Original practice sets for Fluid Mechanics are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Pressure at depth h in a liquid of density ρ (atmospheric pressure P₀) is:
Explanation: Absolute pressure = atmospheric pressure + hydrostatic pressure = P₀ + ρgh.
2. SI unit of pressure is:
Explanation: 1 Pascal (Pa) = 1 N/m². Pressure is force per unit area.
3. Pascal's law states that pressure applied to an enclosed fluid is:
Explanation: Any additional pressure applied to a confined fluid is transmitted equally throughout — the basis of hydraulic machines.
4. Archimedes' principle states that buoyant force equals:
Explanation: F_b = weight of fluid displaced = ρ_fluid × V_submerged × g.
5. A body floats when:
Explanation: Floating equilibrium: F_b = W. For a floating body, ρ_body/ρ_fluid = fraction submerged.
6. A wooden block (density 600 kg/m³) floats in water (density 1000 kg/m³). Fraction above water surface is:
Explanation: Fraction submerged = ρ_block/ρ_water = 0.6. Fraction above = 1 − 0.6 = 0.4.
7. Hydraulic press works on Pascal's law. If input piston area is 5 cm² and output is 100 cm², and input force is 20 N, output force is:
Explanation: P₁ = P₂ → F₁/A₁ = F₂/A₂. F₂ = 20 × 100/5 = 400 N.
8. A U-tube manometer measures gauge pressure. Gauge pressure is:
Explanation: Gauge pressure = Absolute pressure − Atmospheric pressure. Negative gauge = vacuum.
9. When a piece of ice melts in a glass full of water, the water level:
Explanation: Ice floats displacing water equal to its weight. When it melts, it produces exactly that volume of water. Level stays constant.
10. The pressure at the bottom of a lake 10 m deep (ρ_water = 1000 kg/m³, P₀ = 10⁵ Pa, g = 10 m/s²):
Explanation: P = P₀ + ρgh = 10⁵ + 1000×10×10 = 10⁵ + 10⁵ = 2×10⁵ Pa.
11. A hollow sphere of outer radius R, inner radius r, and material density ρ_s just floats in water (density ρ_w). The relation is:
Explanation: Weight of shell = ρ_s × (4π/3)(R³−r³)g. Buoyant force = ρ_w × (4π/3)R³g. Setting equal: ρ_s(R³−r³) = ρ_w R³.
12. A submarine is at depth 100 m. If atmospheric pressure = 10⁵ Pa, seawater density = 1025 kg/m³, g = 10 m/s², gauge pressure on hull is:
Explanation: Gauge P = ρgh = 1025×10×100 = 1.025×10⁶ Pa = 10.25 atm. All three options express the same value.
13. A body is in stable floating equilibrium when:
Explanation: Stable equilibrium requires metacentre (M) above centre of gravity (G). If GM > 0, a tilted body rights itself.
14. Relative density (specific gravity) of a substance is measured using Archimedes' principle as:
Explanation: Relative density = Weight in air / (Weight in air − Weight in water) = Weight in air / Buoyant force.
15. Atmospheric pressure supports a column of mercury of height:
Explanation: 1 atm = 101,325 Pa = 76 cmHg = 10.3 m H₂O = 1013.25 hPa — all equivalent.
16. An iceberg floats with 1/9 of volume above water. Density of ice is (ρ_water = 1000 kg/m³):
Explanation: Fraction submerged = 8/9 = ρ_ice/ρ_water → ρ_ice = 8000/9 ≈ 889 kg/m³.
17. Two connected containers have different cross sections A₁ and A₂ (A₁ < A₂). When liquid is poured in container 1, rise in container 2 is:
Explanation: Volume conservation: A₁h₁ = A₂h₂. Since A₂ > A₁, h₂
18. A balloon filled with helium rises because:
Explanation: Net upward force = Buoyant force − Total weight = (ρ_air − ρ_He)Vg (approximately). Since ρ_air > ρ_He, net upward force exists.
19. A piston of area A at depth h in water exerts pressure P₀ + ρgh on the liquid. If additional force F is applied on piston, pressure at depth h becomes:
Explanation: Pressure at depth = atmospheric + piston force + hydrostatic = P₀ + F/A + ρgh.
