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EAPCET Physics · Intermediate first year

Mechanical Properties of Fluids

Mechanical Properties of Fluids covers pressure, Pascal’s law, Bernoulli’s principle, viscosity and surface tension.

Subject
Physics
Taught in
Intermediate first year
Physics in the paper
40 of 160 questions

Key formulas and results

  • P=P0+ρghP = P_{0} + \rho gh
  • Continuity: A1v1=A2v2A_{1}v_{1} = A_{2}v_{2}
  • Bernoulli: P+12ρv2+ρgh=P + \tfrac{1}{2}\rho v^{2} + \rho gh = constant along a streamline
  • Terminal velocity v=2r2(ρ−σ)g9ηv = \frac{2r^{2}(\rho - \sigma)g}{9\eta}
  • Excess pressure: 2T/r2T/r in a drop, 4T/r4T/r in a soap bubble
  • Capillary rise h=2Tcos⁡θrρgh = \frac{2T\cos\theta}{r\rho g}
  • Surface tension as force per unit length: T=F/lT = F/l
  • Surface energy U=T×areaU = T \times \text{area}; work to increase the area W=T ΔAW = T\,\Delta A (a soap film has two surfaces, so W=2T ΔAW = 2T\,\Delta A)
  • Excess pressure inside an air bubble formed in a liquid: P=2TrP = \frac{2T}{r}
  • Radius of the common interface of two soap bubbles in contact: r=r1r2r2−r1r = \frac{r_{1}r_{2}}{r_{2} - r_{1}}
  • Two soap bubbles that merge isothermally in vacuum form a bubble of radius rr with r2=r12+r22r^{2} = r_{1}^{2} + r_{2}^{2}
  • Reynolds number: Re=ρvDη\text{Re} = \frac{\rho vD}{\eta}; low Re means laminar flow and high Re means turbulent flow
  • Hydraulic lift (Pascal’s law): FA1=WA2\frac{F}{A_{1}} = \frac{W}{A_{2}}, so W=A2A1FW = \frac{A_{2}}{A_{1}}F
  • Capillary rise in terms of meniscus radius RR: h=2TρgRh = \frac{2T}{\rho gR}, where R=rcos⁡θR = \frac{r}{\cos\theta}

What this chapter covers

  • Pressure and Pascal’s law
  • Streamline flow and continuity
  • Bernoulli’s principle
  • Viscosity and Stokes’ law
  • Surface tension and capillarity

How it is usually asked

  • Bernoulli and continuity problems
  • Terminal velocity
  • Capillary rise
  • Excess pressure in drops and bubbles

Understanding Mechanical Properties of Fluids

Pressure in a fluid at rest increases with depth as ρgh, and by Pascal’s law any extra pressure applied to an enclosed fluid is transmitted equally throughout it, which is how a hydraulic lift multiplies force. For an ideal fluid in steady flow, the continuity equation Av = constant says the fluid speeds up where the pipe narrows.

Bernoulli’s principle is energy conservation per unit volume: where the speed is higher, the pressure is lower. Real fluids also have viscosity, an internal friction; a small sphere falling through a viscous liquid reaches a terminal velocity proportional to r². Surface tension makes liquid surfaces behave like stretched membranes, causing excess pressure inside drops and bubbles and the rise of liquid in thin capillary tubes.

EAPCET usually tests continuity with Bernoulli’s equation, terminal velocity, capillary rise, and excess pressure in drops and bubbles. Scaling relations save time: terminal velocity varies as r², while capillary rise and excess pressure both vary as 1/r. When n identical small drops merge into one big drop, volume is conserved, so the big radius is n^(1/3) times the small one, which fixes every other ratio.

Worked example

EAPCET-style question

Water flows through a horizontal pipe whose cross-section narrows from 4 cm24\ \text{cm}^{2} to 1 cm21\ \text{cm}^{2}. If the speed in the wider part is 1 m/s, the pressure difference between the wide and narrow parts is (density of water =1000 kg/m3= 1000\ \text{kg/m}^{3})

  1. (1)7500 Pa
  2. (2)8000 Pa
  3. (3)1500 Pa
  4. (4)15000 Pa

Solution

  1. Continuity: A1v1=A2v2A_{1}v_{1} = A_{2}v_{2}, so v2=41×1=4 m/sv_{2} = \frac{4}{1} \times 1 = 4\ \text{m/s}.
  2. For a horizontal pipe, Bernoulli gives P1−P2=12ρ(v22−v12)P_{1} - P_{2} = \tfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right).
  3. P1−P2=12×1000×(16−1)P_{1} - P_{2} = \tfrac{1}{2} \times 1000 \times (16 - 1).
  4. P1−P2=7500 PaP_{1} - P_{2} = 7500\ \text{Pa}, with higher pressure in the wider part.

