Mechanical Properties of Fluids

Mechanical Properties of FluidsNEET Physics · Class 11 · NCERT Chapter 9

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Mechanical Properties of Fluids. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Pressure with depth

Hydraulic press / lift

Archimedes' principle and buoyancy

Equation of continuity

Bernoulli's equation in two NEET classics

Terminal velocity (Stokes' law)

Capillary rise (or depression)

Pressure and Pascal's law

How pressure grows with depth, plus the hydraulic press that turns small forces into large ones.

Pressure

Pressure with depth

Gauge pressure grows linearly with depth: P = ρ·g·h. Pick a fluid and depth to see the value.

Fluid density ρ: 1000 kg/m³

Depth h: 10 m

Quick presets

Gauge pressure (above atmospheric)

100.0 kPa = 0.99 atm

P_{gauge} = ρ g h = 1000 \times 10 \times 10 = 1.00 e + 5 Pa

Absolute pressure (incl. atmosphere)

201.3 kPa = 1.99 atm

Try this

In water, pressure rises by about 1 atm every 10 m of depth.
Mercury is 13.6× denser than water, so it transmits 13.6× more gauge pressure per metre.
At 1000 m underwater, pressure is roughly 100 atm. Submersibles need very strong hulls.

Hydraulics

Hydraulic press / lift

Pascal's law in action. F₂ = F₁ × (A₂/A₁): a small force on a small area becomes a large force on a large area.

Force on small piston F₁: 50 N

Small piston area A₁: 10 cm²

Large piston area A₂: 500 cm²

Force on large piston

F₂ = 2500 N

Mechanical advantage

50.0×

F_{2} = F_{1} \cdot \frac{A _{2}}{A _{1}} = 50 \cdot \frac{500}{10} = 2500 N

Try this

Apply 50 N to a 10 cm² piston, lift on a 500 cm² piston: 2500 N. Multiplication factor 50×.
The cost: the small piston has to move 50× further than the big one moves. Energy is conserved.
Hydraulic brakes use this — light pedal force becomes large braking force at the wheels.

Buoyancy and continuity

Archimedes' principle for floating bodies and apparent weight, plus the continuity equation.

Archimedes

Archimedes' principle and buoyancy

Compare body density to fluid density. The widget tells you whether the body floats (and how much), or sinks (and what its apparent weight is).

Body density ρ_b: 0.60 g/cm³

Fluid density ρ_f: 1.00 g/cm³

Body volume V: 1000 cm³

Floats partially submerged

60.0% submerged · 40.0% above

\frac{V _{submerged}}{V} = \frac{ρ _{b}}{ρ _{f}} = \frac{0.60}{1.00} = 0.600

Try this

Body density 0.6, fluid density 1.0 → 60% submerged. A typical ice cube floats with about 90% submerged.
Body density > fluid density → sinks. Compute apparent weight in the fluid.
Body density = fluid density → neutral buoyancy: stays at any depth.

Continuity

Equation of continuity

A pipe narrowing from A₁ to A₂. The fluid speeds up by exactly A₁/A₂ to conserve mass.

Wide section A₁: 10 cm²

Narrow section A₂: 2 cm²

Speed in wide section v₁: 1.0 m/s

Speed in narrow section

v₂ = 5.00 m/s

A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = \frac{A _{1}}{A _{2}} v_{1} = \frac{10}{2} \times 1 = 5.00

Try this

Halving the cross-section doubles the speed. Why a thumb on a hose nozzle makes water shoot out faster.
Volume flow rate Q = Av is constant everywhere along a streamline (for incompressible fluid).
Real fluids with friction lose some flow rate to viscosity, but for ideal flow this is exact.

Bernoulli, viscosity and surface tension

Bernoulli's equation in two NEET classics, plus terminal velocity and capillary rise.

Bernoulli

Bernoulli's equation in two NEET classics

Toggle between speed of efflux (Torricelli) and Venturi pressure drop. Both follow from Bernoulli's equation.

Water escapes from a small hole at depth h below the surface of an open tank. Speed of efflux equals the speed of free fall through the same height.

Depth of hole h: 2.0 m

Speed of efflux

v = 6.32 m/s

v = 2 g h = 2 \times 10 \times 2 = 6.32 m/s

Try this

Torricelli: water from a hole 5 m down comes out at √(2·10·5) = 10 m/s. Same as a free fall through 5 m.
Venturi: doubling the speed in the constriction → 4× the kinetic energy density → that drop in pressure feeds the speedup.
Bernoulli applies only to non-viscous, incompressible, steady flow along one streamline.

Stokes & terminal v

Terminal velocity (Stokes' law)

A small sphere falling through a viscous fluid reaches a terminal velocity given by Stokes' law.

Sphere radius r: 1.00 mm

Sphere density ρ_s: 7850 kg/m³

Fluid density ρ_f: 1260 kg/m³ · Viscosity η: 1.490 Pa·s

Terminal velocity

v_t = 0.0096 m/s

v_{t} = \frac{2 r ^{2} ( ρ _{s} - ρ _{f} ) g}{9 η}

Stokes regime — valid for slow, small spheres in laminar flow. Larger spheres reach turbulent drag.

Try this

Doubling r multiplies v_t by 4× (r² scaling). A 2 mm steel ball falls 4× faster than a 1 mm one in glycerine.
Honey has very high η (~10 Pa·s) — even a heavy ball falls slowly through it.
If sphere density equals fluid density, v_t = 0 (neutrally buoyant). If less than fluid density, v_t becomes negative — the body floats up.

Capillarity

Capillary rise (or depression)

A thin tube in a wetting liquid sees the liquid climb. h ∝ T·cos(θ)/(ρ·g·r). Mercury (θ > 90°) shows depression instead.

Surface tension T: 0.072 N/m

Liquid density ρ: 1000 kg/m³

Contact angle θ: 0°

Tube radius r: 0.50 mm

Capillary rise

h = 29.39 mm

h = \frac{2 T cos θ}{ρ g r} = \frac{2 \times 0.072 \times 1.000}{1000 \times 9.8 \times 0.00050}

Wetting liquid (θ < 90°): liquid rises in the capillary.

Try this

Water in a 0.5 mm tube rises about 29 mm. In a 0.05 mm tube it rises 290 mm — h ∝ 1/r.
Click "Mercury" → contact angle 140° → cos θ ≈ −0.77 → h is negative: mercury is pushed DOWN in glass capillaries.
For soap solution, T is much smaller, so the rise is also smaller. Wetting alone is not enough.

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Mechanical Properties of Solids

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