Intermediate

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Mechanical Properties of Fluids

Master Mechanical Properties of Fluids for JEE Main, JEE Advanced, NEET & Board Exams. Covers pressure, Pascal's law, Archimedes' principle, Bernoulli's theorem, viscosity, Stokes' law, surface tension and capillarity with solved examples.

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A fluid is any substance that can flow — this includes both liquids and gases. Unlike solids, fluids cannot sustain a shearing stress and deform continuously under any applied tangential force. The study of fluids at rest is called hydrostatics, while the study of fluids in motion is called hydrodynamics. This topic is highly relevant for JEE (Main & Advanced), NEET, and board exams, with questions frequently appearing on pressure, buoyancy, Bernoulli's equation, and viscosity.

1. Pressure in a Fluid

Pressure is defined as the normal force per unit area exerted by a fluid:

$P = \frac{F}{A}$

SI unit: Pascal (Pa) = N/m². Other units: $1 atm = 1.013 \times 10^{5} Pa$ ; $1 bar = 10^{5} Pa$ ; $1 mmHg = 133.3 Pa$ .

Pressure at a Depth in a Static Fluid

For a fluid of uniform density $ρ$ at rest, the pressure at depth $h$ below the free surface is:

$P = P_{0} + ρ g h$

where $P_{0}$ is the atmospheric pressure at the surface. Key properties:

Pressure depends only on depth, not on the shape or cross-section of the container (Hydrostatic Paradox).
Pressure acts equally in all directions at any point in a static fluid (Pascal's Law basis).
All points at the same horizontal level in a connected static fluid have the same pressure.

Gauge Pressure and Absolute Pressure

Type	Definition	Formula
Absolute Pressure	Total pressure including atmosphere	$P_{a b s} = P_{0} + ρ g h$
Gauge Pressure	Pressure above atmospheric	$P_{g a u g e} = P_{a b s} - P_{0} = ρ g h$
Vacuum Pressure	Pressure below atmospheric	$P_{v a c} = P_{0} - P_{a b s}$

2. Pascal's Law and Its Applications

Pascal's Law: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ $\Rightarrow$ $F_{2} = F_{1} \cdot \frac{A_{2}}{A_{1}}$

Hydraulic Lift / Press: A small force $F_{1}$ on a small piston of area $A_{1}$ generates a large force $F_{2}$ at a large piston of area $A_{2}$ . This is the working principle of hydraulic brakes, hydraulic jacks, and car lifts.
Work done is conserved: $F_{1} d_{1} = F_{2} d_{2}$ , so the smaller piston moves a larger distance.
Hydraulic Brakes: Pressure applied at the brake pedal is transmitted equally to all four wheel cylinders, applying equal braking force at all wheels.

3. Buoyancy and Archimedes' Principle

When a body is partially or fully submerged in a fluid, the fluid exerts an upward force on it called the buoyant force or upthrust.

Archimedes' Principle

The buoyant force on a submerged (or floating) body equals the weight of the fluid displaced by the body:

$F_{b} = ρ_{f l u i d} \cdot V_{d i s p l a c e d} \cdot g$

The buoyant force acts upward through the centre of buoyancy — the centroid of the displaced fluid volume.

Conditions for Floating, Sinking, and Neutral Buoyancy

Condition	Relation	Result
$ρ_{b o d y} < ρ_{f l u i d}$	$F_{b} > W$	Body floats (partially submerged)
$ρ_{b o d y} = ρ_{f l u i d}$	$F_{b} = W$	Neutral buoyancy (suspended anywhere)
$ρ_{b o d y} > ρ_{f l u i d}$	$F_{b} < W$	Body sinks

Law of Flotation

For a freely floating body, weight of body = weight of fluid displaced:

$ρ_{b o d y} \cdot V_{b o d y} \cdot g = ρ_{f l u i d} \cdot V_{s u b m e r g e d} \cdot g$

$\frac{V_{s u b m e r g e d}}{V_{b o d y}} = \frac{ρ_{b o d y}}{ρ_{f l u i d}}$

This ratio gives the fraction of the body submerged. For example, ice ( $ρ = 917$ kg/m³) in water ( $ρ = 1000$ kg/m³): fraction submerged $= \frac{917}{1000} = 0.917$ , i.e., about $91.7 %$ of an iceberg is below water.