20. A body weighs 50 N in air and 30 N in water. Volume of body is (g = 10 m/s², ρ_water = 1000 kg/m³):
Explanation: Buoyant force = 50 − 30 = 20 N. V = F_b/(ρ_water × g) = 20/(1000×10) = 2×10⁻³ m³.
21. Equation of continuity for steady incompressible flow states:
Explanation: Volume flow rate Q = Av = constant for steady incompressible flow (mass conservation).
22. Bernoulli's theorem applies to:
Explanation: Bernoulli's theorem is derived using energy conservation for ideal fluid with no viscous losses.
23. Bernoulli's equation is: P + ½ρv² + ρgh = constant. This represents:
Explanation: Each term has units of J/m³ (pressure energy + kinetic energy density + potential energy density = constant).
24. A venturimeter measures fluid flow rate by measuring:
Explanation: Constriction increases velocity, decreasing pressure. Pressure drop is measured to calculate flow rate via Bernoulli's equation.
25. Velocity of efflux from a hole at depth h below the surface of a liquid is:
Explanation: Torricelli's theorem: v = √(2gh). Derived from Bernoulli's equation between surface and hole.
26. Water flows through a pipe of diameter 4 cm at 2 m/s. Diameter narrows to 2 cm. New speed is:
Explanation: A₁v₁ = A₂v₂. A ∝ d². (4²)×2 = (2²)×v₂ → 32 = 4v₂ → v₂ = 8 m/s.
27. An aerofoil generates lift because:
Explanation: Curved upper surface increases air speed above wing → lower pressure (Bernoulli). Pressure differential creates lift.
28. A tank has water to height H. A hole at height h from bottom: water shoots out horizontally and lands at horizontal distance from hole equal to:
Explanation: v = √(2g(H−h)) (Torricelli). Time to fall h: h = ½gt² → t = √(2h/g). Range = v×t = √(2g(H−h))×√(2h/g) = 2√(h(H−h)).
29. Surface tension is defined as:
Explanation: Surface tension T = Force/Length (N/m). It also equals surface energy per unit area (J/m²).
30. Newton's law of viscosity states that shear stress is proportional to:
Explanation: τ = η(dv/dy). The constant of proportionality η is called dynamic viscosity (unit: Pa·s).
31. Excess pressure inside a soap bubble of radius r and surface tension T is:
Explanation: Soap bubble has two surfaces (inner and outer). Each contributes 2T/r. Total excess pressure = 4T/r.
32. Water rises in a capillary tube of radius r to height h. If r is halved, height becomes:
Explanation: h = 2T cosθ/(ρgr). If r → r/2, then h → 2h.
33. Stokes' law gives drag force on a sphere of radius r moving with velocity v in fluid of viscosity η as:
Explanation: Stokes' law: F = 6πηrv. This drag force acts opposite to velocity.
34. Terminal velocity of a sphere (radius r, density ρ_s) falling in fluid (density ρ_f, viscosity η) is:
Explanation: At terminal velocity, weight = buoyancy + drag. Solving: v_t = 2r²(ρ_s − ρ_f)g / (9η).
35. Water flows from a wide pipe (area 8 cm², velocity 2 m/s, pressure 3×10⁵ Pa) to a narrow horizontal pipe (area 2 cm²). Pressure in narrow pipe is:
Explanation: By continuity: v₂ = 8×2/2 = 8 m/s. Bernoulli: P₁+½ρv₁² = P₂+½ρv₂². P₂ = 3×10⁵ + ½×1000×(4−64) = 3×10⁵−3×10⁴ = 2.7×10⁵ Pa.
36. A liquid drop of radius R splits into 8 equal drops. Change in surface area is:
Explanation: Volume: (4/3)πR³ = 8×(4/3)πr³ → r = R/2. Initial SA = 4πR². Final SA = 8×4π(R/2)² = 8πR². Increase = 4πR².
37. Poiseuille's equation for volume flow rate through a cylindrical pipe is Q = πr⁴ΔP/(8ηL). If radius is halved, flow rate becomes:
Explanation: Q ∝ r⁴. If r → r/2, Q → (1/2)⁴ × Q = Q/16. The r⁴ dependence makes radius critically important.
38. A needle floats on water surface due to:
Explanation: Steel needle (density > water) is supported by the surface tension force along the contact line when placed gently. The curved meniscus provides upward force.
39. Mercury in a glass capillary tube:
Explanation: Mercury-glass contact angle > 90°. Mercury cohesion > mercury-glass adhesion. Result: capillary depression, not rise.
