Chapter 10 — Mechanical Properties of Fluids

Fluids — liquids and gases — flow because they cannot sustain shear stress at rest. We develop the statics (pressure, Pascal's law, atmospheric pressure) and the dynamics (continuity, Bernoulli, viscosity, surface tension) needed to explain hydraulic lifts, aircraft lift, droplet formation and capillary rise.

Concept Map

• 10.1 Fluid concept; pressure
• 10.2 Pascal's law; hydraulic machines
• 10.3 Pressure with depth; barometer, manometer
• 10.4 Streamline & turbulent flow; equation of continuity
• 10.5 Bernoulli's principle; Venturi, Torricelli, dynamic lift
• 10.6 Viscosity; Stokes' law; terminal velocity; Reynolds number; Poiseuille
• 10.7 Surface tension; surface energy; excess pressure; capillarity

10.1 Fluid Concept; Pressure

Definition

A fluid is any substance that flows under shear — it cannot maintain a static shear stress, so a fluid at rest is acted upon by purely normal forces from its container.

Pressure at a point:

$\boxed{\ P = \lim_{\Delta A \to 0} \frac{\Delta F_\perp}{\Delta A}\ }$

SI unit: $\text{N}/\text{m}^2 = \text{Pa}$. Dimensions $[\text{ML}^{-1}\text{T}^{-2}]$. Pressure is a scalar — at a point in a fluid at rest the magnitude is the same in every direction (Pascal's first observation).

Useful units:

Derivation — pressure is isotropic in a static fluid

Consider a small right-angled prism inside the fluid with faces of areas $\Delta A_x, \Delta A_y, \Delta A_z$. Force balance (neglecting weight as $O(\Delta V)$ while areas are $O(\Delta A)$) yields $P_x = P_y = P_z$. As the prism shrinks to a point, isotropy is exact.

Gauge vs Absolute

$P_\text{gauge} = P_\text{abs} - P_\text{atm}.$

A tyre pressure gauge reads gauge pressure. A weather barometer reads absolute pressure.

Worked Example

A force of $200$ N acts on a piston of radius $4$ cm. Pressure?

$P = \frac{200}{\pi (0.04)^2} = 3.98\times 10^4\ \text{Pa} \approx 0.39\ \text{atm}.$

Pitfalls

• Pressure is scalar; the force it produces on a surface is vector — along the inward normal.
• "Pressure of $1$ atm" usually means absolute; "tyre at $30$ psi" means gauge.

10.2 Pascal's Law

Definition

A change of pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.

Derivation

Consider two pistons of areas $A_1$ and $A_2$ in communicating cylinders. Pressing piston-1 by force $F_1$ raises the pressure everywhere by $\Delta P = F_1/A_1$. The same $\Delta P$ acts on piston-2:

$F_2 = \Delta P \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}.$

Mechanical advantage $= A_2/A_1$. Work, of course, is conserved: $F_1 d_1 = F_2 d_2$.

Applications

• Hydraulic lift: small force on small piston lifts a car.
• Hydraulic brakes: pedal force amplified through brake fluid to all four wheels simultaneously.
• Hydraulic press.

Worked Example

In a hydraulic lift, $A_1 = 10\ \text{cm}^2, A_2 = 1000\ \text{cm}^2$, $F_1 = 100$ N. Maximum weight liftable?

F_2 = 100 \cdot \frac{1000}{10} = 10^4\ \text{N} \approx 1$ tonne.

Pitfalls

• Force is amplified; distance is reduced by the same factor — energy is not free.
• The fluid must be (nearly) incompressible; air would simply compress and waste the input.

10.3 Pressure with Depth; Barometer, Manometer

Derivation — hydrostatic equation

Take a cylindrical column of fluid of cross-section $A$, between depths $h$ and $h+dh$. Vertical force balance:

$P(h+dh)A - P(h)A - \rho g A\, dh = 0,$ $\frac{dP}{dh} = \rho g \quad\Longrightarrow\quad \boxed{\ P(h) = P_0 + \rho g h\ }$

(where $h$ is depth below the free surface). Consequences:

• Pressure depends only on depth — same in every direction at the same level.
• The shape of the container is irrelevant — the hydrostatic paradox (Pascal's barrel).

