Mechanical Properties of Fluids Class 11 NCERT Guide & Solutions

Understanding Fluid Hydrostatics: Pressure and Pascal's Law

To understand fluids at rest (hydrostatics), we must define fluid pressure. Pressure ($P$) is the normal force per unit area exerted by a fluid on a surface: $P = F/A$. Unlike solids, fluids exert pressure equally in all directions.

Within a stationary fluid under gravity, pressure increases linearly with depth. Consider a cylindrical column of fluid of density $\rho$, height $h$, and cross-sectional area $A$. By balancing the forces acting on this column—namely the atmospheric pressure $P_a$ from the top, gravity acting downwards ($mg = \rho A h g$), and the upward fluid force ($P A$) from the bottom—we obtain the fundamental hydrostatic equation:

$P = P_a + \rho g h$

Here, $P$ is the absolute pressure at depth $h$, and $\rho g h$ represents the gauge pressure, which is the pressure excess over atmospheric pressure.

This leads directly to Pascal's Law, which states that any change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. This simple rule is the engineering foundation behind hydraulic jacks, brakes, and lifts. In a hydraulic lift, a small force $F_1$ applied on a smaller piston of area $A_1$ generates a much larger force $F_2 = F_1 (A_2 / A_1)$ on a larger piston of area $A_2$. This shows how fluids can act as force multipliers.

Step-by-Step Derivation of Bernoulli's Equation

1. State the Assumptions — We assume the fluid is ideal: it must be incompressible (constant density $\rho$), non-viscous (no internal friction between layers), and the flow must be steady/streamline.
2. Apply the Equation of Continuity — For a pipe of varying cross-section, the volume of fluid entering per second must equal the volume leaving. This gives the continuity equation: $A_1 v_1 = A_2 v_2 = \text{constant}$, showing that velocity increases where the tube narrows.
3. Calculate Work Done by Pressure Forces — As fluid moves through a pipe from position 1 to position 2, the net work done on the fluid by pressure forces is $W = P_1 A_1 \Delta x_1 - P_2 A_2 \Delta x_2 = (P_1 - P_2) \Delta V$, where $\Delta V$ is the volume of fluid elements shifted.
4. Account for Changes in Kinetic and Potential Energy — The change in kinetic energy is $\Delta K = \frac{1}{2} \Delta m (v_2^2 - v_1^2)$ and the change in potential energy is $\Delta U = \Delta m g (y_2 - y_1)$, where $\Delta m = \rho \Delta V$.
5. Formulate the Conservation of Energy — According to the Work-Energy Theorem, $W = \Delta K + \Delta U$. Substituting the terms gives: $(P_1 - P_2) \Delta V = \frac{1}{2} \rho \Delta V (v_2^2 - v_1^2) + \rho \Delta V g (y_2 - y_1)$. Dividing throughout by $\Delta V$ and rearranging terms yields Bernoulli's equation: $P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2$.

CBSE Exam Traps & Conceptual Keys

• Gauge vs. Absolute Pressure: Always read numericals carefully. If it asks for 'absolute pressure', you must add atmospheric pressure ($1.013 \times 10^5 \text{ Pa}$) to the calculated hydrostatic pressure ($\rho g h$).
• Viscosity Temperature Dependency: Remember that viscosity of liquids decreases as temperature rises (cohesive forces weaken), but viscosity of gases increases as temperature rises (molecular collisions increase).
• Surface Tension Bubble Factor: Be highly vigilant about bubbles! When a bubble has two surfaces (like a soap bubble in air), the excess pressure is $4T/r$. If it has only one surface (like an air bubble inside water or a liquid drop), the excess pressure is $2T/r$.
• Equation of Continuity: Fluid speed is inversely proportional to cross-sectional area. This explains why water flows faster from a hose when you partially block the nozzle with your thumb.

Solved Examples on Fluid Mechanics

• Example 1: Hydraulic Lift Calculation A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 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? Step 1: Identify given values. Mass $m = 3000 \text{ kg}$, Area $A = 425 \text{ cm}^2 = 425 \times 10^{-4} \text{ m}^2$, $g = 9.8 \text{ m/s}^2$. Step 2: Calculate the force acting on the piston: $F = m \times g = 3000 \times 9.8 = 29400 \text{ N}$. Step 3: Use Pascal's principle. The pressure $P$ exerted on both pistons is equal: $P = F/A = 29400 / (425 \times 10^{-4}) = 6.92 \times 10^5 \text{ Pa}$. Final Answer: The maximum pressure the smaller piston has to bear is $6.92 \times 10^5 \text{ Pa}$.
• Example 2: Stokes' Law and Terminal Velocity Find the terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at $20^{\circ}\text{C}$. Given: density of copper $\rho_s = 8.9 \times 10^3 \text{ kg/m}^3$, density of oil $\rho_f = 1.5 \times 10^3 \text{ kg/m}^3$, viscosity of oil $\eta = 0.99 \text{ kg m}^{-1}\text{ s}^{-1}$. Step 1: Recall the terminal velocity formula: $v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}$. Step 2: Substitute values: $r = 2 \times 10^{-3} \text{ m}$, $\rho_s - \rho_f = (8.9 - 1.5) \times 10^3 = 7.4 \times 10^3 \text{ kg/m}^3$, $\eta = 0.99 \text{ kg m}^{-1}\text{ s}^{-1}$, $g = 9.8 \text{ m/s}^2$. Step 3: Calculate: $v_t = \frac{2 \times (2 \times 10^{-3})^2 \times 7.4 \times 10^3 \times 9.8}{9 \times 0.99} = \frac{2 \times (4 \times 10^{-6}) \times 7400 \times 9.8}{8.91} \approx 0.065 \text{ m/s}$. Final Answer: The terminal velocity of the ball is $0.065 \text{ m/s}$ (or $6.5 \text{ cm/s}$).