40. A large open tank has hole at its base. Tank empties faster when:
Explanation: Efflux speed v = √(2gh). Higher water level means higher h, higher speed, and faster emptying.
41. Surface energy per unit area is numerically equal to:
Explanation: Surface tension T = Surface energy/Area (J/m² = N/m).
42. The angle of contact (θ) for water-glass is:
Explanation: Water wets glass (adhesion > cohesion), so θ
43. Viscosity of a fluid decreases with temperature for:
Explanation: In liquids, viscosity decreases with temperature (thermal agitation overcomes intermolecular bonds). In gases, viscosity increases with temperature (more momentum transfer via collisions).
44. Reynold's number Re = ρvD/η. Flow is turbulent when Re is:
Explanation: Re 4000: fully turbulent flow.
45. Excess pressure inside a liquid drop (radius r, surface tension T) is:
Explanation: A liquid drop has ONE surface. Excess pressure ΔP = 2T/r. (Soap bubble has two surfaces: ΔP = 4T/r.)
46. Capillary rise formula h = 2T cosθ/(ρgr) shows that rise is greater for:
Explanation: h ∝ 1/r. Narrower tube (smaller r) gives greater capillary rise.
47. Streamline flow means:
Explanation: In steady (streamline/laminar) flow, velocity at any fixed point does not change with time.
48. At terminal velocity, net force on a falling sphere is:
Explanation: Terminal velocity = constant speed → zero acceleration → net force = 0. Weight = buoyancy + drag.
49. Detergents lower surface tension of water to:
Explanation: Detergent molecules reduce surface tension, allowing water to spread over oily surfaces rather than beading up.
50. If capillary tube is not long enough for full capillary rise, the liquid:
Explanation: Liquid doesn't overflow. Instead, meniscus angle adjusts so that the capillary height formula is still satisfied with the available length.
51. Coefficient of viscosity η has SI unit:
Explanation: From τ = η(dv/dy): η = τ/(dv/dy) = (N/m²)/(m/s/m) = N·s/m² = Pa·s.
52. Terminal velocity of a sphere is proportional to r². If radius doubles, terminal velocity:
Explanation: v_t ∝ r². If r → 2r, v_t → 4v_t.
53. Energy needed to blow a soap bubble of radius R (surface tension T) is:
Explanation: Soap bubble has two surfaces. Total surface area = 2 × 4πR² = 8πR². Energy = T × Area = 8πR²T.
54. Blood flow in large arteries remains laminar even at high speed because blood has high:
Explanation: High viscosity increases the denominator in Re = ρvD/η, keeping Re low and flow laminar.
55. A capillary tube is dipped in water at angle 60° to vertical. Capillary rise is h. Rise compared to vertical tube (h₀) is:
Explanation: Vertical component of capillary rise = h cosθ_tilt = h₀. So h = h₀/cos60° = 2h₀. The actual column length increases but vertical height stays same.
56. Two soap bubbles of radii r₁ and r₂ (r₁ > r₂) are connected. Gas flows from:
Explanation: Excess pressure = 4T/r. Smaller bubble has higher pressure → gas flows into larger bubble. Larger bubble grows, smaller shrinks.
57. Viscosity of a gas arises from:
Explanation: In gases, faster layers exchange momentum with slower layers via molecular collisions, creating viscous drag.
58. In oil of viscosity 5 Pa·s, a 1 mm radius steel sphere (ρ = 8000 kg/m³) falls. Terminal velocity is (ρ_oil = 800 kg/m³, g = 10 m/s²):
Explanation: $v_t=2r^2(ρ_s-ρ_f)g/(9η)=2(10^{-3})^2(7200)(10)/(45)= 3.2×10^{-3}\text{ m/s} $.
59. Rise of sap in trees involves:
Explanation: Both cohesion-tension (capillary action in xylem) and osmotic pressure contribute to raising water in tall trees.
60. Temperature coefficient of surface tension is:
Explanation: Surface tension decreases with temperature. At critical temperature of a liquid, surface tension becomes zero.
61. The pressure inside a liquid increases with depth because:
Explanation: Each layer of liquid must support the weight of all liquid above it, increasing pressure with depth.
62. A body sinks in a liquid when:
Explanation: If ρ_body > ρ_liquid, weight > buoyant force, and body sinks.
