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JEE Mains Physics · Formula sheet

Mechanical Properties of Fluids formulas

14 formulas, 1 reference table and 37 common traps for JEE Mains Physics Mechanical Properties of Fluids, grouped by subtopic.

Full notes with worked examples

Pressure, Pascal's Law and Buoyancy

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Pressure at a depth, absolute and gauge

Pressure at depth h

P=P0+ρghP = P_0 + \rho g h

Pascal's law and the hydraulic lift

Hydraulic lift

F1A1=F2A2,F1d1=F2d2\frac{F_1}{A_1} = \frac{F_2}{A_2}, \qquad F_1 d_1 = F_2 d_2

Floating bodies and Archimedes' principle

Submerged fraction

VsubV=ρbρl\frac{V_{\text{sub}}}{V} = \frac{\rho_b}{\rho_l}

Common traps

Doubling the depth does not double the pressure

Only ρgh doubles. The atmosphere's share stays the same, so the absolute pressure grows by less than 100%. Doubling the whole of P counts the atmosphere twice.

Keep P₀ in a force on a base when the stem gives it

For the force a liquid exerts on the bottom of a tube open to the air, JEE has keyed (P₀ + ρgh) × A when it gives P₀. The smaller option, ρgh × A, leaves the atmosphere out.

Across a partition, the atmosphere cancels

Air presses on both free surfaces, so only the difference of the liquid pressures pushes on a door in the partition: (ρ₁ − ρ₂)ghA.

The small piston carries the same pressure

Pressure is the same throughout the enclosed fluid. The small piston has a smaller FORCE on it, but the pressure on it is still the load divided by the large area.

A lift multiplies force, not work

The force gain A₂/A₁ is paid for in distance: the small piston moves A₂/A₁ times farther. Work out equals work in.

Body over liquid, not liquid over body

The submerged height is H × ρ_body/ρ_liquid, which is less than H. Inverting the ratio gives a height bigger than the block itself.

A load adds displaced liquid, not displaced wood

The extra depth is m/(ρ_liquid × A). The density of the floating block plays no part in it.

Continuity, Bernoulli's Equation and Efflux

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Continuity and Bernoulli's equation in a pipe

Continuity and Bernoulli

A1v1=A2v2,P+ρgh+12ρv2=constantA_1 v_1 = A_2 v_2, \qquad P + \rho g h + \tfrac{1}{2}\rho v^{2} = \text{constant}

Speed from a pressure drop: gauges and wings

Lift on a wing

F=12ρ(vtop2−vbottom2)AF = \tfrac{1}{2}\rho\left(v_{\text{top}}^{2} - v_{\text{bottom}}^{2}\right)A

Liquid leaving a hole: Torricelli's law

Torricelli's law and the range

v=2gh,x=2h yv = \sqrt{2gh}, \qquad x = 2\sqrt{h\,y}

Common traps

Put both areas in the same unit

1 cm² = 100 mm². A ratio of 1 cm² to 20 mm² is 5, not 1/20. Convert before using continuity.

The narrow section is at the lower pressure

Faster flow means lower pressure. In a venturi meter with v₁ at the wide part, 2gh = v₂² − v₁². A statement written as v₁² − v₂² has the sign the wrong way round.

Square the speeds, then subtract

The pressure difference is ½ρ(v₂² − v₁²), not ½ρ(v₂ − v₁)². The second form is a common wrong option.

Two wings: add both areas

'Each of its two wings has an area A' means a total of 2A. Using one wing's area halves the lift and the mass.

Convert km/h before squaring

Square the speeds in m/s. Converting after squaring needs (5/18)², and mixing the two orders is a common slip.

Depth is measured from the free surface

h in √(2gh) is the depth of the hole below the water surface: the water's height minus the hole's height. Using the hole's height above the base gives the wrong speed.

A load adds W/A, the atmosphere adds nothing

Air presses on the surface and on the jet alike, so P₀ cancels. Only the extra pressure of the load, its weight over the tank's area, joins ρgh.

Viscosity, Stokes' Law and Reynolds Number

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Newton's law of viscosity and the Reynolds number

Viscous force and Reynolds number

F=ηAvd,Re=ρvdηF = \eta A\frac{v}{d}, \qquad R_e = \frac{\rho v d}{\eta}

Stokes' law and the viscous force at constant velocity

Viscous force at constant velocity

Fv=mg(1−ρ0ρ)=43πr3(ρ−ρ0)gF_v = mg\left(1 - \frac{\rho_0}{\rho}\right) = \tfrac{4}{3}\pi r^{3}(\rho - \rho_0)g

Common traps

Liquids and gases go opposite ways with temperature

Heating a liquid lowers its viscosity, so hot water flows faster. Heating a gas raises its viscosity. A statement that says viscosity rises with temperature is true only for a gas.

Convert km/h and poise first

72 km/h is 20 m/s, and 1 poise is 0.1 Pa s. A stem that mixes units gives an answer off by a power of ten.

Buoyancy is subtracted, not added

Weight = buoyancy + viscous drag, so the drag is the weight MINUS the buoyancy. Adding them gives a force bigger than the weight.

