gas) per unit area on the walls of its container or any surface it touches. Unlike solid pressure, which is exerted in a fixed direction (usually downwards), fluid pressure acts equally in all directions at a given depth due to the random motion of fluid particles. In solids, pressure depends on force and contact area, while in fluids, it depends on depth, density, and gravitational acceleration.

What is the hydrostatic paradox and why does it seem counterintuitive?

The hydrostatic paradox says that pressure at a given depth in a liquid depends only on the height of the liquid column—not the container’s shape or volume. This seems counterintuitive since a wider container holds more liquid, yet the pressure at the bottom is the same as in a narrow one of equal height.

What is atmospheric pressure and how is it measured?

It is the pressure exerted by the weight of air in the Earth's atmosphere on all objects. At sea level, it is approximately: 1.013 times 10 to the power 5 Pa, which is equal to 76 cm of Hg It is measured using a barometer, typically a mercury barometer, where the height of the mercury column represents atmospheric pressure. An aneroid barometer, which uses a sealed metal chamber, is also commonly used.

What is the function of a U-tube manometer and how does it measure pressure?

A U-tube manometer is a device used to measure pressure difference between two points or between a gas and the atmosphere. It consists of a U-shaped tube partially filled with a liquid (like mercury or water). When one end is connected to a gas source, the difference in fluid levels in the two arms reflects the pressure difference, calculated using: P=hg

Why do dams have a broader base compared to the top?

Since fluid pressure increases with depth, the pressure at the bottom of a dam is much greater than at the top. To withstand this increasing pressure, the base of the dam is made wider and stronger. This ensures stability and prevents structural failure due to the high water pressure acting at greater depths.

Pressure in Fluids: Definition, Types & FAQs

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JEE Physics

Pressure In Fluids

Pressure in fluids is a key concept in physics and fluid mechanics. Fluids—liquids and gases—exert pressure in all directions due to the random motion of their particles. This concept helps explain natural phenomena and is widely applied in devices like barometers, hydraulic systems, and fluid pipelines.Important topics include types of pressure, pressure at depth, the hydrostatic paradox, and tools such as the barometer and U-tube manometer—all vital for exam preparation and understanding fluid behavior in static and dynamic conditions.

1.0Definition of Pressure

Pressure (P) is defined as the amount of normal force applied per unit area of surface.

$P = \frac{F _{⊥}}{A}$

here F⊥ = normal force on a surface of area A

SI UNIT : Pascal (Pa) ; $1 P a = 1 N / m^{2}$
Dimensions: $[M L^{- 1} T^{- 2}]$
Practical Units : atmospheric pressure (atm), bar and torr
Pressure is a scalar quantity.

$1 a t m = 1.01325 \times 1 0^{5} P a = 1.01325 ba r = 760 t orr =$ 760 mm of Hg=10.33 of water

$1 ba r = 1 0^{5} P a$ ; 1 torr=1 mm of mercury column=133 Pa

2.0Types of Pressures

Pressure is of three types

Atmospheric Pressure: Force exerted by an atmospheric column on a unit cross-sectional area at mean sea level is called atmospheric pressure (P0).

$P_{0} = 101.3 k N / m^{2} \Rightarrow P_{0} = 1.013 \times 1 0^{5} N / m^{2}$

Note: Barometer is used to measure the atmospheric pressure.

Gauge Pressure: Excess Pressure over the atmospheric pressure ( $P - P_{a t m}$ )measured with the help of pressure measuring instruments is called gauge pressure.

Note: Gauge pressure is always measured with the help of a "manometer".

Absolute Pressure: The sum of the atmospheric and gauge pressure is called absolute pressure.

$P_{ab s} = P_{a t m} + P_{g a ug e}$

$P_{ab s} = P_{0} + h ρ g$

3.0Pressure Due to Liquid At Depth

When a liquid is at rest, it exerts pressure on the walls and base of its container. This pressure escalates with depth due to the weight of the liquid above.

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m = mass of liquid element

A = Area of cross - section of liquid element

Since fluid is at rest,

$P_{1} A + m g = P_{2} A$

$P_{1} A + (A h ρ g) = P_{2} A$

$P_{2} = P_{1} + ρ g h$

Note: Pressure exerted by the same liquid at any point does not depend on the shape and size of the container (it means quantity of liquid). It depends only on the height of the liquid column.

The Hydrostatic Paradox refers to the observation that the pressure at a given depth in a fluid is the same, regardless of the shape or size of the container holding the fluid.

Pressure due to liquid,

$P_{A} = P_{B} = P_{C} = P_{D}$

4.0Pressure Due To Liquid on a Vertical Wall of Container

Pressure due to liquid on a vertical wall is different at different depths, so average fluid pressure on side wall of the container is equal to mean pressure

$P_{a vg} = \frac{ρ g h}{2}$ (h = height of wall)

5.0Barometer

A tube of length 1 m and uniform cross section is taken. It is filled with mercury and inverted into a mercury tray. The height of the mercury column in equilibrium inside the tube is 76 cm.
A mercury barometer measures atmospheric pressure based on the height of a mercury column. It works on the principle that atmospheric pressure can support a column of liquid, and the height of that column indicates the pressure exerted by the atmosphere.