63. In horizontal flow, the Bernoulli equation reduces to:
Explanation: For horizontal flow, h = constant, so the ρgh term cancels and P + ½ρv² = constant.
64. A horizontal pipe has cross section 10 cm² at point A and 5 cm² at point B. If speed at A is 3 m/s, speed at B is:
Explanation: A₁v₁ = A₂v₂. 10×3 = 5×v₂ → v₂ = 6 m/s.
65. A tank has water to height 80 cm. Speed of water at a hole at the base is (g = 10 m/s²):
Explanation: v = √(2gh) = √(2×10×0.8) = √16 = 4 m/s.
66. Work done in blowing a soap bubble from radius R₁ to R₂ is:
Explanation: Soap bubble has 2 surfaces. ΔEnergy = T × Δ(total area) = T × 2 × 4π(R₂² − R₁²) = 8πT(R₂² − R₁²).
67. Which has higher viscosity at room temperature?
Explanation: Glycerine (glycerol) has η ≈ 1.5 Pa·s at 20°C, far higher than water (10⁻³ Pa·s), alcohol (~10⁻³), or mercury.
68. Barometric pressure decreases with altitude because:
Explanation: Atmospheric pressure = weight of air column above. As altitude increases, less air is above, so pressure decreases.
69. A Pitot tube measures fluid speed. Static pressure P₁ and stagnation pressure P₂ give speed as:
Explanation: At stagnation point, all kinetic energy converts to pressure: P₂ = P₁ + ½ρv². Solving: v = √(2(P₂−P₁)/ρ).
70. Two liquid drops of the same mass are combined into one. Change in surface energy:
Explanation: Combined drop has smaller total surface area than two separate drops (volume same, sphere has minimum surface area). Surface energy decreases — the combination is energetically favorable.
71. In zero gravity, capillary rise would be:
Explanation: Capillary rise h = 2Tcosθ/(ρgr). If g → 0, h → ∞ — liquid rises to fill the entire tube regardless of length.
72. In a horizontal Bernoulli flow, pressure energy P is maximum when:
Explanation: P + ½ρv² = constant. When v is minimum, P is maximum.
73. Effect of temperature on viscosity: In engine oil, viscosity decreases sharply on heating. This means oil:
Explanation: Lower viscosity = less resistance to flow = oil flows more easily. Hot oil is thinner, which is why engine oil viscosity ratings (e.g., 5W-30) specify behaviour at different temperatures.
74. In a U-tube with two immiscible liquids (ρ₁, ρ₂) at equilibrium, the heights h₁ and h₂ are related by:
Explanation: Pressure balance at the liquid-liquid interface: ρ₁g h₁ = ρ₂g h₂ → ρ₁h₁ = ρ₂h₂.
75. A solid sphere falls through viscous oil and attains terminal velocity. When radius increases by 10%, terminal velocity increases by approximately:
Explanation: v_t ∝ r². If r → 1.1r, v_t → 1.21v_t. Increase = 21%.
76. Falling raindrops are spherical because:
Explanation: For a given volume, sphere has minimum surface area. Surface tension minimises surface energy → spherical shape.
77. In a siphon working to drain liquid over a hump, the pressure at the top of the siphon is:
Explanation: By Bernoulli's equation, as liquid climbs the hump, pressure decreases. At the top, P
78. A diver at 30 m depth (ρ_water = 1000 kg/m³, P₀ = 10⁵ Pa). Pressure on diver's eardrums equals:
Explanation: P = P₀ + ρgh = 10⁵ + 1000×10×30 = 10⁵ + 3×10⁵ = 4×10⁵ Pa.
79. Reynolds number for a sphere of diameter D falling through fluid at speed v:
Explanation: Re = ρvL/η where L is the characteristic length (diameter for a sphere).
80. A block of wood (mass 100 g, density 500 kg/m³) is held submerged in water (density 1000 kg/m³). Force needed to hold it is (g = 10 m/s²):
Explanation: Volume = 100g/500 = 200 cm³ = 2×10⁻⁴ m³. Buoyancy = 1000×10×2×10⁻⁴ = 2 N. Weight = 0.1×10 = 1 N. Net upward force = 2−1 = 1 N. So 1 N downward force needed to hold it.
81. For ideal fluid flow, the condition that flow lines do not cross is called:
Explanation: In steady flow, velocity at each point is constant in time, and streamlines are fixed in space (they cannot cross).
