Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Mechanical Properties of Fluids (Hydrostatics and Hydrodynamics) for CBSE Class 11 Physics.
Key Concepts
Fluid Mechanics
The physics of liquids and gases at rest and in motion
1
Pressure & Variation with Depth
Formula
Pressure in a fluid at rest increases linearly with depth. Absolute pressure ($P$) at depth $h$ is the sum of atmospheric pressure ($P_a$) and gauge pressure ($\rho g h$).
$$P = P_a + \rho g h \quad | \quad P_{\text{gauge}} = \rho g h$$
2
Pascal’s Law
Formula
If pressure is applied to an enclosed incompressible fluid, it is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. (Principle of Hydraulic Lift).
$$\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_2 = F_1 \left( \frac{A_2}{A_1} \right)$$
3
Archimedes’ Principle & Buoyancy
Formula
When a body is partially or fully immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by it.
$$F_B = \text{Weight of displaced fluid} = V_{\text{submerged}} \cdot \rho_{\text{fluid}} \cdot g$$
4
Equation of Continuity
Formula
Based on the conservation of mass. For an incompressible, non-viscous fluid in streamlined flow, the volume flux (Area $\times$ Velocity) remains constant. Fluid speeds up where the pipe narrows.
$$A_1 v_1 = A_2 v_2 \quad \implies \quad Av = \text{constant}$$
5
Bernoulli’s Principle
Formula
Based on the conservation of energy. For a streamlined fluid flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant.
$$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$$
6
Torricelli’s Law (Speed of Efflux)
Formula
The speed of fluid flowing out of an open tank through a small hole at a depth $h$ below the surface is the same as the speed of a freely falling body dropped from height $h$.
$$v = \sqrt{2gh}$$
7
Viscosity & Stokes’ Law
Formula
Viscosity is fluid friction. Stokes’ law gives the viscous drag force ($F$) acting on a small sphere of radius $r$ falling with velocity $v$ through a fluid of coefficient of viscosity $\eta$.
$$F = 6\pi\eta r v$$
8
Terminal Velocity
Formula
The constant maximum velocity acquired by a body falling through a viscous fluid when the net force (Weight vs. Drag + Buoyancy) becomes zero.
$$v_t = \frac{2r^2 (\rho – \sigma) g}{9\eta}$$
9
Surface Tension ($S$)
Formula
The property of a liquid surface to act like a stretched elastic membrane. It is defined as force per unit length or surface energy per unit area.
$$S = \frac{F}{L} = \frac{\text{Work}}{\Delta \text{Area}}$$
10
Excess Pressure & Capillarity
Case Formulas
Pressure inside a curved liquid surface is greater than outside. Capillary action is the rise or fall of a liquid in a narrow tube due to surface tension.
$$\text{Liquid Drop: } \Delta P = \frac{2S}{R} \quad | \quad \text{Soap Bubble: } \Delta P = \frac{4S}{R}$$
$$\text{Capillary Rise: } h = \frac{2S \cos\theta}{r \rho g}$$
Concept Deep Dive
01
Bernoulli’s Principle & Airplane Wings
Dynamic LiftAn airplane wing (aerofoil) is curved on the top and relatively flat on the bottom. As the plane moves forward, the air flowing over the curved top surface has to travel a longer distance in the same amount of time, meaning its velocity increases compared to the air below.
According to Bernoulli’s equation ($P + \frac{1}{2}\rho v^2 = \text{const}$ for a horizontal pipe), an increase in fluid velocity leads to a decrease in pressure. Therefore, the pressure on top of the wing becomes lower than the pressure below it. This pressure difference creates an upward force called Dynamic Lift!
According to Bernoulli’s equation ($P + \frac{1}{2}\rho v^2 = \text{const}$ for a horizontal pipe), an increase in fluid velocity leads to a decrease in pressure. Therefore, the pressure on top of the wing becomes lower than the pressure below it. This pressure difference creates an upward force called Dynamic Lift!
02
Why Don’t Raindrops Kill Us?
