SF Prep
06 Physics Notes 🕒 Updated 2026-10-06
Page 1
Density, Pressure & Depth

Mechanical Properties of Fluids

1. Fluids – Density & Relative Density

  • Fluid : material which has the property to flow → liquids + gases.
  • Liquid : no fixed shape, but fixed volume. Gas : neither fixed shape nor fixed volume.
ρ = mV  ⇒  m = ρV
  • Density = mass per unit volume ; unit kg/m³. Water ≈ 1000 kg/m³ = 1 g/cm³ (1 litre ≈ 1 kg).
  • 1 cm³ = 10⁻⁶ m³ ;  1 litre = 10⁻³ m³.
Relative density (specific gravity) = density of materialdensity of water
  • A ratio ⇒ no unit. e.g. specific gravity 7.2 ⇒ ρ = 7200 kg/m³ = 7.2 g/cm³.

2. Density of a Mixture

  • Liquids mixed together (total volume = sum of volumes) ; use m = ρV :
ρeq = total masstotal volume = ρ₁V₁ + ρ₂V₂ + …V₁ + V₂ + …
  • Equal volumes : ρeq = ρ₁ + ρ₂ + …n (arithmetic mean).
  • Equal masses : ρeq = n1/ρ₁ + 1/ρ₂ + … (harmonic mean).
Mixing by mass (m₁ : m₂) : ρeq = m₁ + m₂m₁/ρ₁ + m₂/ρ₂

e.g. 1 : 2 by volume ⇒ (ρ₁ + 2ρ₂)/3 ; 1 : 2 by mass ⇒ 3ρ₁ρ₂/(ρ₂ + 2ρ₁).

3. Pressure & Force due to Pressure

  • Pressure : force per unit area on a surface, due to collisions of the fluid molecules with it. Unit N/m² = Pascal (Pa) ; a scalar.
F = PA  ;  F = ∫ P dA

(F = PA when P is uniform over the area ; F = ∫P dA when P varies)

  • Force due to pressure is always perpendicular to the surface and is a push (compressive) force.

Atmospheric pressure :

  • Due to the weight of the air column above a point ; measured by a barometer. Decreases with altitude.
Patm = P₀ ≈ 1.013 × 10⁵ Pa ≈ 10⁵ Pa
  • At the interface of air and liquid, pressure is continuous ⇒ at the free surface P = P₀.

4. Variation of Pressure with Depth

  • Fluid at rest ; liquid incompressible (ρ = constant). Pressure varies due to gravity, acceleration and rotation of the vessel.
P₁A P₂A (dm)g dx AB h air, P₀
  • Thin layer (area A, thickness dx, mass dm = ρA dx) in equilibrium :

P₂A = P₁A + (ρA dx)g ⇒ dP = ρg dx ⇒ ∫AB dP = ∫₀h ρg dx

ΔP = ρgh   ;   P = P₀ + ρgh
  • Pressure increases linearly as we move down.
  • Holds between any two points of the same liquid at rest, along any path (h = vertical depth of B below A).
  • All points at the same horizontal level in the same continuous liquid at rest have the same pressure — independent of the shape of the vessel.
h P₀ P = P₀ + ρgh (all equal)

5. Pressure in a Vertically Accelerating Vessel

ΔP = ρ geff h
Motion of vesselgeffPB − PA
Acceleration a upwardg + aρ(g + a)h
Acceleration a downwardg − aρ(g − a)h
Free fall (a = g, down)00

(B is at depth h below A ; in free fall the pressure is the same everywhere in the liquid.)

Page 2
Layers, Vessel Walls & Pascal

6. Layered Liquids & U-tubes

  • Immiscible layers in an open vessel : add ρgh layer by layer.
Pbottom = P₀ + ρ₁gh₁ + ρ₂gh₂ + ρ₃gh₃
  • U-tube rule : equate pressures at the level of the lower interface, where both points lie in the same liquid ⇒ ρ₁h₁ = ρ₂h₂ (heights measured above that level).
h h₁ ρ2ρ P₀P₀ same level
Ex. Liquids ρ (column h₁, left) and 2ρ in a U-tube. Height difference h of the two surfaces?
Sol. At the interface level (both points in 2ρ) : P₀ + ρgh₁ = P₀ + 2ρg(h₁ − h) ⇒ h = h₁/2 (lighter-liquid side higher).

