CBSE Class 11 Physics • Chapter 09
Chapter Overview: This chapter deals with the properties of fluids (liquids and gases) at rest (Hydrostatics) and in motion (Hydrodynamics). Key topics include Pascal's Law, Bernoulli's Principle, Viscosity, and Surface Tension.
Fluid: A substance that can flow (liquids and gases). It has no definite shape.
Thrust: The normal force exerted by a fluid on a surface.
Pressure (P): Thrust per unit area.
SI Unit: $\text{N/m}^2$ or Pascal (Pa). (Scalar quantity).
Density ($\rho$): Mass per unit volume. $\rho = M/V$. Unit: $\text{kg/m}^3$. Water: $10^3 \text{ kg/m}^3$.
Q: The two thigh bones (femurs), each of cross-sectional area $10 \text{ cm}^2$ support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs. ($g=10$)
Ans: Total area $A = 2 \times 10 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$. Force $F = 40 \times 10 = 400 \text{ N}$.
$P = F/A = 400 / (20 \times 10^{-4}) = 20 \times 10^4 = \mathbf{2 \times 10^5 \text{ Pa}}$.
Statement: The pressure in a fluid at rest is the same at all points if they are at the same height. Also, pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid.
Application: Hydraulic Lift
Figure 9.1: Hydraulic Lift Application
From Pascal's Law: Pressure is transmitted equally.
$$ P = \frac{F_1}{A_1} = \frac{F_2}{A_2} $$
Since $A_2 \gg A_1$, we get $F_2 \gg F_1$. This acts as a force multiplier.
Consider a fluid cylinder of height $h$ and area $A$.
Forces: Weight $mg (\downarrow)$, Force from top $P_1 A (\downarrow)$, Force from bottom $P_2 A (\uparrow)$.
Equilibrium: $P_2 A = P_1 A + mg$. ($m = \rho V = \rho A h$)
$$ P_2 A = P_1 A + \rho A h g \Rightarrow P_2 = P_1 + \rho g h $$
where $P_a$ is atmospheric pressure. Gauge pressure $P_g = P - P_a = \rho g h$.
Streamline Flow: Steady flow where velocity at every point is constant in time.
Turbulent Flow: Irregular flow when velocity exceeds Critical Velocity.
Equation of Continuity: Based on conservation of mass. For an incompressible fluid:
Velocity is higher where area is smaller (e.g., usually specific at nozzle).
Statement: For streamline flow of an ideal fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant.
Derivation (Work-Energy Theorem):
Consider a fluid flowing through a pipe of varying cross-section (Area $A_1$, height $h_1$ to Area $A_2$, height $h_2$).
1. Work done by pressure force: $W_1 = P_1 A_1 v_1 \Delta t$, $W_2 = -P_2 A_2 v_2 \Delta t$.
Net Work $W_P = (P_1 - P_2) \Delta V$.
2. Change in Potential Energy: $\Delta U = \rho \Delta V g (h_2 - h_1)$.
3. Change in Kinetic Energy: $\Delta K = \frac{1}{2} \rho \Delta V (v_2^2 - v_1^2)$.
Work-Energy Theorem: $W_P = \Delta U + \Delta K$
$(P_1 - P_2) = \rho g (h_2 - h_1) + \frac{1}{2} \rho (v_2^2 - v_1^2)$
Rearranging: $P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$.
Figure 9.4: Flow through variable cross-section
Q: A tank is filled with water to height H. A small hole is made at depth $h$ below the surface. Find the distance where the water hits the ground (Range).
Ans: $v = \sqrt{2gh}$. Time to fall height $(H-h)$ is $t = \sqrt{2(H-h)/g}$.
Range $R = v \times t = \sqrt{2gh} \times \sqrt{2(H-h)/g} = 2\sqrt{h(H-h)}$.
Max range occurs at $h = H/2$.
Viscosity: The property of fluid to oppose relative motion between its layers. (Fluid Friction).
Coefficient of Viscosity ($\eta$): $F = - \eta A \frac{dv}{dx}$. Unit: Poiseuille (Pl) or $Pa \cdot s$.
Figure 9.5: Viscous Drag between layers
For a small sphere of radius $r$ falling in viscous medium:
Viscous Drag Force: $F_v = 6\pi \eta r v$.
Terminal Velocity ($v_T$): When Net Force = 0 (Drag + Buoyancy = Weight).
($\rho$ = density of body, $\sigma$ = density of fluid).
Surface Tension (S): The property of a liquid surface to behave like a stretched elastic membrane. It tends to minimize surface area.
Surface Energy (E): Energy stored in the surface per unit area. $E = S \times \Delta A$.
Figure 9.6: Molecular origin of Surface Tension and Excess Pressure
The phenomenon of rise or fall of liquid in a capillary tube.
Q: What is the excess pressure inside a bubble of soap solution of radius $5.00 \text{ mm}$, given that the surface tension of soap solution is $2.50 \times 10^{-2} \text{ N/m}$?
Ans: Soap bubble has 2 free surfaces. Excess pressure $P = 4S/R$.
$P = \frac{4 \times (2.50 \times 10^{-2})}{5.00 \times 10^{-3}} = \frac{10 \times 10^{-2}}{5 \times 10^{-3}} = \mathbf{20 \text{ Pa}}$.
Question: Derive the expression for Terminal Velocity given by Stokes' Law.
Answer: At equilibrium, Net Force = 0.
Weight = Upthrust + Viscous Drag.
$mg = F_B + F_v$
$\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v$
$6\pi \eta r v = \frac{4}{3}\pi r^3 g (\rho - \sigma)$
$v = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.
Question: Why does a small air bubble rise slowly while a large one rises rapidly in a liquid?
Answer: Terminal velocity $v_T \propto r^2$. A large bubble has a larger radius $r$, so its terminal velocity is much higher (square relation) compared to a small bubble.