Mechanical Properties of Fluids

This chapter explores the mechanics of fluids, which are substances that can flow, like liquids and gases. It is divided into two main parts: fluid statics and fluid dynamics. Fluid statics deals with fluids at rest, covering concepts like pressure, density, and how pressure varies with depth. You will learn to apply Pascal's law, which explains how hydraulic systems work, and Archimedes' principle, which governs buoyancy and why objects float or sink.

Fluid dynamics examines fluids in motion. It introduces the ideas of streamline and turbulent flow, and the principle of mass conservation through the equation of continuity. The central concept here is Bernoulli's principle, an energy conservation equation for flowing fluids that explains phenomena like the lift on an airplane wing. The chapter also covers properties of real fluids, including viscosity (internal friction) and surface tension (the cohesive force at a liquid's surface), leading to topics like Stokes' law, terminal velocity, and capillary action.

The application of Bernoulli's theorem to complex systems can be difficult, as can problems that combine fluid dynamics with buoyancy or other forces. A strong foundation in Newton's Laws of Motion and the Work-Energy theorem is necessary, as many problems require setting up and solving force or energy balance equations for fluids.

Formulas for Mechanical Properties of Fluids

Pressure Variation with Depth: $P = P_0 + \rho g h$ P is pressure at depth h, P₀ is surface pressure, ρ is fluid density.
Buoyant Force (Archimedes' Principle): $F_B = V_{sub} \rho_f g$ V_sub is the submerged volume of the object, ρ_f is the fluid density.
Equation of Continuity: $A_1 v_1 = A_2 v_2$ For an incompressible fluid in steady flow; A is cross-sectional area, v is fluid speed.
Bernoulli's Equation: $P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$ For an ideal fluid in streamline flow; P is pressure, v is speed, h is height.
Stokes' Law: $F_v = 6 \pi \eta r v$ Viscous drag force on a sphere of radius r moving at speed v in a fluid of viscosity η.
Terminal Velocity: $v_t = \frac{2}{9} \frac{r^2 (\rho_s - \rho_l) g}{\eta}$ Constant velocity of a sphere (density ρ_s) falling in a fluid (density ρ_l).
Excess Pressure in a Liquid Drop: $\Delta P = \frac{2T}{R}$ ΔP is the pressure difference across the surface; T is surface tension, R is radius.
Excess Pressure in a Soap Bubble: $\Delta P = \frac{4T}{R}$ A soap bubble has two liquid-air interfaces, hence the factor of 4.
Capillary Rise or Fall: $h = \frac{2T \cos\theta}{r \rho g}$ h is the height, T is surface tension, θ is the angle of contact, r is the tube radius.

One of these questions, worked through

JEE Advanced 2015

Two spheres $P$ and $Q$ of equal radii have densities $\rho_1$ and $\rho_2$ , respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities $\sigma_1$ and $\sigma_2$ and viscosities $\eta_1$ and $\eta_2$ , respectively. They float in equilibrium with the sphere $P$ in $L_1$ and sphere $Q$ in $L_2$ and the string being taut (see figure). If sphere $P$ alone in $L_2$ has terminal velocity $\overline{V}_P$ and $Q$ alone in $L_1$ has terminal velocity $\overline{V}_Q$ , then

Mechanical Properties of Fluids Two spheres

P

and

Q

of equal radii have densities

\rho_1

and

\rho_2

, respectively. The spheres are connected by a massless string and placed in liquids

L_1

and

L_2

of densities

\sigma_1

and

\sigma_2

and viscosities

\eta_1

and

\eta_2

, respectively. They float in equilibrium with the sphere

P

L_1

and sphere

Q

L_2

and the string being taut (see figure). If sphere

P

alone in

L_2

has terminal velocity

\overline{V}_P

and

Q

alone in

L_1

has terminal velocity

\overline{V}_Q

, then

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Step 1: Analyze the equilibrium of the connected spheres.

Sphere $P$ of volume $V$ floats in liquid $L_1$ , and sphere $Q$ of volume $V$ floats in liquid $L_2$ . Since the string is taut, the tension $T$ acts downwards on $P$ and upwards on $Q$ .

For sphere $P$ in liquid $L_1$ :

\sigma_1 V g = \rho_1 V g + T

For sphere $Q$ in liquid $L_2$ :

\sigma_2 V g + T = \rho_2 V g

Step 2: Relate the densities using the tension.

Solve both equations for the tension $T$ .

T = (\sigma_1 - \rho_1) V g = (\rho_2 - \sigma_2) V g

Equating the terms gives:

\sigma_1 - \rho_1 = \rho_2 - \sigma_2

Rearranging the terms yields:

\rho_1 - \sigma_2 = -(\rho_2 - \sigma_1)

Taking the absolute values:

|\rho_1 - \sigma_2| = |\rho_2 - \sigma_1|

Step 3: Write the expressions for the terminal velocities.

The terminal velocity of a sphere of radius $r$ and density $\rho$ in a liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law.

For sphere $P$ alone in liquid $L_2$ :

\overline{V}_P = \frac{2}{9}\frac{r^2 g(\rho_1 - \sigma_2)}{\eta_2}

For sphere $Q$ alone in liquid $L_1$ :

\overline{V}_Q = \frac{2}{9}\frac{r^2 g(\rho_2 - \sigma_1)}{\eta_1}

Step 4: Find the ratio of the magnitudes of the terminal velocities.

Using the equality of the density differences $|\rho_1 - \sigma_2| = |\rho_2 - \sigma_1|$ , we find the ratio of the speeds:

\frac{|\overline{V}_P|}{|\overline{V}_Q|} = \frac{\eta_1}{\eta_2}

Step 5: Determine the sign of the dot product of the velocities.

Since the density differences have opposite signs:

\rho_1 - \sigma_2 = -(\rho_2 - \sigma_1)

The terminal velocities $\overline{V}_P$ and $\overline{V}_Q$ point in opposite directions. Thus, their dot product is negative:

\overline{V}_P \cdot \overline{V}_Q < 0

Answer: The correct relations are $\dfrac{|\overline{V}_P|}{|\overline{V}_Q|} = \dfrac{\eta_1}{\eta_2}$ and $\overline{V}_P \cdot \overline{V}_Q < 0$ .

Open this question on its own page →

Past-paper questions on Mechanical Properties of Fluids

JEE Advanced 54 questions