Apparent Weight

When a body of density $ρ_{b}$ is fully submerged in a fluid of density $ρ_{f}$ :

$W_{a p p a r e n t} = W_{a c t u a l} - F_{b} = m g (1 - \frac{ρ_{f}}{ρ_{b}})$

4. Fluid Dynamics — Equation of Continuity

For a fluid in motion, we assume it is ideal (incompressible, non-viscous, irrotational, streamline flow) unless stated otherwise.

Streamlines and Laminar Flow

A streamline is a curve whose tangent at any point gives the direction of fluid velocity at that point. Streamlines never cross each other.
Laminar (streamline) flow: Orderly flow where layers of fluid slide past each other without mixing. Occurs at low speeds.
Turbulent flow: Irregular, chaotic flow with eddies and vortices. Occurs above a critical speed.

Equation of Continuity

For an incompressible fluid flowing through a pipe of varying cross-section, the mass flow rate is constant:

$A_{1} v_{1} = A_{2} v_{2} = constant$ (Volume flow rate $Q$ )

where $A$ is the cross-sectional area and $v$ is the fluid speed. This means:

Where the pipe is narrower ( $A$ smaller), fluid flows faster ( $v$ larger).
Where the pipe is wider ( $A$ larger), fluid flows slower ( $v$ smaller).

5. Bernoulli's Theorem

For an ideal fluid in steady flow, the total mechanical energy per unit volume remains constant along a streamline:

$P + \frac{1}{2} ρ v^{2} + ρ g h = constant$

This is essentially the conservation of energy applied to fluid flow. The three terms represent:

$P$ — pressure energy per unit volume
$\frac{1}{2} ρ v^{2}$ — kinetic energy per unit volume
$ρ g h$ — potential energy per unit volume

Between any two points 1 and 2 along a streamline:

$P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$

Applications of Bernoulli's Theorem

(i) Venturimeter

A device to measure the flow speed of a fluid in a pipe. It has a constriction (throat) where velocity increases and pressure drops. The flow speed is:

$v_{1} = A_{2} \sqrt{\frac{2 (P_{1} - P_{2})}{ρ (A_{1}^{2} - A_{2}^{2})}}$

(ii) Torricelli's Theorem — Speed of Efflux

For a liquid in a large open tank with a small orifice at depth $h$ below the free surface:

$v = \sqrt{2 g h}$

This is the same as the speed of a body falling freely from height $h$ . The liquid behaves like a projectile after leaving the orifice — it follows a parabolic path.

Range of the liquid jet on the ground (orifice at height $H$ above ground, depth $h$ below surface):

$x = 2 \sqrt{h (H - h)}$

Maximum range occurs when $h = H / 2$ , i.e., the orifice is at the midpoint of the tank height.

(iii) Dynamic Lift — Aerofoil and Magnus Effect

Aerofoil (Aircraft Wing): The wing is shaped so that air flows faster over the curved upper surface than the flat lower surface. By Bernoulli's theorem, pressure above is lower, creating a net upward lift force.
Magnus Effect: A spinning ball curves in flight because spin creates unequal air speeds on two sides, leading to a pressure difference and a sideways force (used in cricket, tennis, football).

6. Viscosity

Viscosity is the property of a fluid by virtue of which it opposes relative motion between its adjacent layers. It is the fluid equivalent of friction in solids.

Newton's Law of Viscosity

The viscous force between two adjacent fluid layers is:

$F = - η A \frac{d v}{d x}$

where $η$ is the coefficient of viscosity (dynamic viscosity), $A$ is the area of the layer, and $\frac{d v}{d x}$ is the velocity gradient (rate of shear). SI unit of $η$ : Pa·s (or Poise in CGS: $1 Pa·s = 10 Poise$ ).

Effect of Temperature on Viscosity

Fluid Type	Effect of Increasing Temperature on $η$	Reason
Liquids	$η$ decreases	Intermolecular cohesive forces weaken
Gases	$η$ increases	Molecular momentum transfer increases

Stokes' Law and Terminal Velocity

When a small sphere of radius $r$ moves through a viscous fluid with velocity $v$ , the viscous drag force is:

$F_{d r a g} = 6 π η r v$ (Stokes' Law)

A body falling through a viscous fluid accelerates until the net downward force is zero. At this point it reaches terminal velocity $v_{T}$ :

$W = F_{b} + F_{d r a g}$

$m g = ρ_{f} V g + 6 π η r v_{T}$

$v_{T} = \frac{2 r^{2} (ρ_{b} - ρ_{f}) g}{9 η}$

$v_{T} \propto r^{2}$ — larger droplets fall faster (important in rainfall and cloud physics).
$v_{T} \propto (ρ_{b} - ρ_{f})$ — if $ρ_{b} < ρ_{f}$ , terminal velocity is upward (bubble rising in liquid).
$v_{T} \propto \frac{1}{η}$ — more viscous fluids give lower terminal velocity.