7. Force on the Bottom & Walls of a Vessel

  • Ignore P₀. At the bottom (depth h) pressure is uniform = ρgh ⇒ F₁ = ρgh × (base area).
  • On slant walls pressure is variable ⇒ F₂ = ∫P dA (always ⊥ wall).
  • FBD of the liquid : weight W (down), F₁ from bottom (up), net vertical force F₂ from the walls ⇒ F₁ ± F₂ = W.
Vessel (filled, height h)Result
CylinderF₁ = W ; wall forces horizontal, cancel ⇒ net 0
Wider at topW > F₁ ; liquid pushes walls down with W − F₁
Narrower at topW < F₁ ; liquid pushes walls up with F₁ − W

e.g. frustum, bottom radius R, top 2R : W = 73πR²ρgh, F₁ = πR²ρgh ⇒ walls pushed down by 43πR²ρgh. Top R/2, bottom R : W = 712πR²ρgh ⇒ walls pushed up by 512πR²ρgh.

8. Hydrostatic Paradox

  • Vessels of different shapes, same base area A, filled to the same height h with the same liquid :
h FFF W₁W₂W₃
F₁ = F₂ = F₃ = (ρgh)A

but the weights differ : W₂ > W₁ > W₃

  • Force on the base depends only on h and A, not on the amount (weight) of liquid.
  • Weighing machine reads the total weight (vessel + liquid), not the force on the base — the liquid also pushes on the walls.

9. Force on a Vertical Wall

dx x P = ρgx ρ wall
  • Strip of area l dx at depth x (cube of side l, filled) : P = ρgx

F = ∫₀l (ρgx)(l dx) = ρgl³2 = mg2  (bottom : ρgl · l² = mg)

Fwall = PC × Aprojected

PC = pressure at the centroid C of the (projected) wall area — valid because P varies linearly with depth.

Rectangular wall (width b, liquid depth h) : F = ρgbh²2
  • Centre of pressure (line of action of F) lies below the centroid : at depth 2h3 for a rectangular wall (torque about the bottom edge = F·h/3 = ρgbh³/6).
  • Layered liquids : apply the rule separately to the part of the wall in contact with each layer.
  • Common wall between liquids ρ₂ > ρ₁ (same h, width a) : net force = (ρ₂ − ρ₁)gah²/2 towards the lighter side.
  • Curved wall : liquid in equilibrium ⇒ force on the curved wall = force on the flat face it closes, e.g. half-cylinder (radius R, height h) : F = (ρgh/2)(2Rh) = ρgRh².
Ex. Square base a × a ; liquid ρ (height h) floats on 2ρ (height h). Force on a side wall?
Sol. Upper part : (ρg h2)(ah) = ½ρgah². Lower part : centre pressure = ρgh + 2ρgh2 = 2ρgh ⇒ 2ρgah².
Total = 52ρgah² ; bottom : (ρgh + 2ρgh)a² = 3ρgha².

10. Pascal's Law & Hydraulic Lift

  • Pascal's law : a pressure change applied at any point of a confined incompressible fluid is transmitted undiminished to every point of the fluid and to the walls of the container.
F₁ F₂ A₁A₂ P₁ = P₂

P₁ = P₂ ⇒ F₁A₁ = F₂A₂

F₂F₁ = A₂A₁
  • Small force on the small piston lifts a large load on the big piston — force multiplier.
  • Force is multiplied, work is not : A₁d₁ = A₂d₂ ⇒ F₁d₁ = F₂d₂.
  • Pistons at different levels (lower one at depth h below the other) : FlowAlow = FhighAhigh + ρgh.
Page 3
Barometer, Accelerated & Rotating Liquids

11. Gauge Pressure & Manometer

  • Gauge pressure = pressure measured with respect to atmospheric pressure (P itself is the absolute pressure).
Pgauge = P − Patm
  • May be positive, negative or zero : P = 1.5 × 10⁵ Pa ⇒ +0.5 × 10⁵ Pa ; P = 0.4 × 10⁵ Pa ⇒ −0.6 × 10⁵ Pa.
  • At depth h in an open liquid : Pgauge = ρgh.

Manometer (U-tube measuring gas pressure) :

Gas, P h P₀
P = P₀ + ρgh  ⇒  Pgauge = ρgh
  • Several liquids : start from one end, add ρgh going down and subtract going up, until the other end ; e.g. specific gravity 4, h = 6 cm ⇒ Pgauge = 2400 Pa.

12. Barometer & Faulty Barometer

  • Tube closed at the top, filled with liquid and inverted in a trough ; space above the column is (nearly) vacuum, P = 0.
vacuum, P = 0 h P₀ (air) P₀ ρ
P₀ = ρgh  ⇒  h = P₀ρg
  • Mercury (13.6 g/cm³) : h ≈ 76 cm ; water : h ≈ 10.3 m (too tall ⇒ mercury is used).
  • P₀ ≈ 10⁵ Pa ≈ 76 cm of Hg. In mercury 5 cm below the surface ⇒ 81 cm of Hg ; 2 cm above the outside level inside a column ⇒ 74 cm of Hg.