Atmospheric Pressure & the Barometer

Torricelli's mercury barometer: a glass tube of mercury inverted into a dish. At the dish surface the atmosphere pushes mercury up the tube; equilibrium gives column height $h$ with

$P_\text{atm} = \rho_\text{Hg}\, g\, h, \quad h \approx 0.76\ \text{m at sea level}.$

Manometer

A U-tube manometer measures gauge pressure of a gas:

$P_\text{gas} - P_\text{atm} = \rho g\, \Delta h.$

Worked Example

A diver is $30$ m below the ocean surface ($\rho_\text{sea} = 1030\ \text{kg/m}^3$). Absolute pressure?

$P = 1.013\times 10^5 + (1030)(9.8)(30) = 1.013\times 10^5 + 3.03\times 10^5 = 4.04\times 10^5\ \text{Pa} \approx 4\ \text{atm}.$

Pitfalls

• Atmospheric pressure pushes up on a barometer, not down; mercury column is held up by it.
• Variation $\rho g h$ assumes incompressible $\rho$; for the atmosphere $\rho$ falls with height — exponential atmosphere.

10.4 Streamline & Turbulent Flow; Continuity

Streamline (laminar) flow

In steady flow, every fluid particle passing through a given point follows the same path — a streamline. Streamlines never cross; their tangent gives the velocity direction. A tube of flow bundles adjacent streamlines.

Turbulent flow: irregular, chaotic, eddying. Above a critical Reynolds number, laminar flow breaks down.

Derivation — Equation of Continuity

For an incompressible fluid in steady flow through a tube whose cross-section changes from $A_1$ to $A_2$ with corresponding speeds $v_1, v_2$: mass entering = mass leaving in time $\Delta t$:

$\rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t \quad\Longrightarrow\quad \boxed{\ A v = \text{const}\ }$

Volume flow rate $Q = Av$ is constant along a tube of flow.

Worked Example

Water flows through a pipe whose radius halves. By how much does speed change?

$A \propto r^2 \implies A \to A/4 \implies v \to 4v.$

Pitfalls

• $Av =$ const assumes incompressibility — for gases at large Mach numbers $\rho Av =$ const instead.
• Streamlines crowd together where flow is fastest, not slowest.

10.5 Bernoulli's Principle

Derivation

Consider a tube of flow between sections 1 and 2 (heights $h_1, h_2$; areas $A_1, A_2$; speeds $v_1, v_2$; pressures $P_1, P_2$). Apply work-energy theorem to a mass element $\Delta m = \rho A v \Delta t$ that moves from section 1 to section 2:

Work done by pressure forces: $W_p = P_1 A_1 v_1 \Delta t - P_2 A_2 v_2 \Delta t = (P_1 - P_2)\frac{\Delta m}{\rho}$.

Gravitational PE change: $\Delta U = \Delta m\, g (h_2 - h_1)$.

KE change: $\Delta K = \tfrac{1}{2}\Delta m (v_2^2 - v_1^2)$.

Set $W_p = \Delta K + \Delta U$ and divide by $\Delta m/\rho$:

$\boxed{\ P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const along a streamline}\ }$

Assumptions

1. Incompressible fluid.
2. Non-viscous (no friction).
3. Steady, streamlined flow.
4. Along a single streamline.

Applications

(a) Venturi meter — narrow throat speeds flow up, drops pressure; pressure difference measured by a manometer gives the flow rate:

$Q = A_1 \sqrt{\frac{2g\Delta h}{(A_1/A_2)^2 - 1}}.$

(b) Torricelli's theorem — efflux speed from a small hole at depth $h$ below the free surface of an open tank:

$v = \sqrt{2 g h}.$

(Same as a body falling through height $h$ — "speed of efflux equals speed of free fall.")

(c) Dynamic lift on an aircraft wing — wing shape forces air over the top to travel faster (longer path). Faster $\Rightarrow$ lower pressure (Bernoulli) $\Rightarrow$ net upward force.

(d) Atomiser / sprayer — air blown across the top of a vertical tube lowers pressure there; liquid is pushed up the tube by atmospheric pressure and sprayed.

(e) Magnus effect — a spinning ball drags air around it; combined with translational airflow, one side moves faster than the other, lowering pressure on that side and curving the trajectory. Explains the swing in cricket and curveball in baseball.