Working:

A glass tube (closed at one end) is filled with mercury and inverted into a mercury-filled container.
Mercury flows down slightly, creating a vacuum at the top of the tube (called a Torricellian vacuum).
The atmospheric pressure on the open mercury surface supports the column of mercury inside the tube.
The height (h) of the mercury column shows the atmospheric pressure.

Standard atmospheric pressure = 76 cm (760 mm) of Hg at sea level

Atmospheric Pressure

$P_{0} = ρ g h = 13.6 \times 1 0^{3} \times 9.81 \times 76 \times 1 0^{- 2}$

$= 1.013✕1 0^{5} N / m^{2}$

Barometer in lift

Case (1) : Lift accelerating upwards with acceleration 'a'

$P_{a t m} = ρ (g + a) h$

Case (2) : Lift accelerating downwards with acceleration 'a'

$P_{a t m} = ρ (g - a) h$

6.0U- Tube Concept

The U-tube is a simple device used to compare fluid pressures and measure pressure differences using a manometer.
It consists of a U-shaped glass tube partially filled with liquid (often water or mercury).
When both arms are open to the atmosphere, the liquid levels stay the same.
When one side is connected to a gas or liquid under pressure, the difference in height (Δh) of the liquid columns indicates the pressure difference.

U-Tube Manometer

For the same stationary liquid, pressure at the same horizontal level remains the same.

$P_{A} = P_{B}$

$P_{0} + h_{H g} ρ_{H g} g = P_{0} + h_{w} ρ_{w} g = h_{H g} ρ_{H g} = h_{w} ρ_{w}$

Illustration-1: An open U-tube contains water and unknown liquid separated by mercury. The mercury columns in two arms are level with 8 cm of water in one arm and 10 cm of unknown liquid in the other. Find the specific gravity of unknown liquid.

Solution:

$= h_{w} ρ_{w} = h_{l i q} . ρ_{l i q} .$

$(S . G)_{l i q} . = \frac{ρ _{l i q .}}{ρ _{W}} = \frac{h _{w}}{h _{l i q .}}$

$(S . G)_{l i q .} = \frac{8}{10} = 0.8$

Illustration-2: A cuboid (a × a × 2a) is filled with a liquid of density ' $ρ$ ' as shown in figure. Neglecting atmospheric pressure, find out (a) Force on base wall of the cuboid (b) Force on side wall of the cuboid.

Solution:

(a) Force on base wall= $P_{ba se w a ll} \times A_{ba se w a ll} = ρ g (2 a) \times a^{2} = 2 ρ g a^{3}$

(b) Force on side wall=

$P_{s i d e w a ll} \times A_{s i d e w a ll} = [\frac{0 + ρ g ( 2 a )}{2}] \times 2 a^{2} = 2 ρ g a^{3}$

Illustration-3. A vertical tank is 5 meters tall. It is first filled halfway with water, and then the remaining half is filled with oil having a density of 0.85 g/cm³. What is the total pressure exerted by the liquids at the bottom of the tank?

Solution:

Pressure at the bottom,

$P = (h_{1} d_{1} + h_{2} d_{2}) \frac{g - wt}{c m ^{2}} = [250 \times 1 + 250 \times 0.85]$ $= 250 [1.85] \frac{g - wt}{c m ^{2}} = 462.5 \frac{g - wt}{c m ^{2}}$

Illustration-4. A girl weighing 50 kg stands balanced on a single high-heeled shoe. The heel has a circular base with a diameter of 1 cm. Calculate the pressure exerted by the heel on the floor. g=10 m/s2

Solution:

$m = 50 k g, D = 1 c m = 1 0^{- 2} m, g = 10 m / s^{2}$

Pressure delivered by the heel on the horizontal floor is

$p = \frac{F}{A} = \frac{m g}{π ( \frac{D}{2} ) ^{2}} = \frac{4 m g}{π D ^{2}} = \frac{4 \times 50 K g \times 10 m / s ^{2}}{3.14 \times ( 1 0 ^{- 2} m ) ^{2}} = 6.4 \times 1 0^{6} P a$

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Pressure In Fluids

1.0Definition of Pressure

2.0Types of Pressures

3.0Pressure Due to Liquid At Depth

4.0Pressure Due To Liquid on a Vertical Wall of Container

5.0Barometer

6.0U- Tube Concept

Table of Contents

Frequently Asked Questions

What is fluid pressure and how is it different from solid pressure?

What is the hydrostatic paradox and why does it seem counterintuitive?

What is atmospheric pressure and how is it measured?

What is the function of a U-tube manometer and how does it measure pressure?

Why do dams have a broader base compared to the top?

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