Faulty barometer (air trapped at the top) :

P + ρgh = P₀  ⇒  P = P₀ − ρgh
  • Reading h is less than the true value P₀/ρg.
  • Trapped gas at constant temperature : P₁V₁ = P₂V₂ ; uniform tube ⇒ V = A × (length of air column).
  • Air column closed by a mercury pellet of length l : open end down ⇒ P = P₀ − l ; open end up ⇒ P = P₀ + l ; horizontal ⇒ P = P₀ (in cm of Hg).

13. Pressure in a Horizontally Accelerated Liquid

  • Horizontal column (length l, area A) between B (rear) and A (front), acceleration a :

PBA − PAA = (ρAl)a

PB − PA = ρal
  • Gravity : pressure increases along mg ⇒ ΔP = ρgh (vertical distance h).
  • Acceleration : pressure increases along the pseudo force, i.e. opposite to a ⇒ ΔP = ρal (horizontal distance l).

14. Free Surface of an Accelerated Liquid

  • Open vessel with horizontal acceleration a : at steady state the free surface is inclined.
α a air, P₀ AC B yx

A, B on the free surface (P₀) ; C below A and behind B : P₀ + ρgy = P₀ + ρax

tanα = ag
  • Surface rises towards the rear ; it is perpendicular to geff.
  • U-tube (horizontal arm l) accelerated along the arm : h = alg, rear arm higher.

Vessel moving up a smooth incline (θ) with acceleration a :

tanα = a + g sinθg cosθ

(α = angle of the surface with the incline ; per unit mass : a + g sinθ along the incline, g cosθ normal to it.)

Closed vessel completely filled :

  • Closed vessel (length l, height h), completely filled, acceleration a towards the front : Prear top − Pfront top = ρal ; Pfront bottom − Pfront top = ρgh ; Prear bottom − Pfront top = ρgh + ρal (rear bottom maximum, front top minimum).
  • General acceleration : pressure is maximum at the corner farthest along geff = g − a and minimum at the opposite corner. Cube of side l, a = a₁î + a₂ĵ + a₃k̂ (ĵ up) : Pmax − Pmin = ρ(g + a₂)l + ρa₁l + ρa₃l.

15. Pressure in a Rotating Liquid

  • Horizontal tube of liquid rotating with angular velocity ω about a vertical axis through O.
ω dx OA x l

Element (dm = ρA dx) needs centripetal force dm ω²x from the pressure difference :

(P + dP)A − PA = (ρA dx)ω²x ⇒ ∫P₀PA dP = ∫₀l ρω²x dx

PA − P₀ = ½ ρω²l²

(P₀ = pressure on the axis ; A at distance l from the axis.)

PB − PA = ½ ρω²(l₂² − l₁²)
  • Pressure increases away from the axis (l₂ > l₁ measured from the axis).
  • Rotating U-tube, arms at distances r₁ and r₂ from the axis : h = ω²(r₂² − r₁²)2g, the farther arm higher (e.g. 2l and l ⇒ 3ω²l²/2g).
  • Tube closed at both ends, full of liquid of mass m, rotated about one end : force on the far end = ½ρω²L²A = mω²L/2 (= net centripetal force on the liquid).
Page 4
Rotating Vessel, Archimedes & Floating

16. Rotating Vessel – Paraboloid Surface

ω ABC h R
Ex. Cylinder of radius R rotates about its axis with ω. Height difference of the surface between centre and wall? Shape of the surface?
Sol. A (centre) and B (wall) at P₀ ; C at A's level below B : P₀ + ρgh = P₀ + ½ρω²R² ⇒ h = ω²R²/2g.
Any surface point (x, y) from A : ρgy = ½ρω²x² ⇒ y = ω²x²/2g — a parabola (surface is a paraboloid of revolution).
h = ω²R²2g  ;  y = ω²x²2g

e.g. R = 5 cm, 2 rev/s (ω = 4π rad/s) ⇒ h ≈ 2 cm.

17. Archimedes' Principle & Buoyant Force

  • A body immersed in a fluid, fully or partially, experiences an upward buoyant force equal to the weight of the fluid displaced by it.

Spring balance : 5 kg in air, 3 kg immersed ; the overflow water weighs 2 kg = loss in weight.

Fb = mliquid displaced g = ρ V g
  • ρ → density of the liquid ; V → immersed volume of the body.
  • Accelerating frame (lift / vessel) : Fb = ρ V geff.
  • Mass of the body M = σV₀ (σ = density of body, V₀ = total volume).

18. Buoyancy from Pressure

  • Buoyant force = net force exerted by the liquid on the body due to pressure.
P₀ A F₁ F₃ F₄ F₂ ρ h₁ h

Side forces cancel (F₂ = F₄). Fnet = F₃ − F₁ = [P₀ + ρg(h + h₁)]A − [P₀ + ρgh₁]A = ρ(hA)g

Fnet = Fbottom − Ftop = ρ V g = Fb
  • Independent of P₀ and of the depth h₁ of the body.
  • Any shape : replace the body by liquid of the same shape ; that liquid is in equilibrium, so the net pressure force on it = its weight = ρVg.
  • Trick : Fb = ρVg is a short-cut to the net pressure force ; conversely Fbottom − Ftop = Fb gives an unknown pressure force on one surface.