Worked Example — Torricelli

A tank is filled to a height of $5$ m; a hole is punched at the bottom. Efflux speed?

$v = \sqrt{2(9.8)(5)} = \sqrt{98} = 9.9\ \text{m/s}.$

Worked Example — Venturi

Pipe area shrinks from $10\ \text{cm}^2$ to $5\ \text{cm}^2$; water enters at $2$ m/s. Pressure drop?

$v_2 = v_1 \frac{A_1}{A_2} = 4\ \text{m/s},$ $\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}(1000)(16 - 4) = 6000\ \text{Pa}.$

Pitfalls

• Bernoulli holds along a single streamline. Two distinct streamlines need not share the same constant.
• It does NOT hold for viscous flow — extra friction term needed.
• "Faster $\Rightarrow$ lower pressure" is only valid at the same height.

10.6 Viscosity

Definition — Newton's law

In laminar shear flow between two parallel plates separated by $y$, with the top moving at speed $v$, the shear stress required is

$\boxed{\ \tau = \eta\,\frac{dv}{dy}\ }$

where $\eta$ is the coefficient of viscosity (dynamic). SI unit: $\text{Pa·s}$. CGS unit: poise = $0.1\ \text{Pa·s}$.

Typical values:

Viscosity of a liquid decreases with temperature; viscosity of a gas increases with temperature.

Stokes' Law

A sphere of radius $r$ moving at speed $v$ through a viscous fluid of coefficient $\eta$ experiences a drag

$\boxed{\ F_v = 6\pi\eta r v\ }$

(valid for low Reynolds number, $\text{Re}\lesssim 1$). Stokes derived this by solving the slow-viscous-flow equations around a sphere.

Derivation — terminal velocity

A sphere of density $\rho$ falling through a fluid of density $\sigma$ experiences gravity, buoyancy and viscous drag. At terminal velocity $v_t$ the net force is zero:

$\tfrac{4}{3}\pi r^3 \rho g = \tfrac{4}{3}\pi r^3 \sigma g + 6\pi\eta r v_t,$ $\boxed{\ v_t = \frac{2 r^2 (\rho - \sigma) g}{9\eta}\ }$

Consequences:

• $v_t \propto r^2$ — fine raindrops fall slowly; fog hangs in air.
• If $\rho < \sigma$, $v_t < 0$ — the sphere rises (bubble in water).

Reynolds Number

$\boxed{\ \text{Re} = \frac{\rho v d}{\eta}\ }$

A dimensionless ratio of inertial to viscous forces. In pipe flow:

• $\text{Re} \lesssim 2000$: laminar.
• $\text{Re} \gtrsim 3000$: turbulent.
• $2000 - 3000$: transitional / unstable.

Poiseuille's Formula (statement only)

Volume flow rate of a viscous incompressible fluid through a horizontal pipe of radius $a$, length $L$, under pressure difference $\Delta P$:

$\boxed{\ Q = \frac{\pi a^4 \Delta P}{8 \eta L}\ }$

The $a^4$ dependence is severe — halving the radius cuts flow by a factor of 16.

Worked Example — Terminal velocity of a raindrop

$r = 0.1$ mm, $\rho = 1000$ kg/m³, $\sigma_\text{air} = 1.2$ kg/m³, $\eta_\text{air} = 1.8\times 10^{-5}$ Pa·s.

$v_t = \frac{2(10^{-4})^2(1000)(9.8)}{9(1.8\times 10^{-5})} \approx 1.2\ \text{m/s}.$

A heavy drop ($r = 1$ mm): $v_t \approx 120$ m/s — but at this size, Stokes' law fails (high Re); actual terminal speed is $\sim 6$ m/s due to a quadratic drag regime.

Pitfalls

• $v_t \propto r^2$ only in the Stokes regime; large drops follow $v_t \propto \sqrt{r}$ (Newtonian drag).
• Reynolds number is dimensionless — the famous trap is to check units.
• $\eta$ for liquids drops with temperature (motor oil works better when hot); for gases it rises (counter-intuitive).

10.7 Surface Tension

Molecular origin

A molecule deep inside a liquid is pulled equally in all directions by its neighbours. A molecule at the surface has neighbours only on the liquid side, so it experiences a net inward pull. Bringing a molecule to the surface costs energy — the surface has an associated surface energy per unit area, equal to the surface tension $T$.