The saving grace of Terminal VelocityClouds form at altitudes of 2 to 10 kilometers. If a raindrop fell from that height purely under gravity ($v = \sqrt{2gh}$), it would hit your head at the speed of a bullet (over $200 \text{ m/s}$), causing lethal damage.
Fortunately, as the drop falls and accelerates, air resistance (viscous drag) increases ($F = 6\pi\eta r v$). Eventually, this upward drag perfectly balances the downward weight of the drop. The net acceleration becomes zero, and the drop falls at a constant, gentle speed of about $9 \text{ m/s}$. This safe speed is its Terminal Velocity.
Fortunately, as the drop falls and accelerates, air resistance (viscous drag) increases ($F = 6\pi\eta r v$). Eventually, this upward drag perfectly balances the downward weight of the drop. The net acceleration becomes zero, and the drop falls at a constant, gentle speed of about $9 \text{ m/s}$. This safe speed is its Terminal Velocity.
Compare & Contrast
✗ Streamline (Steady) Flow
- Velocity of fluid at any specific point remains constant over time.
- Fluid layers glide smoothly over each other.
- Occurs at low velocities (below Critical Velocity).
- Reynolds Number ($R_e$) is typically less than 1000.
- Bernoulli’s equation is strictly valid here.
✓ Turbulent Flow
- Velocity of fluid fluctuates randomly and chaotically.
- Characterized by eddies, whirlpools, and mixing of layers.
- Occurs at high velocities (above Critical Velocity).
- Reynolds Number ($R_e$) is typically greater than 2000.
- Lots of energy is dissipated as heat/sound.
Common Mistakes to Avoid
Mistake 1
Soap Bubble vs. Liquid Drop: When calculating surface energy or excess pressure, students forget that a soap bubble has TWO free surfaces (inside and outside) exposed to air. Therefore, its excess pressure is $\Delta P = \frac{4S}{R}$ and its surface energy is $W = S \times (2 \Delta A)$. A water drop only has ONE free surface ($\frac{2S}{R}$).
Mistake 2
Misinterpreting the Equation of Continuity: When water flows from a wider pipe into a narrower pipe, many students guess that “pressure increases” because it’s being squeezed. Actually, the Equation of Continuity ($A_1 v_1 = A_2 v_2$) says velocity increases. Bernoulli’s principle then tells us that because velocity increased, the pressure must actually DECREASE in the narrow section!
Mistake 3
Gauge vs Absolute Pressure: Absolute pressure is the actual total pressure. Gauge pressure is the difference between absolute pressure and atmospheric pressure ($P – P_a$). Blood pressure, tire pressure, and scuba depth gauges all measure Gauge pressure. Don’t add $1 \text{ atm} (1.013 \times 10^5 \text{ Pa})$ unless the question specifically asks for Absolute pressure.
Exam Tips
Tip 1
Terminal Velocity Ratio Shortcut: From the formula $v_t \propto r^2$, if two identical raindrops of radius $r$ combine to form a larger drop of radius $R$, use volume conservation ($\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3 \implies R = 2^{1/3} r$) to quickly find the new terminal velocity.
Tip 2
Insufficient Capillary Tube: If a capillary tube is too short for the water to rise to its calculated height $h$, the liquid will NOT overflow. Instead, the meniscus flattens out to increase its radius of curvature ($R’$) such that $h R = h’ R’ = \text{constant}$.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Mechanical Properties of Fluids · CBSE Exam1
A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000 \text{ kg}$. The area of cross-section of the piston carrying the load is $425 \text{ cm}^2$. What maximum pressure would the smaller piston have to bear? [2 marks]
Answer
$6.92 \times 10^5 \text{ Pa}$
📝
Explanation
By Pascal’s Law, the pressure is transmitted equally throughout the fluid. Therefore, the pressure the smaller piston bears is equal to the pressure exerted by the larger piston.
Force on large piston $F = mg = 3000 \times 9.8 = 29400 \text{ N}$.
Area $A = 425 \text{ cm}^2 = 425 \times 10^{-4} \text{ m}^2$.