19. Sinking, Floating & Free Float

DensityForces (fully immersed)Result
σ < ρ (wood)Fb > mgmoves up, floats partly immersed
σ = ρFb = mgstays at rest anywhere inside
σ > ρ (iron)Fb < mgsinks
V V₀ Fb mg ρ P₀
  • Free float (only Fb and mg act) : ρVg = σV₀g
V = σρ V₀   (fraction immersed = σ/ρ)
Ex. Ice (σ = 900 kg/m³) in water : V = 9001000V₀ = 0.9 V₀ ⇒ 90 % of the ice is immersed, 10 % above water. A body floating half-immersed in A and ⅔ immersed in B (ρA = 2σ, ρB = 3σ/2) floats ⅗ immersed in a 1 : 2 (by volume) mixture, since ρeq = 5σ/3.

20. Floating Results & Apparent Weight

  • Extra load to just submerge a floating body : mg = ρ(V₀ − V)g (extra buoyancy of the part above the liquid).
  • Boat (base area A) sinks further by Δh when load m is added : m = ρAΔh.
  • Hollow body : displaced volume uses the outer volume, weight uses the volume of material : ρVdisp = σVmaterial (e.g. shell of radii R, 2R floating half-immersed ⇒ σ = 4ρ/7).
  • Body across two liquids (length h₁ in ρ₁, h₂ in ρ₂, h₁ + h₂ = h) :
σh = ρ₁h₁ + ρ₂h₂

Apparent weight :

T = mg − Fb  (string / spring balance)
  • The body pushes the liquid down with Fb (Newton's 3rd law) ⇒ weighing machine under the beaker reads more by Fb ; total of both readings is unchanged.

21. Vertical Oscillation of a Floating Block

  • Block (height h, area A, density σ) floats in liquid ρ with immersed depth h₁ : ρh₁ = σh.

Push down by x : Fnet = extra Fb = (ρAg)x (restoring) ⇒ SHM with k = ρAg

T = 2π √σhρg = 2π √h₁g

(Viscous drag and liquid motion neglected ; block stays partly immersed.)

Page 5
Buoyancy Cases, Flow & Continuity

22. Force on Curved Surfaces

  • Treat the curved part as one face of a closed body : Fcurved (net, vertical) and Fflat differ by the buoyant force.
Fup, bottom − Fdown, top = Fb = ρV g
  • Cone (base area πR², height H) hung vertex down, base at depth h : Fcurved = ρgπR²(h + H3) upward — independent of the cone's density.
  • Sphere (radius R) with top at depth h, force on the lower hemisphere (flat face at depth h + R) :
F = [P₀ + ρg(h + R)]πR² + 23πR³ρg  (upward)

23. Two Balls & Two-Liquid Floating

  • Two balls (same V) joined by a taut string, in equilibrium inside liquid ρ : upper σ₂ < ρ < σ₁ lower.
σ₁ + σ₂ = 2ρ  ;  T = (σ₁ − ρ)Vg = (ρ − σ₂)Vg
  • Block floating across two liquids : the upper liquid presses only on the top face, the lower only on the bottom face ; Fbottom − Ftop = Fb1 + Fb2 = mg.

24. Buoyancy in an Accelerating Vessel

  • Vessel with horizontal acceleration a : in its frame a pseudo force σVa acts backward ; the liquid gives an extra buoyant force ρVa along a (replace g by a).
  • W.r.t. the vessel : heavier body (iron) moves backward, lighter body (wood) moves forward.
Horizontal : Fb = ρVa (along a)  ;  pseudo force = σVa (opposite to a)
a θ T ρVg ρVa σVg σVa

Ball (σ > ρ) on a string in a closed, filled vessel : T sinθ = (σ − ρ)Va , T cosθ = (σ − ρ)Vg

tanθ = ag  ;  T = (σ − ρ)V√(g² + a²)

(String lies along geff ; for a light ball σ < ρ it tilts forward.)

25. Ball Dropped into a Liquid

  • Ball (density σ < ρ) released from height h above the surface ; no loss of speed on entry, no viscosity.

In air : v₀² = 2gh. In liquid (upward retardation) : ρVg − σVg = σVa ⇒ a = (ρσ − 1)g

Work–energy : σVg(h + s) − ρVgs = 0 ⇒

smax = σhρ − σ

e.g. σ = 800, ρ = 1000 kg/m³ ⇒ a = g/4, s = 4h.

