Skip to main content
Chapters 30 total chapters

NEET UG · Physics

Mechanical properties of fluids

Revise the concepts. Try a question. Understand your next step.

21 hard questions available · Try answers free. Sign in to save progress or take a quiz.

NEET UG · Physics

Continue studying Mechanical Properties Of Fluids

Understand the concept, practise questions and check your learning.

Understand

Distinguish gauge and absolute pressure, apply continuity, and check Bernoulli’s assumptions before NEET fluid-mechanics chapter practice. Pressure at rest and in flow: choose the reference and the model

About 6 min · NEET UG

Read the guide

Practise

Try 21 hard chapter MCQs. Untimed; check answers and available explanations as you go.

Free answers · Sign in to save progress

Practise chapter MCQs

Check your learning

Prepare a timed chapter quiz. Choose the available question count and pace before starting.

Sign-in required · Explanations after submission

Prepare a chapter quiz

This is chapter practice, not a full-length NEET mock paper.

Choose your practice

All Physics chapters →

These are playable question counts after quality and duplicate checks.

How practice and quizzes work

The practice player uses single-correct-answer MCQs. Chapter quizzes are not full-length NEET mock papers. Check the practice guide and official syllabus sources.

Revise before you practise

Study resources 39

Open full resource hub →

PDF notes 13

View all 13 →

Mind maps 13

View all 13 →

Pressure at rest and in flow: choose the reference and the model

Distinguish gauge and absolute pressure, apply continuity, and check Bernoulli’s assumptions before NEET fluid-mechanics chapter practice.

What you will understand

For NEET UG students: choose a pressure reference for water at rest, use continuity before calculating flow pressure, and identify when the ideal Bernoulli model applies. The worked examples keep depth, area, speed and pressure from becoming interchangeable shortcuts.

Jump to chapter practice →

Quick answer: first decide whether the fluid is at rest

For a constant-density liquid at rest in uniform gravity, pabs=psurface+ρghp_\mathrm{abs}=p_\mathrm{surface}+\rho gh, with hh measured downward from the free surface. Gauge pressure is pabs−patmp_\mathrm{abs}-p_\mathrm{atm}; it equals ρgh\rho gh when the surface is at that same atmospheric pressure.

For steady incompressible flow in a filled tube with no leaks, continuity gives A1v1=A2v2A_1v_1=A_2v_2 using mean section speeds. Under a one-dimensional ideal-flow model with uniform section speeds, negligible viscosity, no pump or energy loss and points on the same streamline, use p+12ρv2+ρgy=constantp+\tfrac12\rho v^2+\rho gy=\text{constant}. Use one consistent pressure reference.

Compare pressure at equal depths, not equal branch widths

Two open branches are joined underneath and contain the same stationary water. Both free surfaces have the stated pressure 100 kPa. Points A and B are each 2 metres below those surfaces; their branch widths differ. With density 1000 kilograms per cubic metre and g=10 metres per second squared, both points have gauge pressure 20 kPa relative to the 100 kPa atmosphere in this model, and absolute pressure 120 kPa. The drawing is schematic and neglects surface-curvature effects.
Two open branches are joined underneath and contain the same stationary water. Both free surfaces have the stated pressure 100 kPa. Points A and B are each 2 metres below those surfaces; their branch widths differ. With density 1000 kilograms per cubic metre and g=10 metres per second squared, both points have gauge pressure 20 kPa relative to the 100 kPa atmosphere in this model, and absolute pressure 120 kPa. The drawing is schematic and neglects surface-curvature effects.

Open full-size diagram ↗

Worked example: gauge and absolute pressure are different

Two branches contain the same connected stationary water. In this model both surfaces are exposed to patm=100 kPap_\mathrm{atm}=100\,\mathrm{kPa}. Take ρ=1000 kg m−3\rho=1000\,\mathrm{kg\,m^{-3}}, g=10 m s−2g=10\,\mathrm{m\,s^{-2}} and depth h=2.0 mh=2.0\,\mathrm m.

ρgh=20000 Pa=20 kPa\rho gh=20000\,\mathrm{Pa}=20\,\mathrm{kPa} is the gauge pressure at each point. Absolute pressure is 100+20=120 kPa100+20=120\,\mathrm{kPa}. Different branch widths do not change this equality at the same depth in the same connected liquid. The stated 100 kPa is the model’s atmospheric value, not a definition of one standard atmosphere.

Pressure is not force, and a pressure difference needs a cause

Pressure is a scalar; the pressure force on a surface element acts normally to that surface. For uniform pressure on a flat area, its normal force magnitude is pApA, so equal pressure does not mean equal force on different areas. Hydrostatic buoyancy is the net result of pressure forces; for a displaced volume VV in a constant-density liquid, its upward magnitude is ρgV\rho gV.

In an ideal confined liquid at rest, an imposed pressure change is transmitted throughout the liquid. Account for height differences when comparing pressures. In flow, do not use “faster means lower pressure” without the streamline, height and energy assumptions.