82. Bernoulli's theorem fails in viscous flow because:
Explanation: Bernoulli assumes no energy loss. Viscous friction converts kinetic energy to heat — total mechanical energy is not conserved.
83. For an incompressible fluid in hydrostatic equilibrium in a rotating cylindrical container (angular velocity ω), the pressure varies as:
Explanation: In rotating frame, centrifugal pressure dP/dr = ρω²r. Integrating: P = P₀ + ½ρω²r². Pressure increases with r².
84. In the absence of gravity, liquid in a container forms a:
Explanation: Without gravity, surface tension dominates. Liquid adopts the shape that minimises surface area for its volume — a sphere — positioned wherever it was placed.
85. Non-Newtonian fluid has viscosity that:
Explanation: Newtonian fluids have constant η. Non-Newtonian fluids (ketchup, blood, cornstarch in water) have viscosity that changes with applied shear rate or stress.
86. The shape of meniscus (concave vs convex) depends on:
Explanation: Adhesion > cohesion → concave meniscus (water-glass). Cohesion > adhesion → convex meniscus (mercury-glass).
87. A horizontal converging nozzle discharges water. If nozzle exit area is 1/4 of inlet area and inlet speed is 2 m/s, the exit kinetic energy per unit volume is:
Explanation: v_exit = 4×2 = 8 m/s (continuity). KE/volume = ½ρv² = ½×1000×64 = 32000 J/m³.
88. Aneroid barometer works without liquid because it uses:
Explanation: Aneroid means 'no fluid'. A sealed evacuated metal capsule compresses/expands with atmospheric pressure, mechanical amplification converts this to a reading.
89. A tank empties through a hole. Time to empty from height H₀ to zero is proportional to:
Explanation: Rate of fall: −dh/dt ∝ √h. Integrating gives t ∝ √H₀.
90. Contact angle of 90° means:
Explanation: At θ = 90°, cosθ = 0. Capillary rise h = 2Tcosθ/(ρgr) = 0. No rise or depression. Liquid meniscus is flat.
91. Blood viscosity is about 3–4 times higher than water. This high viscosity is mainly due to:
Explanation: Red blood cells (about 45% by volume) and plasma proteins like fibrinogen significantly increase blood's apparent viscosity over water.
92. Magnus effect (spinning ball curving in flight) is explained by:
Explanation: Spinning ball drags air: one side has higher speed (lower pressure) and other side has lower speed (higher pressure). Net pressure force causes the ball to curve.
93. Venturimeter throat pressure may fall below atmospheric. This is a problem because:
Explanation: If throat pressure falls below vapour pressure, liquid vaporises forming bubbles — cavitation. This damages pipes and pumps.
94. The compressibility of an ideal gas at constant temperature (isothermal) is:
Explanation: Isothermal bulk modulus = P. Compressibility = 1/B = 1/P. (Compare adiabatic compressibility = 1/(γP).)
95. A boat with heavy cargo is in a swimming pool. Cargo is thrown into pool. Water level:
Explanation: Floating boat displaces water equal to its total weight (boat + cargo). Cargo in water (if denser than water) displaces only its own volume. If cargo density > water density, total displaced volume decreases → level falls.
96. Soap film on a frame: when the film is stretched, work done against surface tension is stored as:
Explanation: Stretching a soap film increases its area, increasing surface energy = T × ΔArea (for two surfaces). This energy is recoverable — the film contracts if released.
97. In laminar flow through a pipe, velocity profile is:
Explanation: Poiseuille flow: v(r) = v_max(1 − r²/R²). Velocity is maximum at centreline and zero at walls (no-slip condition).
98. A pitot tube on an aircraft measures air speed. If indicated air speed doubles, dynamic pressure (½ρv²):
Explanation: Dynamic pressure ∝ v². Doubling speed quadruples dynamic pressure. This is why aircraft structures face 4× the load at twice the speed.
99. Jurin's law states that capillary rise is inversely proportional to tube radius. This fails for very narrow tubes because:
Explanation: In very narrow tubes, the simple cylindrical meniscus assumption breaks down and the contact angle changes, deviating from Jurin's law.
100. Karman vortex street (periodic vortex shedding) forms when Reynold's number is:
Explanation: Vortex shedding from a cylinder occurs for Re ~ 40–1000. At very high Re, flow becomes fully turbulent without regular vortex street.