26. Melting of Floating Ice

before (h₁) after (h₂ < h₁)
Ex. Ice floats in water ; does the level change when it melts? What if a stone (σs > ρ) is inside the ice?
Sol. Floating : displaced volume V = σV₀/ρ. Melting conserves mass : melt-water volume V′ = σV₀/ρ = V ⇒ level unchanged.
With a stone : while floating it displaces ms/ρ ; after melting it sinks and displaces only ms/σs < ms/ρ ⇒ level falls.
  • Similarly : a steel block on a floating wooden block thrown into the water ⇒ level decreases.

27. Centre of Buoyancy (COB)

  • Fb acts at the centre of buoyancy = centroid of the displaced volume.
  • Weight acts at the COM. If COB and COM are not on one vertical line, Fb and mg form a couple ⇒ the body rotates.

Wood–iron rod (equal volumes) fully immersed : COB at the geometric centre, COM towards iron ⇒ iron end goes down.

  • Rod hinged at the bottom (length l, density σ < ρ, liquid depth h) : immersed length L = h secθ ; torque about the hinge ⇒ ρL² = σl² :
cosθ = hl√ρσ  (θ with the vertical)

(Valid when h < l√(σ/ρ) ; otherwise the rod stays vertical.)

28. Fluid Dynamics – Streamlines & Types of Flow

  • Streamlines : curves that are, at every instant, tangent to the velocity of the flow at each point.

Laminar (streamline) flow :

  • Streamlines do not change with time (steady flow) ; every particle through a given point follows the same path.

Turbulent flow :

  • Streamlines change with time ; particles through a point do not follow the same path (eddies form).

29. Reynolds Number

  • Dimensionless ratio of inertial force to viscous force ; decides whether pipe flow is laminar or turbulent.
Re = ρ v dη = 4ρQπηd

ρ → density of liquid, v → speed, d → diameter of tube, η → coefficient of viscosity (Pa·s), Q → volume flow rate (v = 4Q/πd²)

  • Re < ≈ 2000 : laminar ; Re > ≈ 3000 : turbulent ; in between : unstable (transition). Exact limits vary slightly from book to book.
  • Large ρ, v, d ⇒ turbulence more likely ; large η ⇒ laminar.
  • For the same Q : Re ∝ 1/d (wider tube ⇒ smaller Re).

30. Flow Rate & Equation of Continuity

  • Pipe of area A, speed v : in time dt the liquid crossing a section fills length v dt.
Q = dVdt = Av  ;  dmdt = ρAv = ρQ

Units : m³/s (or L/s) and kg/s. Water : 1 L/s ↔ 1 kg/s ; 1 cm³/s ↔ 1 g/s.

  • Mass is conserved (steady flow, no leak) : mass in per second = mass out per second, at any section or junction.
v₁ v₂ A₁A₂ ρ₁ρ₂
ρ₁A₁v₁ = ρ₂A₂v₂
  • Incompressible liquid (ρ constant) :
A₁v₁ = A₂v₂  ⇒  Av = constant
  • Narrower section ⇒ faster flow (v ∝ 1/A ∝ 1/r²) ; streamlines crowd where speed is large.
  • Junction : Av = A₁v₁ + A₂v₂ + … (volume in = volume out).
Page 6
Bernoulli, Siphon, Efflux & Lift

31. Bernoulli's Equation

  • For steady, streamline flow of an incompressible, non-viscous liquid, the total energy per unit volume is the same at all points of a streamline.
v₁ v₂ P₁, A₁ P₂, A₂ h₁ h₂ reference

Energy of a liquid element (mass dm, volume dV) :

EnergyTotalPer unit massPer unit volume
Kinetic½(dm)v²½v²½ρv²
Potential(dm)ghghρgh
PressureP dVP/ρP

(Pressure energy per unit volume equals the pressure P itself.)

P + ½ρv² + ρgh = constant
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

together with continuity : A₁v₁ = A₂v₂

  • Horizontal tube : where speed is more, pressure is less.
  • Uniform tube ⇒ same v everywhere ⇒ Plower − Pupper = ρgh (as in statics).
  • Note : P = P₀ + ρgh holds only for a liquid at rest ; for moving liquid use Bernoulli.
  • Point open to atmosphere (free jet / open end) ⇒ P = P₀.
  • Valve in a horizontal pipe : closed reading P₁, open P₂ ⇒ v = √(2(P₁ − P₂)/ρ).