Use continuity before Bernoulli

An ideal horizontal tube carries steady incompressible water of density 1000 kilograms per cubic metre with uniform section speeds. At section 1, area is 4 square centimetres, speed 2 metres per second and absolute static pressure 140 kPa. At section 2, area is 2 square centimetres, so speed is 4 metres per second and static pressure is 134 kPa. Volume flow is 0.8 litres per second. The kinetic pressure terms are 2 and 8 kPa; each static-plus-kinetic sum is 142 kPa. There are no pumps, viscous losses or height changes in this model. The tube profile and arrows are schematic.
An ideal horizontal tube carries steady incompressible water of density 1000 kilograms per cubic metre with uniform section speeds. At section 1, area is 4 square centimetres, speed 2 metres per second and absolute static pressure 140 kPa. At section 2, area is 2 square centimetres, so speed is 4 metres per second and static pressure is 134 kPa. Volume flow is 0.8 litres per second. The kinetic pressure terms are 2 and 8 kPa; each static-plus-kinetic sum is 142 kPa. There are no pumps, viscous losses or height changes in this model. The tube profile and arrows are schematic.

Open full-size diagram ↗

Worked example: a narrower section changes speed and static pressure

Use the stated horizontal ideal-flow model with uniform section speeds. Let A1=4.0 cm2A_1=4.0\,\mathrm{cm^2}, A2=2.0 cm2A_2=2.0\,\mathrm{cm^2} and v1=2.0 m s−1v_1=2.0\,\mathrm{m\,s^{-1}}. Continuity gives v2=(A1/A2)v1=4.0 m s−1v_2=(A_1/A_2)v_1=4.0\,\mathrm{m\,s^{-1}}. The common volume flow is Q=(4.0×10−4)(2.0)=8.0×10−4 m3 s−1=0.80 L s−1Q=(4.0\times10^{-4})(2.0)=8.0\times10^{-4}\,\mathrm{m^3\,s^{-1}}=0.80\,\mathrm{L\,s^{-1}}.

With ρ=1000 kg m−3\rho=1000\,\mathrm{kg\,m^{-3}} and equal heights, p1−p2=12ρ(v22−v12)=6000 Pa=6.0 kPap_1-p_2=\tfrac12\rho(v_2^2-v_1^2)=6000\,\mathrm{Pa}=6.0\,\mathrm{kPa}. If p1=140 kPap_1=140\,\mathrm{kPa} absolute, p2=134 kPap_2=134\,\mathrm{kPa}. Static-plus-kinetic pressure terms are 140+2=142 kPa140+2=142\,\mathrm{kPa} and 134+8=142 kPa134+8=142\,\mathrm{kPa}. A pressure in Pa is an energy per unit volume, not an energy in joules.

Switch models when viscosity or an interface matters

Viscosity describes resistance to shear flow and has unit Pa s; it can produce mechanical-energy losses that the simple Bernoulli equation above omits. Surface tension is force per unit length, with unit N/m; curved interfaces can create pressure jumps. A flat free-surface hydrostatic example does not supply a universal rule for droplets, bubbles or capillary tubes. Choose the model from the conditions stated in the question.

Common mistakes to catch

Do not report gauge pressure as absolute pressure or add atmospheric pressure twice. Equal depths in the stated connected static liquid give equal pressure, regardless of branch width. In a tube, area and diameter are different: a doubled circular diameter gives four times the area. The ideal-flow pressure difference includes the gravitational term if heights differ, and a real pump or viscous loss changes the balance.

Try it yourself 1: choose the pressure reference

In the water-at-rest example, surface pressure is 100 kPa and the depth contribution is 20 kPa. What is the absolute pressure? A. 20 kPa; B. 100 kPa; C. 120 kPa.

Reveal the answer

C. Absolute pressure includes the surface pressure. Twenty kilopascals is the gauge pressure relative to the stated 100 kPa atmosphere in this model.

Try it yourself 2: apply continuity before guessing pressure

In the horizontal ideal-flow example, the area changes from 4 cm² to 2 cm². The initial speed is 2 m/s and static pressure 140 kPa. Which result follows? A. 4 m/s and 134 kPa; B. 1 m/s and 146 kPa; C. 4 m/s and 140 kPa.

Reveal the answer

A. Continuity doubles the speed. With the stated uniform-speed, equal-height, no-loss model, the kinetic pressure term rises by 6 kPa, so the static pressure falls by 6 kPa. Faster flow does not mean a larger volume flow rate here.

A useful check on your next visit

Cover the diagrams and distinguish gauge from absolute pressure, then explain why the narrower flow section carries the same volume per second. Before each question, write “at rest” or “flowing” and state the pressure reference and assumptions.

Choose your next chapter task

Explain the distinction you missed, then apply it to a fresh chapter question.

Return to chapter practice and quiz choices →

Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, numerical examples and self-checks are original teaching material. Human subject review is not claimed.

Sources and reading
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.