Definitions

• Surface tension $T$: force per unit length on an imaginary line in the surface, perpendicular to the line, tangent to the surface. SI unit: N/m.
• Surface energy $\sigma$: work done to create unit area of surface; numerically $\sigma = T$ in J/m² = N/m.

Typical values at $20^\circ$C:

Surface tension decreases with temperature, and vanishes at the critical temperature.

Excess Pressure

Liquid drop (one surface): isolating a hemispherical part, force balance gives

$P_\text{in} - P_\text{out} = \frac{2T}{R}.$

Soap bubble (two surfaces, inner and outer):

$P_\text{in} - P_\text{out} = \frac{4T}{R}.$

Air bubble inside a liquid (one surface):

$P_\text{in} - P_\text{out} = \frac{2T}{R}.$

Smaller bubbles have higher internal pressure — when two bubbles merge, the smaller one collapses into the larger.

Derivation — excess pressure in a drop

Imagine cutting a spherical drop of radius $R$ along a great circle. Each hemisphere is held to the other by the surface tension acting along the rim (length $2\pi R$), giving a pulling force $T\cdot 2\pi R$. The excess pressure on the cross-section ($\pi R^2$) supplies the balancing push:

$\Delta P\cdot \pi R^2 = T\cdot 2\pi R \quad\Longrightarrow\quad \Delta P = \frac{2T}{R}.$

Angle of contact

The angle $\theta$ between the tangent to the liquid surface and the solid surface at the line of contact, measured through the liquid:

• $\theta < 90^\circ$: liquid wets the surface (water on clean glass, $\theta \approx 0^\circ$).
• $\theta > 90^\circ$: liquid does not wet (mercury on glass, $\theta \approx 140^\circ$).

Derivation — capillary rise

A capillary tube of radius $r$ dipped vertically in a liquid of surface tension $T$, density $\rho$, contact angle $\theta$. The surface tension pulls upward along the perimeter at angle $\theta$ from vertical:

$F_\text{up} = T\cos\theta \cdot 2\pi r.$

This balances the weight of the lifted column of height $h$:

$\pi r^2 h\,\rho g = 2\pi r T\cos\theta,$ $\boxed{\ h = \frac{2T\cos\theta}{r\rho g}\ }$

• For water on glass ($\theta \approx 0$, $\cos\theta=1$): rise (positive $h$).
• For mercury on glass ($\cos\theta < 0$): depression (negative $h$).
• $h \propto 1/r$ — thinner capillaries lift higher.

Worked Example — capillary

Water in a capillary of radius $0.2$ mm: $T = 0.073$ N/m, $\rho = 1000$ kg/m³, $\theta = 0$.

$h = \frac{2(0.073)(1)}{(2\times 10^{-4})(1000)(9.8)} = 7.4\times 10^{-2}\ \text{m} \approx 7.4\ \text{cm}.$

Worked Example — drop merging

Eight droplets of radius $1$ mm merge into one big drop. Surface energy released? ($T = 0.073$ N/m.)

Volume conservation: $8 \cdot \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi R^3 \Rightarrow R = 2r$.

Initial area: $8\cdot 4\pi r^2 = 32\pi r^2$. Final area: $4\pi R^2 = 16\pi r^2$. Reduction $16\pi r^2$.

$\Delta U = T\cdot \Delta A = 0.073 \cdot 16\pi(10^{-3})^2 = 3.67\times 10^{-6}\ \text{J}.$

This appears as heat — large drops are warmer than the parent droplets.

Pitfalls

• Soap bubble: $4T/R$, not $2T/R$ — two surfaces.
• "Surface tension is force per unit length", not force per area.
• Capillary rise reverses for non-wetting liquids ($\cos\theta < 0$).
• $h \propto 1/r$ — but if the capillary is shorter than $h$, the liquid rises to the top and the meniscus simply flattens; it does not overflow.

Solved Problems

1. A cylindrical tank of radius $1$ m is filled to height $4$ m. A small hole of area $1\ \text{cm}^2$ at the bottom. Initial efflux rate?