32. Venturi Meter

  • Device to measure the volume flow rate in a pipe.
h A₁ , v₁ , P₁ A₂ , v₂ , P₂

Continuity : v₂ = A₁v₁/A₂ ; Bernoulli (same height) : P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ρv₁²A₁² − A₂²A₂²

Manometer tubes : P₁ − P₂ = ρgh ⇒ v₁ = A₂ √(2gh)√(A₁² − A₂²)

Q = A₁v₁ = A₁A₂ √2ghA₁² − A₂²
  • Wide part : v less, P more ; throat : v more, P less.
Ex. Horizontal tube, areas 40 cm² and 20 cm² ; pressure difference 700 Pa (water). Flow rate?
Sol. vB = 2v ⇒ ΔP = ½ρ(4v² − v²) = 32ρv² ⇒ 700 = 1500 v² ⇒ v ≈ 0.683 m/s
Q = (40 × 10⁻⁴)(0.683) ≈ 2.73 × 10⁻³ m³/s (≈ 2.73 L/s).

33. Siphon

  • Uniform tube draining a vessel : A = inlet (depth h₁), B = highest point (h₂ above the surface), C = open outlet (h₃ below A).
A B C (P₀) P₀ h₁ h₂ h₃
  • Uniform tube ⇒ vA = vB = vC = v ; C open to air ⇒ PC = P₀.
  • Liquid at A is moving, so PA ≠ P₀ + ρgh₁ ; Bernoulli with C :
PA = P₀ − ρgh₃  ;  PB = P₀ − ρg(h₁ + h₂ + h₃)
  • Flow condition : PB > 0 ⇒ P₀ > ρg(h₁ + h₂ + h₃) ; C must be below the free surface.
  • Free surface → C (tube area ≪ vessel area) : v = √(2g(h₁ + h₃)).

34. Speed of Efflux (Torricelli) & Range

  • Small hole (area a) in the wall of a vessel (area A), depth h below the free surface.
v h h₁ H A , P₀ R

Continuity : surface speed vs = av/A ; Bernoulli : P₀ + ½ρvs² + ρgh = P₀ + ½ρv²

v = √2gh1 − a²/A² ≈ √(2gh)  (a ≪ A)
  • Same speed as a body falling freely through h.

Range on the ground (hole at height h₁ = H − h) :

time of fall t = √(2h₁/g) ⇒ R = vt = √(2gh)·√(2h₁/g)

R = 2√(hh₁) = 2√(h(H − h))
  • R is maximum at h = h₁ = H/2 ; Rmax = H.
  • Same R if h and h₁ are interchanged (hole at depth d from the top and at height d from the bottom).

Two liquid layers (ρ₁, h₁ over ρ₂, h₂), hole at the bottom :

v = √2g(ρ₁h₁ + ρ₂h₂)ρ₂
  • Piston pushing liquid out of a nozzle : Ppiston − P₀ = ½ρ(v₂² − v₁²) ⇒ force F = (Ppiston − P₀)A₁.
  • A falling stream narrows (RC = R(H/(H + h))1/4 at distance h below the tap) ; a rising jet slows and widens.

35. Thrust Force

  • Liquid flowing out with speed v (relative to the vessel) pushes the vessel the opposite way.
Fth = vreldmdt = ρav² = 2ρagh
  • v = √(2gh) (a ≪ A). Smooth ground : acceleration = Fth/M ; rough ground : on the verge of sliding when Fth = μMg (M = vessel + liquid).
  • Light vessel, hole at the bottom : μmin = 2ρaghρAhg = 2aA.

36. Time to Empty a Tank

A H a v = √(2gh)

A(−dhdt) = a√(2gh) ⇒ ∫Hh dh√h = −aA√(2g) ∫₀t dt

t = Aa√2g(√H − √h)
Complete emptying :  T = Aa√2Hg
  • T ∝ √H : the lower half takes longer ; t2nd half/t1st half = 1/(√2 − 1) = √2 + 1 (e.g. 200 s → ≈ 483 s).
  • Steady level with inflow Q (volume per second) : Q = a√(2gh) ⇒ h = Q²2ga².

37. Dynamic Lift (Bernoulli in Air)

Wing v₂ (faster), P₂ (less) v₁ , P₁
  • Air above the wing moves faster (v₂ > v₁) ⇒ pressure above is less ⇒ net upward force.
Lift = (P₁ − P₂)A = ½ρ(v₂² − v₁²)A
  • Take-off when lift = Mg.
  • Wind of speed v over a roof (still air inside) : ΔP = ½ρv² ⇒ force ½ρv²A upward (roof blown off).
Page 7
Viscosity & Terminal Velocity

38. Viscosity & Viscous Force

  • Viscosity : property of a fluid by which it resists flow — opposes relative motion between its layers and of bodies moving through it (honey : high, water : low).