7. Surface Tension

Surface tension ( $T$ or $S$ ) is the property of a liquid surface that makes it behave like a stretched elastic membrane. It arises due to the net inward cohesive force on surface molecules.

$T = \frac{F}{l}$ (force per unit length)

SI unit: N/m. It can also be expressed as surface energy per unit area: J/m².

Excess Pressure Inside Curved Surfaces

Surface	Excess Pressure ( $Δ P$ )	Reason
Liquid drop (1 surface)	$Δ P = \frac{2 T}{R}$	One liquid-air interface
Soap bubble (2 surfaces)	$Δ P = \frac{4 T}{R}$	Two soap-air interfaces (inner & outer)
Air bubble in liquid (1 surface)	$Δ P = \frac{2 T}{R}$	One liquid-air interface

Capillarity — Capillary Rise and Fall

When a capillary tube of radius $r$ is dipped in a liquid of surface tension $T$ , density $ρ$ , and contact angle $θ$ :

$h = \frac{2 T \cos θ}{ρ g r}$

If $θ < 90 °$ (e.g., water in glass): liquid rises — meniscus is concave.
If $θ > 90 °$ (e.g., mercury in glass): liquid falls — meniscus is convex.
Capillary rise is independent of the shape of the tube — depends only on radius at the meniscus.
$h \propto \frac{1}{r}$ — narrower tubes show greater rise.

Effect of Temperature on Surface Tension

Surface tension decreases with increasing temperature and becomes zero at the critical temperature (boiling point for most liquids). This is why hot water cleans better than cold water.

8. Reynolds Number

The Reynolds number ( $R e$ ) is a dimensionless quantity that predicts whether fluid flow will be laminar or turbulent:

$R e = \frac{ρ v d}{η}$

where $ρ$ = fluid density, $v$ = flow speed, $d$ = diameter of pipe, $η$ = coefficient of viscosity.

Reynolds Number	Type of Flow
$R e < 1000$	Laminar (streamline) flow
$1000 < R e < 2000$	Unstable / transitional flow
$R e > 2000$	Turbulent flow

Applying Bernoulli's Theorem — Step-by-Step Strategy

Step 1: Identify two points along the same streamline where you know (or want to find) pressure, velocity, or height.
Step 2: Write Bernoulli's equation: $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$ .
Step 3: Apply the continuity equation ( $A_{1} v_{1} = A_{2} v_{2}$ ) to relate velocities if cross-sections are given.
Step 4: Simplify. For a horizontal pipe, $h_{1} = h_{2}$ , so $ρ g h$ terms cancel. For a large tank with a small orifice, $v_{t a n k} \approx 0$ (continuity: $A_{t a n k} ≫ A_{o r i f i c e}$ ).
Step 5: Solve for the unknown. Always check units and the physical reasonableness of the answer.

JEE/NEET Power Tips — Avoid These Common Mistakes

Pressure depends only on depth, not on the amount of fluid or the shape of the container. Two containers of completely different shapes but with the same fluid depth have the same pressure at the bottom — this is the Hydrostatic Paradox.
Buoyant force depends on the volume of fluid displaced, NOT the volume of the body. A hollow object displaces fluid equal to its outer volume when fully submerged.
Soap bubble has TWO surfaces — inner and outer. Always use $Δ P = \frac{4 T}{R}$ for soap bubbles, not $\frac{2 T}{R}$ .
Bernoulli's theorem applies along a streamline in steady, ideal (non-viscous, incompressible) flow. Do not apply it across streamlines or in turbulent flow.
Terminal velocity is reached asymptotically — the body never instantaneously jumps to $v_{T}$ . It accelerates and gradually approaches $v_{T}$ as drag increases.
Surface tension decreases with temperature — never say it is independent of temperature.
Capillary rise $h \propto \frac{1}{r}$ — do not confuse radius of the tube with radius of the meniscus. For a cylindrical tube, they are equal.

Practice Questions (JEE / NEET / Board Level)

Q1: A hydraulic lift has two pistons of cross-sectional areas $30 {cm}^{2}$ and $1500 {cm}^{2}$ . A force of $50$ N applied on the smaller piston can lift a load of:

A) $1000$ N
B) $2500$ N
C) $3500$ N
D) $5000$ N

Answer: B) $2500$ N. By Pascal's law: $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ $\Rightarrow$ $F_{2} = 50 \times \frac{1500}{30} = 50 \times 50 = 2500$ N.