$v = \sqrt{2(9.8)(4)} = 8.85\ \text{m/s},$ $Q = Av = (10^{-4})(8.85) = 8.85\times 10^{-4}\ \text{m}^3/\text{s}.$

2. In a hydraulic press the master piston has radius $1$ cm; slave piston radius $20$ cm. If $50$ N is applied to the master, what force is exerted by the slave?

$F_2 = F_1 \cdot \frac{A_2}{A_1} = 50 \cdot (20)^2 = 2\times 10^4\ \text{N}.$

3. An airplane wing has top-surface airflow at $90$ m/s and bottom at $80$ m/s. Density of air $1.2$ kg/m³. Lift per unit area?

$\Delta P = \tfrac{1}{2}(1.2)(90^2 - 80^2) = \tfrac{1}{2}(1.2)(1700) = 1020\ \text{Pa}.$

4. A steel ball of radius $2$ mm and density $7800$ kg/m³ falls through glycerine ($\rho = 1260,\ \eta = 1.5$ Pa·s). Terminal velocity?

$v_t = \frac{2(2\times 10^{-3})^2(7800-1260)(9.8)}{9(1.5)} = \frac{2(4\times 10^{-6})(6540)(9.8)}{13.5} = 0.0379\ \text{m/s}.$

5. Two soap bubbles of radii $3$ cm and $5$ cm coalesce; assume isothermal merger and that the combined enclosed air remains at atmospheric pressure (approx). New bubble radius?

Volume conservation: $R^3 = R_1^3 + R_2^3 = 27 + 125 = 152 \Rightarrow R \approx 5.34$ cm.

(In NCERT a different version uses Boyle's law with $P+4T/R$ on each side — try that as an extension.)

6. A drop of radius $R$ splits isothermally into $n$ identical droplets. Energy required?

$r = R/n^{1/3},\quad \text{new area} = n\cdot 4\pi r^2 = 4\pi R^2 n^{1/3},$ $\Delta U = T(4\pi R^2)(n^{1/3} - 1).$

7. Water rises $4$ cm in a capillary of radius $0.5$ mm. Surface tension? ($\theta=0$.)

$T = \frac{r\rho g h}{2} = \frac{(5\times 10^{-4})(1000)(9.8)(0.04)}{2} = 0.098\ \text{N/m}.$

JEE/NEET Edge Cases

• Bernoulli applied to a falling liquid stream: cross-section narrows as it speeds up — same as $Av=$ const. Asked routinely in JEE.
• Hydrostatic paradox: total downward force on the bottom of a container = $\rho g h A$, independent of container shape. The walls supply or absorb the rest.
• Manometer in an accelerating frame: $g$ replaced by $g_\text{eff}$.
• Stokes vs Newton drag: small slow objects → Stokes ($F\propto v$); large fast objects → quadratic drag ($F\propto v^2$).
• Two soap bubbles connected by a tube: the smaller (higher pressure) empties into the larger. Counter-intuitive but a direct consequence of $\Delta P = 4T/R$.
• Capillary tube shorter than rise: liquid does not overflow; the meniscus radius adjusts so $h_\text{actual} = L$ and $r_\text{meniscus} = 2T/(\rho g L)$.
• Excess pressure for an air bubble inside a liquid: $2T/R$ (one surface), not $4T/R$.
• Viscous flow vs ideal flow: Bernoulli fails; pressure drops linearly along the pipe.

Quick Recap

• Pressure is scalar, isotropic in static fluid; $P = P_0 + \rho g h$.
• Pascal's law $\Rightarrow$ hydraulic lift, brakes, press; force amplified, displacement reduced.
• Continuity: $Av =$ const for incompressible steady flow.
• Bernoulli: $P + \tfrac12\rho v^2 + \rho g h =$ const along a streamline (ideal fluid).
• Torricelli efflux: $v = \sqrt{2gh}$.
• Viscosity: $\tau = \eta (dv/dy)$; Stokes drag $6\pi\eta r v$.
• Terminal velocity $\propto r^2(\rho-\sigma)/\eta$.
• Reynolds number marks laminar–turbulent transition.
• Surface tension = force per unit length; equal to surface energy per unit area.
• Excess pressure: drop $2T/R$, bubble in air $4T/R$, bubble in liquid $2T/R$.
• Capillary rise $h = 2T\cos\theta/(r\rho g)$.

Formula Sheet