Newton's law of viscosity :

v v + dv dy fixed surface
Fv = − η A dvdy
  • η = coefficient of viscosity ; (−) : the force opposes relative motion — faster layer dragged back, slower layer dragged forward.
SystemUnit of η
SIPa·s (= N s/m²) = 1 decapoise = 1 poiseuille
CGSpoise ;  1 poise = 0.1 Pa·s (10 poise = 1 Pa·s)

Water (room temperature) : η ≈ 1 mPa·s = 10−3 Pa·s (= 0.01 poise) ;  dimensions [ML⁻¹T⁻¹]

Linear velocity profile (e.g. river of depth h) :

  • v varies linearly from 0 at the bed to v at the top ⇒ dvdy = Δvh :
|Fv| = η A Δvh
  • No-slip condition : the fluid layer touching a solid surface moves with that surface (layer at the river bed is at rest).
Viscous forceKinetic friction
Fv ∝ A (area of contact)independent of area
Fv ∝ Δv and ∝ 1/hindependent of velocity
independent of normal forcefk = μN ∝ N

39. Viscous Force between Two Plates

  • Plates of area A separated by a liquid layer of thickness h : only the relative velocity matters.
Fv = η A vrelh
Top plateBottom platevrelFv
v →at restvηAv/h
3v →v →2v2ηAv/h
3v →← v4v4ηAv/h
  • Constant velocity ⇒ applied force = total viscous drag, e.g. boat (area A) on a river of depth D : F = ηAv₀/D ; block sliding down an oiled incline : mg sinθ = ηAv/h.
  • Plate between two others : add the drags from both liquid layers.
  • Cylinder (radius R, length l, mass m) falling in a pipe with lubricant gap r ≪ R : mg = η(2πRl)v/r ⇒ v = mgr2πηRl.

40. Stokes' Law & Terminal Velocity

  • Sphere of radius r moving with velocity v through a liquid of viscosity η (small speed, streamline flow) :
Fv = 6πηrv  (opposite to v)

Terminal velocity :

σ Fv Fb mg v ρ, η

Fnet = mg − Fb − Fv ; Fv grows with v until Fnet = 0 ⇒ constant terminal speed :

6πηrv = (σ − ρ)43πr³g

vT = 2r²g(σ − ρ)9η  ⇒  vT ∝ r²
  • σ > ρ (steel ball in water) : moves down ; σ < ρ (air bubble in water) : vT negative ⇒ rises with terminal speed.
  • Same r, ρ, η : vT ∝ (σ − ρ). At terminal speed, viscous force = mg − Fb.
  • Height of fall for which the speed does not change on entering the liquid : √(2gh) = vT ⇒ h ∝ r⁴.

Graphs (released from rest) :

atO a₀ = g(1 − ρ/σ) vtO vT
  • Acceleration falls from g(1 − ρ/σ) to zero ; speed rises and approaches vT asymptotically.

Rate of heat production at terminal velocity :

P = Fvv = 6πηrvT²

vT ∝ r² ⇒ P ∝ r⁵ ;  P ∝ vT5/2

41. Merging or Breaking of Drops

Volume conservation : 43πR³ = N · 43πr³

R = N1/3 r
Ex. 8 identical rain drops fall with terminal speed 10 cm/s and merge into one drop. New terminal speed?
Sol. R = 81/3r = 2r ; vT ∝ r² ⇒ v′ = 4 × 10 = 40 cm/s. In general v′ = N2/3v ; a sphere broken into 27 drops ⇒ each drop's vT is 1/9 of the original.
Page 8
Surface Tension & Summary

42. Surface Tension & Soap Film

  • Surface tension : tendency of a liquid surface at rest to shrink to the minimum possible area — it behaves like a stretched elastic membrane.
  • e.g. spherical drops (minimum area), brush hair clinging together out of water, insects walking on water, capillary rise ; soap / detergents decrease it (cleaning).
  • Surface tension decreases with increase in temperature and with impurities like soap.
  • Force acts along (tangent to) the surface, perpendicular to the line, and is attractive :
F = T l  ;  Surface energy = T × (surface area)

T in N/m (= J/m², energy per unit area) — fixed for a liquid at a given temperature.

T(2l) F l a soap film (2 surfaces)
  • A soap film has two free surfaces :
Force on the movable wire = 2Tl  ;  Surface energy = 2Tla

43. Excess Pressure inside Drops & Bubbles

Soap bubble (two surfaces) :

T(2πR)·2 P₀A PA A = πR²

Hemisphere in equilibrium : PπR² = P₀πR² + T(2πR)·2

P = P₀ + 4TR  ;  Surface energy = T(4πR²)·2

Liquid drop (one surface) :

P = P₀ + 2TR  ;  Surface energy = T(4πR²)

Air bubble at depth h in a liquid (one surface) :

