Fluid Statics & Fluid Dynamics Practice Set: NEB 12 Physics

Worked problem 1

Pressure at depth and force on a window

A small horizontal inspection window of area 0.020 m² lies 8.0 m below a water surface. Find gauge pressure and approximate fluid force normal to the window, using ρ=1000 kg m⁻³ and g=9.8 m s⁻².

Pg=ρgh=(1000)(9.8)(8.0)=7.84×10⁴ Pa. Over a small window at essentially one depth, F=PA=(7.84×10⁴)(0.020)=1.57×10³ N. Absolute pressure would add atmospheric pressure before multiplying if external reference required it.

Check pressure variation with depth. Pressure depends on vertical depth, not tank shape.

Worked problem 2

Hydraulic press and displacement tradeoff

Input piston area is 4.0 cm² and output area 200 cm². An input force 120 N is applied ideally. Pressure change F₁/A₁=120/(4.0×10⁻⁴)=3.0×10⁵ Pa. Output force F₂=PA₂=(3.0×10⁵)(0.0200)=6000 N.

If input piston moves 0.50 m, volume conservation A₁d₁=A₂d₂ gives d₂=(4/200)(0.50)=0.010 m. Ideal work: 120×0.50=6000×0.010=60 J. Force multiplication does not create energy.

See Pascal’s principle.

Worked problem 3

Buoyant force and apparent weight

A 2.0×10⁻³ m³ object is fully submerged in water. FB=ρgV=(1000)(9.8)(2.0×10⁻³)=19.6 N. If its true weight is 30.0 N and it is held at rest by a scale above, scale reading is 30.0−19.6=10.4 N.

If released, net force is downward 10.4 N initially before drag changes. Do not set buoyancy equal to weight unless floating/suspended equilibrium applies. Use the Archimedes resource.

Worked problem 4

Continuity in a narrowing pipe

Water flows at 2.0 m s⁻¹ through diameter 6.0 cm and enters diameter 3.0 cm. Areas scale with diameter squared, so A₁/A₂=(6/3)²=4. Therefore v₂=8.0 m s⁻¹.

A common error uses diameter ratio 2 and obtains 4.0 m s⁻¹. Continuity uses cross-sectional area. If volume flow rate is needed, Q=A₁v₁=π(0.030)²(2.0)=5.65×10⁻³ m³ s⁻¹.

Check flow rate and velocity.

Worked problem 5

Bernoulli with height and speed change

Water moves from point 1 at v₁=2.0 m s⁻¹, h₁=0 to point 2 at v₂=4.0 m s⁻¹, h₂=3.0 m. For ideal steady flow, P₁−P₂=½ρ(v₂²−v₁²)+ρg(h₂−h₁).

Substitute: 0.5(1000)(16−4)+(1000)(9.8)(3.0)=6000+29400=3.54×10⁴ Pa. Point 1 must have greater pressure to supply both kinetic and gravitational energy per volume.

State assumptions and see Bernoulli’s equation. Do not add a hidden pump or frictionless claim without evidence.

Worked problem 6

Poiseuille radius sensitivity

For the same pressure difference, viscosity and length, one tube has radius 1.5 mm and another 3.0 mm. In the laminar Poiseuille model, Q∝r⁴, so Q₂/Q₁=(3.0/1.5)⁴=16. Doubling radius multiplies ideal flow by sixteen.

If length also doubles, the ratio becomes 16/2=8 because Q∝r⁴/L. Temperature changes viscosity and can alter the comparison, so controlled conditions matter.

Use OpenStax viscosity and Poiseuille for assumptions. Never substitute diameter as r.

Independent set

Solve eight model-switching tasks

1. Find gauge pressure 12 m below freshwater surface.
2. A hydraulic system has area ratio 25; find output force and displacement ratio.
3. Find buoyant force on 0.010 m³ fully submerged in oil density 800 kg m⁻³.
4. A density 600 kg m⁻³ block floats in water; find fraction submerged.
5. Pipe area halves; find speed factor for steady incompressible flow.
6. Use horizontal Bernoulli to find pressure difference when speed changes 1 to 3 m s⁻¹.
7. A tube radius decreases by 20%; find Poiseuille Q factor.
8. Design a terminal-speed experiment and identify two controlled variables.

Prompts: use ρgh; equal pressure change and volume conservation; displaced-fluid volume; density ratio; A v; ½ρΔv²; 0.8⁴; and require steady terminal region, radius measurement and temperature control.

Use the study guide after attempting.

Review protocol

Audit the first unsupported fluid decision

Classify errors as model, pressure reference, depth, area/radius, displaced volume, point selection, continuity, energy/loss, viscosity condition, units or interpretation. Redo without the solution and retest with a fresh context after two days and one week.

Keep the topic inside the Class 12 roadmap. For online or physical tuition call 9846662070. The MKS Education panel gives test-preparation and study-abroad pre-counselling contacts after Grade 12.

Checked sources

References and related learning

• CDC Nepal: Physics Grade 12
• CDC Nepal: Secondary Level Curriculum
• OpenStax: Fluid Statics
• OpenStax: Pressure with Depth
• OpenStax: Archimedes’ Principle
• OpenStax: Flow Rate and Continuity
• OpenStax: Bernoulli’s Equation
• OpenStax: Viscosity and Poiseuille’s Law

Rebuild concepts with the full fluid guide before a fresh practice set. Curriculum and institutional pages were checked on 2 August 2026; follow current CDC, NEB and school notices if requirements change.

Worked problem 7

Terminal speed from force balance

A small sphere of radius r and density ρs falls slowly through a fluid of density ρf and viscosity η. At terminal speed vt, weight equals buoyancy plus Stokes drag: (4/3)πr³ρs g=(4/3)πr³ρf g+6πηrvt. Therefore vt=2r²g(ρs−ρf)/(9η).

The expression predicts zero terminal speed if densities match, greater speed for larger sphere or density difference, and smaller speed for greater viscosity. These are useful limiting checks. It applies only in the small-sphere, slow laminar Stokes regime—not automatically to raindrops, stones or turbulent motion.

Worked problem 8

Capillary relation and radius sensitivity

For a wetting liquid with contact angle near zero in a clean capillary, upward surface-tension force around circumference 2πr balances liquid-column weight ρgπr²h. Thus h=2γ/(ρgr). If radius halves while other conditions stay fixed, rise doubles. The relation uses internal radius and assumes a suitable equilibrium/contact-angle model.

A soap film question is different: force on a movable length l can be 2γl because the film has two surfaces. Draw interfaces rather than importing the capillary equation.

Mixed case

Diagnose an irrigation-line pressure problem

Water enters a pipe, rises to a higher field and passes through a narrower section. First use continuity to relate speeds. Then ideal Bernoulli estimates the pressure change from kinetic and gravitational terms. If the measured downstream pressure is lower than the ideal prediction, attribute the difference to head loss only after checking instruments, heights, areas and flow steadiness. A pump would add energy and must be represented separately.

Finish by changing one condition: double flow rate, halve tube radius, reverse height difference or replace water with a more viscous liquid. Predict direction before calculation. Record whether continuity, ideal energy or viscous resistance controls the new result.