Answer: 7500 Pa

Shortcuts and results to remember

  • Surface tension decreases as temperature rises, so hot water has lower surface tension than cold water.
  • When nn identical drops of radius rr merge into one, the big drop has radius n1/3rn^{1/3}r and the energy released is 4πr2T(n−n2/3)4\pi r^{2}T\left(n - n^{2/3}\right).
  • Of two bubbles in contact, the smaller one has the higher pressure, so the common film bulges into the bigger bubble.
  • Loss of weight in water (in grams) equals the total volume in cm3\text{cm}^{3}, which lets you find hidden cavities.

More practice questions

Practice question 1

1000 identical water drops, each of radius rr, coalesce into one big drop. If TT is the surface tension, the energy released is

  1. (1)400πr2T400\pi r^{2}T
  2. (2)3600πr2T3600\pi r^{2}T
  3. (3)3960πr2T3960\pi r^{2}T
  4. (4)4000πr2T4000\pi r^{2}T

Solution

  1. Volume is conserved: 1000×43πr3=43πR31000 \times \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi R^{3}, so R=10rR = 10r
  2. Initial surface energy =1000×4πr2T=4000πr2T= 1000 \times 4\pi r^{2}T = 4000\pi r^{2}T
  3. Final surface energy =4π(10r)2T=400πr2T= 4\pi(10r)^{2}T = 400\pi r^{2}T
  4. Energy released =4000πr2T−400πr2T=3600πr2T= 4000\pi r^{2}T - 400\pi r^{2}T = 3600\pi r^{2}T

Answer: 3600πr2T3600\pi r^{2}T

Practice question 2

A body with a cavity weighs 50 g in air and 40 g in water. If the density of its material is 10 g/cm310\ \text{g/cm}^{3}, the volume of the cavity is

  1. (1)4 cm34\ \text{cm}^{3}
  2. (2)5 cm35\ \text{cm}^{3}
  3. (3)10 cm310\ \text{cm}^{3}
  4. (4)15 cm315\ \text{cm}^{3}

Solution

  1. Loss of weight =50−40=10 g= 50 - 40 = 10\ \text{g}, so the total volume displaced is 10 cm310\ \text{cm}^{3}
  2. Volume of material =50/10=5 cm3= 50/10 = 5\ \text{cm}^{3}
  3. Cavity volume =10−5=5 cm3= 10 - 5 = 5\ \text{cm}^{3}

Answer: 5 cm35\ \text{cm}^{3}

Common mistakes to avoid

  • Using the excess pressure of a drop (2T/r) for a soap bubble — a bubble has two surfaces, so it is 4T/r.
  • Using radius ratios directly in the continuity equation — area scales as r², so halving the radius makes the speed four times larger.
  • Writing ½ρ(v₂ − v₁)² in Bernoulli’s equation — the correct term is ½ρ(v₂² − v₁²).
  • Forgetting that terminal velocity scales as r² — doubling the radius of a drop makes its terminal velocity four times larger.

How to prepare it

A soap bubble has two surfaces, so its excess pressure is double that of a drop — a classic trap.

Once the formulas feel familiar, test them the way the exam does: practise Mechanical Properties of Fluids questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Mechanical Properties of Fluids chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Mechanical Properties of Fluids part of the EAPCET Physics syllabus?
Yes. Mechanical Properties of Fluids is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Mechanical Properties of Fluids for EAPCET?
A soap bubble has two surfaces, so its excess pressure is double that of a drop — a classic trap.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Mechanical Properties of Fluids?
Using the excess pressure of a drop (2T/r) for a soap bubble — a bubble has two surfaces, so it is 4T/r. Using radius ratios directly in the continuity equation — area scales as r², so halving the radius makes the speed four times larger. Writing ½ρ(v₂ − v₁)² in Bernoulli’s equation — the correct term is ½ρ(v₂² − v₁²). Forgetting that terminal velocity scales as r² — doubling the radius of a drop makes its terminal velocity four times larger.