The ball's density goes in the denominator

F = mg(1 − ρ_liquid/ρ_ball). The option mg(ρ_liquid/ρ_ball − 1) is negative for any ball that sinks.

Terminal Velocity

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The terminal-velocity formula

Terminal velocity

vT=2r2(ρ−σ)g9ηv_T = \frac{2r^{2}(\rho - \sigma)g}{9\eta}

How terminal velocity scales with the radius

Merging drops

vT∝r2,R=n1/3r⇒v′=n2/3vv_T \propto r^{2}, \qquad R = n^{1/3}r \Rightarrow v' = n^{2/3}v

Common traps

Diameter or radius

Stems usually give the diameter. Halve it before squaring, or the answer is four times too big.

Do not mix CGS and SI

With η in poise, use cm, g/cm³ and g = 1000 cm/s². With η in Pa s, use m, kg/m³ and g = 10 m/s². A g of 10 with lengths in cm gives an answer 100 times too small.

A faster launch does not change η

A ball thrown in with some speed still settles to the same terminal velocity, so the measured viscosity is the same.

Same mass is not same material

For one material, v ∝ r². For one mass, a bigger radius means a lighter material and more drag: v ∝ 1/r, so doubling the radius halves the speed.

Merging multiplies by n^(2/3), not n

27 drops merging make the speed 9 times larger, not 27 times. The radius grows only by the cube root of n.

v grows as r², not inversely with r

A reason that says terminal velocity is inversely proportional to the radius is false for a given material. The squared radius is also why its error is doubled.

Surface Energy of Drops and Bubbles

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Drops that merge or split

Splitting a drop into n

W=4πR2T(n1/3−1)W = 4\pi R^{2}T\left(n^{1/3} - 1\right)

Work to blow a soap bubble

Soap bubble

W=8πT(r22−r12)W = 8\pi T\left(r_2^{2} - r_1^{2}\right)

Common traps

A diameter in the stem

A drop of diameter 2 mm has radius 1 mm. Squaring the diameter makes every energy four times too big.

The ratio of energies is n^(1/3), not n

1000 droplets have 10 times the surface energy of the drop they form, not 1000 times. Each droplet is smaller, so the surfaces do not simply add up n-fold.

Comparing two splittings uses n^(1/3) − 1

Splitting one drop into 64 instead of 27 needs (4 − 1)/(3 − 1) = 3/2 times the work, not 64/27 times.

Count both faces of a soap film

A soap bubble needs 8πT(r₂² − r₁²). The 4π form is for a single surface and gives half the answer, which is always one of the options.

Diameters in the stem

Halve diameters before squaring. Using diameters as radii makes the work four times too big.

Excess Pressure and Capillary Rise

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Excess pressure inside drops and bubbles

Excess pressure

ΔP=2Tr (one surface),ΔP=4Tr (soap bubble)\Delta P = \frac{2T}{r}\ \text{(one surface)}, \qquad \Delta P = \frac{4T}{r}\ \text{(soap bubble)}

Capillary rise

h=2Tcos⁡θρgrh = \frac{2T\cos\theta}{\rho g r}

Contact angle, meniscus shape and surface-tension facts

CaseContact angleMeniscus and capillary
Water on clean glassAbout 0°, acuteConcave; the water rises
Mercury on glassAbout 140°, obtuseConvex; the mercury falls below the outside level
Water on grease or waxObtuseWater forms beads and does not wet the surface, so washing with water alone cannot remove a grease stain
Soapy water on greaseMade acute by the detergentThe water spreads and wets the grease; detergent lowers T
Adhesion and cohesion in balance90°Flat surface; no rise and no fall
A liquid that neither rises nor falls has a contact angle of 90°, not 0°.
In h=2Tcos⁡θ/ρgrh = 2T\cos\theta/\rho g r, the sign of cos⁡θ\cos\theta gives the direction: positive for an acute angle (rise), zero at 90°, negative for an obtuse angle (fall).

Common traps

An air bubble in a liquid has ONE surface

Only a soap bubble in air has a film with two faces. An air bubble under water is bounded by water on one side only, so its excess pressure is 2T/r, not 4T/r.

Add the depth term

At depth h, the liquid just outside the bubble is already at P₀ + ρgh. The pressure inside exceeds atmospheric by ρgh + 2T/r.

The common surface uses the difference of the radii

It is r₁r₂/(r₂ − r₁). The sum r₁r₂/(r₁ + r₂) looks like a parallel-resistor formula and is a common wrong option.

Bores are often given as diameters

Halve each diameter before using 1/r₁ − 1/r₂. Using diameters halves the answer.

A tilted tube keeps the vertical height

The liquid still rises to the same vertical height h; only the length along the tube grows, to h/cos α.

The contact angle belongs to the pair

The same water makes about 0° with clean glass and an obtuse angle with wax. A statement that it depends on the liquid alone, or on the solid alone, is false.

Soap water has LOWER surface tension

Detergents are added to water because they lower its surface tension. A statement that soap water has the higher surface tension is false.

Gases are less viscous than liquids

Statement pairs here often slip in a viscosity fact. The viscosity of a gas is far smaller than that of a liquid, so a statement saying the opposite is false.

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