Q2: An iceberg floats in sea water with a fraction $\frac{9}{10}$ of its volume submerged. The density of sea water is $1080 {kg/m}^{3}$ . The density of ice is:

A) $9720 {kg/m}^{3}$
B) $972 {kg/m}^{3}$
C) $900 {kg/m}^{3}$
D) $1080 {kg/m}^{3}$

Answer: B) $972 {kg/m}^{3}$ . $\frac{ρ_{i c e}}{ρ_{s e a}} = \frac{V_{s u b}}{V_{t o t a l}} = \frac{9}{10}$ $\Rightarrow$ $ρ_{i c e} = \frac{9}{10} \times 1080 = 972 {kg/m}^{3}$ .

Q3: Water flows through a horizontal pipe. At a section where the diameter is $4$ cm, the velocity is $2$ m/s. What is the velocity at a section where the diameter is $2$ cm?

A) $1$ m/s
B) $4$ m/s
C) $8$ m/s
D) $16$ m/s

Answer: C) $8$ m/s.
By continuity: $A_{1} v_{1} = A_{2} v_{2}$ . $π (2)^{2} \times 2 = π (1)^{2} \times v_{2}$ $\Rightarrow$ $v_{2} = 4 \times 2 = 8$ m/s. (radii are $2$ cm and $1$ cm respectively).

Q4: A tank is filled with water to a height of 12.5 m. An orifice is made at a height of 2.5 m from the bottom. The speed of water flowing out is ( $g = 10 {m/s}^{2}$ ):

A) 5 m/s
B) 10 m/s
C) 14 m/s
D) 20 m/s

Answer: C) 14 m/s.

Explanation: According to Torricelli's law, the speed of efflux ( $v$ ) is determined by the depth of the orifice below the free surface of the liquid ( $h$ ).

Depth of the orifice ( $h$ ) = Total height - Height from bottom
$h = 12.5 - 2.5 = 10 m$

Using Torricelli's theorem ( $v = \sqrt{2 g h}$ ):
$v = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10 \sqrt{2} m/s$

Since $\sqrt{2} \approx 1.414$ , the speed is approximately $14.1 m/s$ . The closest answer is 14 m/s.

Q5: A spherical ball of radius $r$ and density $ρ_{b}$ falls through a liquid of density $ρ_{f}$ and viscosity $η$ . If the radius is doubled, the terminal velocity becomes:

A) Half
B) Double
C) Four times
D) Eight times

Answer: C) Four times. $v_{T} = \frac{2 r^{2} (ρ_{b} - ρ_{f}) g}{9 η}$ , so $v_{T} \propto r^{2}$ . Doubling $r$ makes $v_{T}$ four times larger.

Q6: The excess pressure inside a soap bubble of radius $R$ is $P$ . If the radius is halved, the new excess pressure is:

A) $P / 2$
B) $P$
C) $2 P$
D) $4 P$

Answer: C) $2 P$ .
$Δ P = \frac{4 T}{R}$ , so $Δ P \propto \frac{1}{R}$ .
Halving $R$ doubles the excess pressure: new pressure $= 2 P$ .

Q7 (JEE Advanced type): Water flows steadily through a horizontal pipe that tapers from a cross-section of $40 {cm}^{2}$ to $10 {cm}^{2}$ . The pressure at the wider section is $4 \times 10^{4}$ Pa and flow speed there is $1$ m/s. What is the pressure at the narrower section? ( $ρ_{w a t e r} = 1000 {kg/m}^{3}$ )

A) $4 \times 10^{4}$ Pa
B) $3.25 \times 10^{4}$ Pa
C) $2.5 \times 10^{4}$ Pa
D) $1 \times 10^{4}$ Pa

Answer: C) $2.5 \times 10^{4}$ Pa.
By continuity:
$v_{2} = v_{1} \times \frac{A_{1}}{A_{2}} = 1 \times \frac{40}{10} = 4$ m/s.
By Bernoulli (horizontal pipe):
$P_{2} = P_{1} + \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2})$
$= 4 \times 10^{4} + \frac{1}{2} (1000) (1 - 16)$
$= 4 \times 10^{4} - 7500 = 3.25 \times 10^{4}$ Pa.
Answer: C.

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DifficultyIntermediate

Est. Read Time28 minutes

CategoryFluid Mechanics

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