Practice Questions with Solutions

• Q: A fully loaded Boeing aircraft has a mass of $3.3 \times 10^5 \text{ kg}$. Its total wing area is $500 \text{ m}^2$. It is in a level flight with a speed of $960 \text{ km/h}$. Estimate the pressure difference between the lower and upper surfaces of the wings. A: Step 1: In a level flight, the upward aerodynamic lift force balances the downward force of gravity (weight) of the plane. Step 2: Lift Force $F_L = m \times g = (3.3 \times 10^5 \text{ kg}) \times 9.8 \text{ m/s}^2 = 3.234 \times 10^6 \text{ N}$. Step 3: The pressure difference $\Delta P$ across the wings is given by the lift force divided by the wing area $A$: $\Delta P = F_L / A = 3.234 \times 10^6 / 500 = 6.468 \times 10^3 \text{ N/m}^2$. Final answer: The estimated pressure difference between the lower and upper surfaces of the wings is $6.47 \times 10^3 \text{ Pa}$.
• Q: Calculate the capillary rise of water in a clean glass tube of radius $0.05 \text{ cm}$ dipped vertically in water. Take surface tension of water $T = 0.073 \text{ N/m}$, angle of contact $\theta = 0^{\circ}$, density of water $\rho = 1000 \text{ kg/m}^3$, and $g = 9.8 \text{ m/s}^2$. A: Step 1: Identify the capillary rise formula: $h = \frac{2 T \cos\theta}{r \rho g}$. Step 2: Convert radius to meters: $r = 0.05 \text{ cm} = 5 \times 10^{-4} \text{ m}$. Step 3: Substitute the known values into the equation: $h = \frac{2 \times 0.073 \times \cos(0^{\circ})}{5 \times 10^{-4} \times 1000 \times 9.8}$. Step 4: Since $\cos(0^{\circ}) = 1$, evaluate the expression: $h = \frac{0.146}{4.9} \approx 0.0298 \text{ m}$. Final answer: The water rises to a height of approximately $2.98 \text{ cm}$.
• Q: A soap bubble has a radius of $4.0 \text{ mm}$. If the surface tension of the soap solution is $0.03 \text{ N/m}$, calculate the excess pressure inside the bubble. A: Step 1: Recall that a soap bubble suspended in air has two liquid-gas interfaces (outer and inner surfaces). Therefore, the excess pressure formula is $\Delta P = \frac{4 T}{r}$. Step 2: Write down the given values and convert radius to SI units: $r = 4.0 \text{ mm} = 4.0 \times 10^{-3} \text{ m}$ and $T = 0.03 \text{ N/m}$. Step 3: Substitute these values into the formula: $\Delta P = \frac{4 \times 0.03}{4.0 \times 10^{-3}} = \frac{0.12}{0.004} = 30 \text{ Pa}$. Final answer: The excess pressure inside the soap bubble is $30 \text{ Pa}$.
• Q: Calculate the absolute pressure at a depth of $100 \text{ m}$ below the surface of the ocean. Take the density of seawater as $1030 \text{ kg/m}^3$ and atmospheric pressure as $1.01 \times 10^5 \text{ Pa}$. A: Step 1: Identify the absolute pressure formula: $P = P_a + \rho g h$. Step 2: Set up the parameters: $P_a = 1.01 \times 10^5 \text{ Pa}$, $\rho = 1030 \text{ kg/m}^3$, $g = 9.8 \text{ m/s}^2$, and depth $h = 100 \text{ m}$. Step 3: Calculate the gauge pressure: $P_{\text{gauge}} = \rho g h = 1030 \times 9.8 \times 100 = 1009400 \text{ Pa} = 1.0094 \times 10^6 \text{ Pa}$. Step 4: Add the atmospheric pressure to get the absolute pressure: $P = 1.01 \times 10^5 + 10.094 \times 10^5 = 11.104 \times 10^5 \text{ Pa}$. Final answer: The absolute pressure at a depth of 100 meters is $1.11 \times 10^6 \text{ Pa}$.

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