Under what conditions is Bernoulli's equation applicable, and what are its limitations?

Bernoulli's equation ($P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$) is a powerful tool, but it's crucial to remember its ideal conditions: it applies to incompressible, non-viscous (ideal) fluids, undergoing steady, laminar flow, and along a single streamline. Its limitations arise when these conditions are violated, such as in turbulent flow, highly viscous fluids (where energy is lost due to friction), or when external work is done on the fluid (e.g., by a pump) or extracted from it (e.g., by a turbine).

What is the fundamental difference between surface tension and viscosity, and how do they manifest in real-world phenomena?

Surface tension ($T$) is a property of a liquid's surface, arising from the cohesive forces between liquid molecules. It causes the liquid surface to behave like a stretched elastic membrane, minimizing its surface area. This manifests in phenomena like spherical liquid drops, capillary action, and insects walking on water. Viscosity ($\eta$), on the other hand, is a measure of a fluid's resistance to flow, representing the internal friction between adjacent layers of the fluid. It manifests in how easily a liquid pours (e.g., honey is more viscous than water) and the drag force experienced by objects moving through the fluid (Stokes' Law, $F_v = 6 \pi \eta r v$). Surface tension is about surface energy, while viscosity is about fluid friction.

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Key Formulas & High-Yield Equations

Pressure in Fluids

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

Defines pressure (

P

) as force (

F

) applied perpendicularly per unit area (

A

). Valid for both static and dynamic fluids, but for static fluids, pressure at a depth

h

P = P_0 + \rho g h

, where

P_0

is atmospheric pressure and

\rho

is fluid density.

Archimedes' Principle (Buoyant Force)

$F_B = \rho_{fluid} V_{submerged} g$

The buoyant force (

F_B

) on an object submerged in a fluid is equal to the weight of the fluid displaced.

\rho_{fluid}

is the density of the fluid,

V_{submerged}

is the volume of the object submerged, and

g

is acceleration due to gravity. Applies to both floating and submerged objects.

Equation of Continuity

$A_1 v_1 = A_2 v_2 = \text{constant}$

For an incompressible, non-viscous fluid in steady flow, the product of the cross-sectional area (

A

) and the fluid speed (

v

) remains constant along a streamline. This principle is a direct consequence of mass conservation.

Bernoulli's Principle

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

States that for an ideal (incompressible, non-viscous), steady, laminar flow of a fluid along a streamline, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.

P

is pressure,

\rho

is fluid density,

v

is fluid speed,

g

is acceleration due to gravity, and

h

is height.

Stokes' Law (Viscous Drag)

$F_v = 6 \pi \eta r v$

The viscous drag force (

F_v

) experienced by a small spherical object of radius

r

moving with speed

v

through a fluid of viscosity

\eta

. Valid for laminar flow and small Reynolds numbers, typically for spheres moving in an infinite fluid medium.

Excess Pressure in a Liquid Drop/Bubble

$\Delta P_{drop} = \frac{2T}{R}, \Delta P_{bubble} = \frac{4T}{R}$

The excess pressure (

\Delta P

) inside a liquid drop (single surface) or a soap bubble (two surfaces) of radius

R

due to surface tension

T

. This pressure acts inwards, balancing the surface tension forces.

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2023

A block of wood floats in water with $\frac{4}{5}$ of its volume submerged. If the same block is placed in an oil, it floats with $\frac{9}{10}$ of its volume submerged. The density of the wood is $0.8 \text{ g/cm}^3$ . What is the density of the oil?

Solution Locked

Step-by-Step Solution & Concept

Step 1

For an object floating, its weight equals the buoyant force. Let $V$ be the total volume of the block and $\rho_w$ be the density of wood. In water, $W = F_{B,water} \Rightarrow \rho_w V g = \rho_{water} (\frac{4}{5}V) g$ . Given $\rho_w = 0.8 \text{ g/cm}^3$ and $\rho_{water} = 1 \text{ g/cm}^3$ , this equation is consistent: $0.8 V g = 1 \times (\frac{4}{5}V) g \Rightarrow 0.8 = 0.8$ .

Step 2

In oil, the weight of the block remains the same, and it equals the buoyant force in oil. Let $\rho_{oil}$ be the density of the oil. So, $W = F_{B,oil} \Rightarrow \rho_w V g = \rho_{oil} (\frac{9}{10}V) g$ .

Step 3

Substitute the known values: $0.8 V g = \rho_{oil} (\frac{9}{10}V) g$ . We can cancel $V g$ from both sides: $0.8 = \rho_{oil} \times \frac{9}{10}$ . Solving for $\rho_{oil}$ : $\rho_{oil} = 0.8 \times \frac{10}{9} = \frac{8}{9} \approx 0.888... \text{ g/cm}^3$ . Rounding to two decimal places, $\rho_{oil} \approx 0.89 \text{ g/cm}^3$ .

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Question #2JEE Main 2022

Water flows through a horizontal pipe. The diameter of the pipe at the wider section is $20 \text{ cm}$ and at the narrower section is $10 \text{ cm}$ . If the speed of water in the wider section is $5\sqrt{2} \text{ m/s}$ , what is the pressure difference between the wider and narrower sections of the pipe? (Density of water $\rho = 1000 \text{ kg/m}^3$ )

Solution Locked

Step-by-Step Solution & Concept

Step 1

Apply the Equation of Continuity ( $A_1 v_1 = A_2 v_2$ ) to find the speed in the narrower section. Given diameters $D_1 = 20 \text{ cm} = 0.2 \text{ m}$ and $D_2 = 10 \text{ cm} = 0.1 \text{ m}$ . Areas are $A = \pi (D/2)^2$ . So, $(\frac{D_1}{2})^2 v_1 = (\frac{D_2}{2})^2 v_2 \Rightarrow D_1^2 v_1 = D_2^2 v_2$ . Substituting values: $(0.2)^2 (5\sqrt{2}) = (0.1)^2 v_2 \Rightarrow 0.04 \times 5\sqrt{2} = 0.01 v_2 \Rightarrow v_2 = \frac{0.2\sqrt{2}}{0.01} = 20\sqrt{2} \text{ m/s}$ .

Step 2

Apply Bernoulli's Principle for a horizontal pipe ( $h_1 = h_2$ ). $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ . We need to find the pressure difference $\Delta P = P_1 - P_2$ . So, $P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$ .

Step 3

Substitute the values: $\rho = 1000 \text{ kg/m}^3$ , $v_1 = 5\sqrt{2} \text{ m/s}$ , $v_2 = 20\sqrt{2} \text{ m/s}$ . $\Delta P = \frac{1}{2} (1000) ((20\sqrt{2})^2 - (5\sqrt{2})^2) = 500 ( (400 \times 2) - (25 \times 2) ) = 500 (800 - 50) = 500 (750) = 375000 \text{ Pa}$ . This is $3.75 \times 10^5 \text{ Pa}$ .

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Mastering Mechanical Properties of Fluids: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Pressure in Fluids

Archimedes' Principle (Buoyant Force)

Equation of Continuity

Bernoulli's Principle

Stokes' Law (Viscous Drag)

Excess Pressure in a Liquid Drop/Bubble

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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