Pressure $P = \frac{F}{A} = \frac{29400}{425 \times 10^{-4}} = \frac{29400}{0.0425} \approx 6.92 \times 10^5 \text{ Pa}$.
2
Water is flowing through a horizontal pipe of varying cross-section. At a certain point where the velocity is $0.2 \text{ m/s}$, the pressure is $0.01 \text{ m}$ of mercury column. What is the pressure at a point where the velocity is $0.4 \text{ m/s}$? (Density of water $= 10^3 \text{ kg/m}^3$, Density of Hg $= 13600 \text{ kg/m}^3$, $g = 10 \text{ m/s}^2$). [3 marks]
Answer
$1300 \text{ Pa}$ or $0.0095 \text{ m}$ of Hg
📝
Explanation
Apply Bernoulli’s Theorem for a horizontal pipe ($h_1 = h_2$):
$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$
First, convert $P_1$ to Pascals: $P_1 = \rho_{Hg} g h_{Hg} = 13600 \times 10 \times 0.01 = 1360 \text{ Pa}$.
$1360 + \frac{1}{2}(1000)(0.2)^2 = P_2 + \frac{1}{2}(1000)(0.4)^2$
$1360 + 500(0.04) = P_2 + 500(0.16)$
$1360 + 20 = P_2 + 80 \implies 1380 = P_2 + 80 \implies P_2 = 1300 \text{ Pa}$.
To convert back to Hg column: $h_{Hg} = \frac{1300}{13600 \times 10} \approx 0.0095 \text{ m}$ of Hg.
3
Calculate the work done in blowing a soap bubble from a radius of $2 \text{ cm}$ to $3 \text{ cm}$. (Surface tension of soap solution $S = 0.03 \text{ N/m}$). [2 marks]
Answer
$\approx 3.77 \times 10^{-3} \text{ Joules}$
📝
Explanation
Work done $W = \text{Surface Tension} \times \text{Increase in Area} = S \times \Delta A$.
Since it is a soap bubble, it has 2 free surfaces!
$\Delta A = 2 \times 4\pi (r_2^2 – r_1^2)$
$r_1 = 0.02 \text{ m}$, $r_2 = 0.03 \text{ m}$.
$\Delta A = 8 \pi [(0.03)^2 – (0.02)^2] = 8 \pi [0.0009 – 0.0004] = 8 \pi [0.0005] = 0.004\pi \text{ m}^2$.
$W = 0.03 \times 0.004\pi = 0.00012 \pi \approx 0.000377 \text{ J} = 3.77 \times 10^{-3} \text{ J}$.
Concept Map
Fluid Mechanics connects to →
Macroscopic Properties
Work & Energy (Bernoulli)
Thermal Properties of Matter
Kinetic Theory of Gases
Newton’s Laws (Stokes’ Law)
Related Posts
- Work Energy and Power – Concept Booster | Class 11 Physics CBSE
- Waves – Concept Booster | Class 11 Physics CBSE
- Units and Measurements – Concept Booster | Class 11 Physics CBSE
- Thermodynamics – Concept Booster | Class 11 Physics CBSE
- Thermal Properties of Matter – Concept Booster | Class 11 Physics CBSE
- System of Particles and Rotational Motion – Concept Booster | Class 11 Physics CBSE
- Significant Figures – Concept Booster | Class 11 Physics CBSE
- Oscillations – Concept Booster | Class 11 Physics CBSE
- Motion in a Straight Line – Concept Booster | Class 11 Physics CBSE
- Motion in a Plane – Concept Booster | Class 11 Physics CBSE
- Mechanical Properties of Solids – Concept Booster | Class 11 Physics CBSE
- Mechanical Properties of Fluids – Concept Booster | Class 11 Physics CBSE
- Laws of Motion (Without Friction) – Concept Booster | Class 11 Physics CBSE
- Laws of Motion (With Friction) – Concept Booster | Class 11 Physics CBSE
- Kinetic Theory – Concept Booster | Class 11 Physics CBSE
- Gravitation – Concept Booster | Class 11 Physics CBSE
- Circular Motion – Concept Booster | Class 11 Physics CBSE
