Class 11 Physics · NCERT notes for NEET

Mechanical Properties of Fluids: NCERT Notes, Formulas and Tricks for NEET

How fluids push, flow and stick: pressure and buoyancy, continuity and Bernoulli, viscosity and terminal velocity, surface tension and capillary rise. NEET regularly asks from it, most often on terminal velocity, Bernoulli or continuity, excess pressure and capillary rise.

1. Pressure and Pascal's law
2. Streamline flow and Bernoulli's principle 🔒
3. Viscosity and Stokes' law 🔒
4. Surface tension and capillarity 🔒

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Topic 1

Pressure and Pascal's law

A dam wall is much thicker at the bottom than at the top: water pressure grows with depth, not with how much water is stored behind it.

Fluids (liquids and gases) can flow. A fluid at rest cannot hold a shear force, so it pushes only normal (perpendicular) to any surface it touches.
Pressure $p = F / A$ (normal force per unit area). It is a scalar; SI unit pascal, $1 Pa = 1 N m^{- 2}$ . Density $ρ = m / V$ , unit $kg m^{- 3}$ (water $1 0^{3}$ , mercury $13.6 \times 1 0^{3}$ ).
At depth $h$ below an open surface: $p = p_{a} + ρ g h$ . The excess $p - p_{a} = ρ g h$ is the gauge pressure.
Pressure depends only on depth, not on the shape or base area of the vessel (hydrostatic paradox). Points at the same level in a connected liquid at rest have equal pressure.
Atmospheric pressure = weight of the air column of unit area above the point. A mercury barometer at sea level stands at about 76 cm: $1 atm = 1.013 \times 1 0^{5} Pa$ .
Other units: $1 bar = 1 0^{5} Pa$ , $1 torr = 1 mm of Hg \approx 133 Pa$ . An open-tube manometer measures pressure differences (gauge pressure).
Pascal's law: a change in pressure applied to an enclosed fluid at rest is transmitted undiminished to every part of the fluid and to the walls.
Hydraulic lift and hydraulic brakes use Pascal's law. Force is multiplied by $A_{2} / A_{1}$ , but the big piston moves less ( $A_{1} d_{1} = A_{2} d_{2}$ ), so work is not multiplied.
Archimedes' principle: a body fully or partly immersed in a fluid gets an upward buoyant force equal to the weight of the fluid it displaces, so it appears lighter.
Floating body: weight = buoyant force, so fraction submerged $V_{sub} / V = ρ_{body} / ρ_{fluid}$ .

Three vessels, same liquid height h and same base area: the base pressure p_a + rho g h (and base force) is the same, though the liquid weights differ. — Three vessels, same liquid height $h$ and same base area: the base pressure $p_{a} + ρ g h$ (and base force) is the same, though the liquid weights differ.

Pressure F_1/A_1 is passed on unchanged, so the big piston gets F_2 = F_1 A_2/A_1. The small piston moves a long way d_1, the big one only d_2. — Pressure $F_{1} / A_{1}$ is passed on unchanged, so the big piston gets $F_{2} = F_{1} A_{2} / A_{1}$ . The small piston moves a long way $d_{1}$ , the big one only $d_{2}$ .

Absolute pressure is a straight line starting at p_a with slope rho g; gauge pressure is the parallel line through the origin. — Absolute pressure is a straight line starting at $p_{a}$ with slope $ρ g$ ; gauge pressure is the parallel line through the origin.

Formulas

$p = \frac{F}{A}$
Normal force over area; unit Pa
$p = p_{a} + ρ g h$
Absolute pressure at depth $h$ below an open surface; gauge pressure $= ρ g h$
$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}, A_{1} d_{1} = A_{2} d_{2}$
Hydraulic lift: equal pressure, equal volume displaced
$F_{B} = ρ_{f} V_{sub} g$
Buoyant force; $ρ_{f}$ = fluid density, $V_{sub}$ = displaced volume
$\frac{V _{sub}}{V} = \frac{ρ _{s}}{ρ _{f}}$
Floating body of density $ρ_{s}$ in fluid of density $ρ_{f}$

Tips, tricks and hacks

10 m water = 1 atm

ρ g h = 1000 \times 9.8 \times 10.3 \approx 1.01 \times 1 0^{5}

Pa. So every 10 m of water adds about 1 atm. A diver at 20 m feels about 3 atm absolute (1 from air + 2 from water).

Diameter ratio, square itHydraulic force ratio

= A_{2} / A_{1} = (d_{2} / d_{1})^{2}

. Pistons of diameter 2 cm and 20 cm: force is multiplied by

1 0^{2} = 100

, not 10.

Fraction under = RDFor a floating body, fraction submerged = relative density (body in water). Wood of RD 0.6 floats 60% under water; ice (RD about 0.9) floats about 90% under.

RD from loss of weightRelative density = weight in air / loss of weight in water. A stone weighs 10 N in air and 8 N in water: RD

= 10/2 = 5

Common mistakes

✗ A wider or bigger vessel has more pressure at the bottom because it holds more liquid.

✓ Pressure at the base depends only on depth: $p = p_{a} + ρ g h$ . Shape and volume do not matter.

✗ Using $ρ g h$ when the question asks for total (absolute) pressure.

✓ Absolute $= p_{a} + ρ g h$ ; gauge $= ρ g h$ . Read which one is asked.

✗ A hydraulic lift gives more work out than you put in.

✓ It multiplies force only. $F_{1} d_{1} = F_{2} d_{2}$ : the big piston rises a small distance.

✗ A heavier (denser) body gets a bigger buoyant force.

✓ Buoyant force depends only on the displaced volume and the fluid density: $ρ_{f} V_{sub} g$ . Equal-volume iron and aluminium balls, fully dipped, get equal buoyancy.

NEET traps

Pressure is a scalar even though force is a vector: at a point in a fluid at rest it acts equally in all directions.

Hydrostatic paradox: vessels of different shapes, same height and base area, have the same force on the base even though the liquid weights differ. Options that say "the heavier vessel has more base pressure" are wrong.

A body floating in water: if it is moved to a denser liquid (salt water), it floats higher, but the buoyant force stays equal to its weight.

Quick check: In a hydraulic lift the small piston has area 2 cm² and the large piston 200 cm². What force on the small piston holds up a 1000 kg car? (g = 10 m s⁻²)

10 N
100 N
1000 N
10⁴ N

Answer: B. $F_{1} = F_{2} A_{1} / A_{2} = 1 0^{4} \times 2/200 = 100$ N.

Topic 2 · 🔒

Streamline flow and Bernoulli's principle

Key points, graphs, every formula, tricks, common mistakes and NEET traps for this topic are inside Dhruv.

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Topic 3 · 🔒

Viscosity and Stokes' law

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Topic 4 · 🔒

Surface tension and capillarity

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