P = P₀ + ρgh + 2TR
CaseFree surfacesPin − Pout
Soap bubble in air24T/R
Liquid drop in air12T/R
Air bubble in liquid12T/R
  • Smaller radius ⇒ larger excess pressure (higher on the concave side) ⇒ two connected bubbles : air flows from the smaller to the larger.
  • Bubble of radius R inside one of 2R : Pouter = P₀ + 2T/R, Pinner = P₀ + 6T/R.
  • Work to expand a soap bubble R₁ → R₂ : W = 8πT(R₂² − R₁²) (R → 2R : 24πR²T).
  • Drop of radius R broken into N drops : ΔE = 4πR²T(N1/3 − 1) increase ⇒ the liquid cools ; merging drops releases energy (heats).
R₂R₁ common film
Ex. Soap bubbles of radii R₁ < R₂ in contact. Radius r of the common film?
Sol. The film is concave towards the smaller (higher-pressure) bubble : 4TR₁ − 4TR₂ = 4Tr ⇒ r = R₁R₂R₂ − R₁

44. Formula Summary

QuantityFormula
Density ; mixtureρ = m/V ; ρeq = ΣρiVi/ΣVi
Pressure at depth hP = P₀ + ρgh
Vertical acceleration aΔP = ρ(g ± a)h ; free fall ΔP = 0
Horizontal acceleration aPrear − Pfront = ρal ; tanα = a/g
Up an incline (θ)tanα = (a + g sinθ)/(g cosθ)
Rotation ωΔP = ½ρω²(l₂² − l₁²) ; h = ω²R²/2g ; y = ω²x²/2g
Force on base / wallF = ρgh × base area ; Fwall = Pcentroid × projected area ; rectangle ρgbh²/2 at depth 2h/3
Hydraulic liftF₁/A₁ = F₂/A₂
BarometerP₀ = ρgh (76 cm Hg ≈ 10.3 m water) ; faulty : P = P₀ − ρgh
Buoyant forceFb = ρVimmgeff
Free float ; two liquidsVimm = (σ/ρ)V₀ ; σh = ρ₁h₁ + ρ₂h₂
Two balls on a stringσ₁ + σ₂ = 2ρ ; T = (σ₁ − ρ)Vg
Floating block (SHM)T = 2π√(σh/ρg)
Ball dropped into liquidsmax = σh/(ρ − σ)
Reynolds numberRe = ρvd/η = 4ρQ/πηd
Flow rate ; continuityQ = Av ; dm/dt = ρAv ; A₁v₁ = A₂v₂
BernoulliP + ½ρv² + ρgh = constant
Venturi meterQ = A₁A₂√(2gh/(A₁² − A₂²))
Efflux ; rangev = √(2gh) ; R = 2√(h h₁) ; Rmax = H at h = H/2
Two-layer effluxv = √(2g(ρ₁h₁ + ρ₂h₂)/ρ₂)
ThrustFth = ρav² = 2ρagh
Time to emptyT = (A/a)√(2H/g)
Dynamic lift(P₁ − P₂)A = ½ρ(v₂² − v₁²)A
Viscous forceF = ηA dv/dy ; plates F = ηAvrel/h
Stokes ; terminal velocityF = 6πηrv ; vT = 2r²g(σ − ρ)/9η
Heat rate at vTP = 6πηrvT² ∝ r⁵
N drops → 1 dropR = N1/3r ; vT → N2/3vT
Surface tensionF = Tl (film : 2Tl) ; energy = T × area
Excess pressuresoap bubble 4T/R ; drop / air bubble 2T/R

ρ = liquid density, σ = body density, η = viscosity, T = surface tension (in the SHM row, T = time period).

Key Points :

  • Pressure force is always ⊥ to the surface ; F = PA (uniform P), F = ∫P dA (variable P).
  • Same level + same continuous liquid at rest ⇒ same pressure (any vessel shape) ; in U-tubes equate pressures at the lower interface level.
  • Pressure increases along g, opposite to the acceleration and away from the rotation axis.
  • Force on the base = ρghA, independent of the liquid's weight ; a weighing scale reads the total weight.
  • Gauge pressure = P − Patm (can be negative). Hydraulic lift multiplies force, not work.
  • Fb uses the liquid density and the immersed volume ; it acts at the centroid of the displaced volume.
  • Floating ice melting : level unchanged ; ice with a stone : level falls. In an accelerating vessel light bodies move along a, heavy ones opposite.
  • Laminar below Re ≈ 2000, turbulent above ≈ 3000 (limits vary slightly by book).
  • Bernoulli : steady, streamline, incompressible, non-viscous flow ; moving liquid ⇒ do not use P = P₀ + ρgh ; open end ⇒ P = P₀. Faster flow ⇒ lower pressure.
  • Viscous force depends on area and relative velocity, not on normal force ; no-slip at solid surfaces ; 1 Pa·s = 10 poise, water η ≈ 10⁻³ Pa·s.
  • vT ∝ r² ; σ < ρ ⇒ the body rises (air bubble).
  • Surface tension falls with temperature and soap ; smaller bubble ⇒ larger pressure ; breaking drops cools, merging heats.