Mechanical Properties of Fluids
1. Fluid Statics
Study of fluids at rest.
Pascal's Law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the
fluid and walls.
Hydraulic Lift: Force multiplier. F2 = (A2/A1) F1.
Archimedes' Principle: Buoyant force = Weight of fluid displaced.
Hydraulic Lift: Force multiplier. F2 = (A2/A1) F1.
Archimedes' Principle: Buoyant force = Weight of fluid displaced.
(Asked in NEET 2018, 2020)
2. Fluid Dynamics
Study of fluids in motion.
Equation of Continuity: A1v1 =
A2v2 (Conservation of Mass).
Bernoulli's Principle: P + ρgh + ½ρv2 = Constant (Conservation of Energy).
Applications: Venturimeter, Lift on Aerofoil, Blood flow.
Bernoulli's Principle: P + ρgh + ½ρv2 = Constant (Conservation of Energy).
Applications: Venturimeter, Lift on Aerofoil, Blood flow.
(Asked in NEET 2016, 2019, 2022)
3. Viscosity
Internal friction between fluid layers.
Stokes' Law: Viscous drag force F = 6πηrv.
Terminal Velocity: Constant velocity attained by falling body when Drag + Buoyancy = Weight.
vt = [2r2(ρ - σ)g] / 9η
Terminal Velocity: Constant velocity attained by falling body when Drag + Buoyancy = Weight.
vt = [2r2(ρ - σ)g] / 9η
4. Surface Tension
Tendency of liquid surface to shrink to minimum area.
Surface Energy: Work done to increase area. W = T ΔA.
Excess Pressure:
Bubble (Air in Liquid): 2T/R.
Soap Bubble (Air-Liquid-Air): 4T/R.
Capillarity: Rise/Fall of liquid in narrow tube. h = 2Tcosθ / rρg.
Excess Pressure:
Bubble (Air in Liquid): 2T/R.
Soap Bubble (Air-Liquid-Air): 4T/R.
Capillarity: Rise/Fall of liquid in narrow tube. h = 2Tcosθ / rρg.
(Asked in NEET 2017, 2021, 2023)
Numericals: Fluids
Q1. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. Area of
cross-section of piston carrying the load is 425 cm2. What maximum pressure would the smaller
piston have to bear?
Solution:
P = F / A.
F = mg = 3000 × 9.8 N. (Use g=9.8)
A = 425 × 10-4 m2.
P = (3000 × 9.8) / (425 × 10-4)
P = 29400 / 0.0425 ≈ 6.92 × 105 Pa.
P = F / A.
F = mg = 3000 × 9.8 N. (Use g=9.8)
A = 425 × 10-4 m2.
P = (3000 × 9.8) / (425 × 10-4)
P = 29400 / 0.0425 ≈ 6.92 × 105 Pa.
Q2. A U-tube contains water and methylated spirit separated by mercury. Mercury columns in the two arms are
in level with 10 cm of water in one arm and 12.5 cm of spirit in other. Find specific gravity of spirit.
Solution:
Pressure at same level (mercury interface) is equal.
Pw = Ps → hw ρw g = hs ρs g.
10 × 1 = 12.5 × ρrel (Relative density).
ρrel = 10 / 12.5 = 100 / 125 = 0.8.
Pressure at same level (mercury interface) is equal.
Pw = Ps → hw ρw g = hs ρs g.
10 × 1 = 12.5 × ρrel (Relative density).
ρrel = 10 / 12.5 = 100 / 125 = 0.8.
Q3. Water flows through a horizontal pipe which has radii 1 cm and 0.5 cm at two sections. If velocity at
first section is 4 m/s, find velocity at second.
Solution:
A1v1 = A2v2
πr12v1 = πr22v2
(1)2 × 4 = (0.5)2 × v2
4 = 0.25 v2 → v2 = 4 / 0.25 = 16 m/s.
A1v1 = A2v2
πr12v1 = πr22v2
(1)2 × 4 = (0.5)2 × v2
4 = 0.25 v2 → v2 = 4 / 0.25 = 16 m/s.
Q4. In Q3, if pressure at first section is 105 Pa, find pressure at second section. (ρ = 1000
kg/m3)
Solution:
P1 + ½ρv12 = P2 + ½ρv22 (h same).
105 + 500(4)2 = P2 + 500(16)2
100000 + 500(16) = P2 + 500(256)
100000 + 8000 = P2 + 128000
108000 = P2 + 128000 → P2 = -20,000 Pa (Impossible: Cavitation/Error in logic, but standard calculation follows this. Negative pressure implies suction/vacuum). Ans: -2 × 104 Pa (Gauge).
P1 + ½ρv12 = P2 + ½ρv22 (h same).
105 + 500(4)2 = P2 + 500(16)2
100000 + 500(16) = P2 + 500(256)
100000 + 8000 = P2 + 128000
108000 = P2 + 128000 → P2 = -20,000 Pa (Impossible: Cavitation/Error in logic, but standard calculation follows this. Negative pressure implies suction/vacuum). Ans: -2 × 104 Pa (Gauge).
Q5. Two drops of same radius fall. If they coalesce into one large drop, what is ratio of new terminal
velocity to old?
Solution:
Volume Conserved: 2 × (4/3)πr3 = (4/3)πR3
R3 = 2r3 → R = 21/3 r.
Velocity v ∝ r2.
Ratio V/v = R2/r2 = (21/3)2 = 22/3 : 1.
Volume Conserved: 2 × (4/3)πr3 = (4/3)πR3
R3 = 2r3 → R = 21/3 r.
Velocity v ∝ r2.
Ratio V/v = R2/r2 = (21/3)2 = 22/3 : 1.
Q6. What is the excess pressure inside a soap bubble of radius 5 mm, if T = 2.5 × 10-2 N/m?
Solution:
Excess P = 4T / R (Soap Bubble has 2 surfaces).
P = 4 × 2.5 × 10-2 / (5 × 10-3)
P = 10 × 10-2 / 5 × 10-3
P = 0.1 / 0.005 = 20 Pa.
Excess P = 4T / R (Soap Bubble has 2 surfaces).
P = 4 × 2.5 × 10-2 / (5 × 10-3)
P = 10 × 10-2 / 5 × 10-3
P = 0.1 / 0.005 = 20 Pa.
Q7. A body floats with 1/3 of its volume outside water. Find density of body.
Solution:
Volume submerged Vsub = 1 - 1/3 = 2/3 V.
Weight = Buoyant Force
V ρb g = Vsub ρw g
ρb = (2/3) ρw = (2/3) × 1000 ≈ 667 kg/m3.
Volume submerged Vsub = 1 - 1/3 = 2/3 V.
Weight = Buoyant Force
V ρb g = Vsub ρw g
ρb = (2/3) ρw = (2/3) × 1000 ≈ 667 kg/m3.
Q8. Water rises 3 cm in a capillary. If tube is cut such that length is only 1.5 cm, what happens to water?
Solution:
It will NOT overflow.
Reason: Radius of curvature of meniscus R changes such that hR = Constant.
Initially h = 3cm. New h' = 1.5cm.
Meniscus becomes flatter (Results in new R' = 2R). Water stands at top.
It will NOT overflow.
Reason: Radius of curvature of meniscus R changes such that hR = Constant.
Initially h = 3cm. New h' = 1.5cm.
Meniscus becomes flatter (Results in new R' = 2R). Water stands at top.
Q9. Calculate work done in blowing a soap bubble of radius 2 cm. (T = 0.03 N/m)
Solution:
W = T × ΔA. Soap bubble has 2 surfaces.
ΔA = 2 × 4πR2 = 8π (0.02)2 = 8π (4 × 10-4) = 32π × 10-4 m2.
W = 0.03 × 32π × 10-4 = 0.96π × 10-4
W ≈ 3 × 10-4 J.
W = T × ΔA. Soap bubble has 2 surfaces.
ΔA = 2 × 4πR2 = 8π (0.02)2 = 8π (4 × 10-4) = 32π × 10-4 m2.
W = 0.03 × 32π × 10-4 = 0.96π × 10-4
W ≈ 3 × 10-4 J.
Q10. A tank is filled to height H. A hole is made at depth h from top. Find range R on ground.
Solution:
Velocity v = √(2gh).
Time to fall (H-h): t = √[2(H-h)/g].
Range R = v × t = √(2gh) × √[2(H-h)/g]
R = 2 √[h(H-h)].
Velocity v = √(2gh).
Time to fall (H-h): t = √[2(H-h)/g].
Range R = v × t = √(2gh) × √[2(H-h)/g]
R = 2 √[h(H-h)].
Important Formulae
20 NEET Golden Facts
- 1. Pascal's Law: Valid only for fluids at rest. Basis for hydraulic brakes, lifts, press.
- 2. Gauge Pressure: Difference between absolute pressure and atmospheric pressure (P - Pa = hρg).
- 3. Streamline: Tangent to streamline gives velocity. Streamlines never cross. Crowded region = Higher velocity.
- 4. Bernoulli: Based on Energy conservation. High Velocity → Low Pressure (e.g., blowing roof off during storm).
- 5. Viscosity: Liquid viscosity decreases with temperature. Gas viscosity increases with temperature.
- 6. Surface Tension: Decreases with rise in temperature. Becomes zero at Critical Temperature.
- 7. Contact Angle: θ < 90° (Acute) → Wets surface (Water-Glass). θ> 90° (Obtuse) → Does not wet (Mercury-Glass).
- 8. Excess Pressure: Always higher on concave side. Pressure inside bubble > Outside.
- 9. Terminal Velocity: Indep of height fallen. Proportional to square of radius (Large drops fall much faster).
- 10. Venturimeter: Measuring device for flow speed. Uses Bernoulli's principle.
- 11. Magnus Effect: Dynamic lift on spinning ball due to pressure difference arising from Bernoulli's principle.
- 12. Detergents: Reduce surface tension of water, allowing it to penetrate pores of cloth better.
- 13. Ice Floats: Density of ice < Density of water. About 1/10th of valume is above water surface.
- 14. Reynolds Number: Re < 1000 (Laminar). Re> 2000 (Turbulent). Ratio of Inertial forces to Viscous forces.
- 15. Ideal Fluid: Incompressible and Non-viscous. (No friction energy loss).
- 16. Height of Atmosphere: Cannot be calculated by P=ρgh because density ρ of air is not constant (varies with height).
- 17. Blood Pressure: Measured by Sphygmomanometer. Gauge pressure is measured (excess over atm pressure).
- 18. Shape of Meniscus: Determined by adhesive (liquid-solid) vs cohesive (liquid-liquid) forces.
- 19. Coalescence: When two drops merge, Surface Area decreases, Energy is released (Temperature rises).
- 20. Aircraft Lift: Air moves faster over upper curved surface of wing → Low pressure. High pressure below pushes wing up.
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