Take session-wise tests or a full 100-question mock on Mechanical Properties of Fluids for NEET — with 90-second per-question timer, answer review, subtopic breakdown, and detailed solutions.
4 sectional sessions (25 Qs each) + 1 Full Mock (100 Qs). NEET 4/−1 marking. 90 sec per question timer.
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1. Pressure is defined as:
Explanation: P = Force / Area = F/A. SI unit: Pascal (Pa) = N/m².
2. Pressure at depth h in liquid (density ρ, g):
Explanation: P = ρgh. Pressure increases linearly with depth.
3. Pressure at depth 10 m in water (ρ=1000 kg/m³, g=10 m/s², atm P₀=10⁵ Pa). Total pressure:
Explanation: P = P₀ + ρgh = 10⁵ + 1000×10×10 = 10⁵+10⁵ = 2×10⁵ Pa.
4. Pascal's law states that pressure applied to enclosed fluid is:
Explanation: Pascal's law: pressure is transmitted equally in all directions throughout an enclosed fluid.
5. Hydraulic press: small piston area 10 cm², large piston area 100 cm². Force on small piston 100 N. Force on large piston:
Explanation: P = F₁/A₁ = F₂/A₂. F₂ = F₁×A₂/A₁ = 100×100/10 = 1000 N.
6. Gauge pressure is:
Explanation: Gauge pressure = Absolute pressure − Atmospheric pressure = pressure above atmospheric.
7. A U-tube manometer has oil (ρ=800) in one arm and water (ρ=1000) in other. Heights: oil 15 cm, water h. For pressure balance:
Explanation: ρ_oil×g×0.15 = ρ_water×g×h. 800×0.15 = 1000×h. h = 120/1000 = 0.12 m = 12 cm.
8. Hydraulic brakes work on:
Explanation: Hydraulic brakes transmit pressure equally through brake fluid — Pascal's law.
9. Atmospheric pressure ≈ 10⁵ Pa can support a column of water of height:
Explanation: h = P/ρg = 10⁵/(1000×10) = 10 m.
10. Pressure at a point in liquid does NOT depend on:
Explanation: P = ρgh. Depends on depth, density, g — but NOT on shape of container (Hydrostatic paradox).
11. 1 atmospheric pressure = 760 mm Hg. In Pa:
Explanation: 1 atm = 760 mm Hg = 1.013×10⁵ Pa.
12. In a hydraulic lift, work input = work output (ideal). This demonstrates:
Explanation: Hydraulic machine provides mechanical advantage but total work is conserved — conservation of energy.
13. In a tilted container, the pressure at a point depends on its:
Explanation: Pressure depends only on vertical depth below the free surface, regardless of container orientation.
14. Applications of Pascal's law include:
Explanation: Pascal's law applications: hydraulic jack, brakes, and press.
15. Pressure of gas enclosed in piston increases when:
Explanation: PV = const (isothermal). If V decreases, P increases.
16. Pressure at bottom of a 2 m tall cylinder vs 4 m tall cylinder (same liquid, same area):
Explanation: P = ρgh. Twice the height → twice the pressure at bottom.
17. In hydraulic press with A₁:A₂ = 1:20 and force F₁=100 N. F₂:
Explanation: F₂ = F₁×A₂/A₁ = 100×20 = 2000 N.
18. Barometer measures:
Explanation: A barometer measures atmospheric pressure.
19. Specific gravity of a liquid is its density relative to:
Explanation: Specific gravity = density of substance / density of water at 4°C = relative to water at 4°C.
20. Blood pressure is measured in mm Hg. 120 mm Hg in Pa (ρ_Hg = 13600 kg/m³, g = 10):
Explanation: P = ρgh = 13600×10×0.12 = 16320 Pa.
21. Pressure in fluid acts in:
Explanation: Fluid pressure acts equally in all directions at a given point (isotropic).
22. In a hydraulic system, if the large piston moves down by 1 cm (area = 100A), small piston (area A) moves:
Explanation: Volume conservation: A₁d₁ = A₂d₂. A×d_small = 100A×1. d_small = 100 cm.
23. As altitude increases, atmospheric pressure:
Explanation: Atmospheric pressure decreases with altitude (less air column above).
24. Equivalent height of atmosphere at sea level pressure (avg density ρ_air=1.2 kg/m³, P₀=10⁵ Pa, g=10):
Explanation: h = P/(ρg) = 10⁵/(1.2×10) = 10⁵/12 ≈ 8333 m ≈ 8.33 km.
25. Pascal's principle applies to:
Explanation: Pascal's law applies to fluids (liquids and gases).
26. Archimedes' principle states that buoyant force equals:
Explanation: Buoyant force = weight of fluid displaced by the submerged object.
27. A 5 kg object (volume 2000 cm³) is fully submerged in water (ρ=1000 kg/m³, g=10 m/s²). Apparent weight:
Explanation: Buoyant force = ρ_water×V×g = 1000×0.002×10 = 20 N. Apparent W = 50−20 = 30 N.
28. A floating body displaces fluid equal in weight to:
Explanation: Floating equilibrium: buoyant force = weight. Fluid displaced = its own weight.
29. Object (density ρ_obj) floats in fluid (density ρ_f). Fraction submerged:
Explanation: At equilibrium: ρ_obj×V×g = ρ_f×V_sub×g. V_sub/V = ρ_obj/ρ_f.
30. An object with density greater than fluid will:
Explanation: If ρ_object > ρ_fluid, weight > buoyant force → object sinks.
31. Ice (density ~917 kg/m³) floats in water (1000 kg/m³). Fraction above water:
Explanation: Fraction submerged = 917/1000 = 0.917. Fraction above = 1−0.917 = 8.3%.
32. A balloon of volume 1 m³ is filled with gas (ρ_gas=0.1 kg/m³). Balloon mass = 0.2 kg. Net upward force in air (ρ_air=1.2 kg/m³, g=10):
Explanation: Buoyant force = 1.2×1×10 = 12 N. Weight = (0.2+0.1)×10 = 3 N. Net upward = 12−3 = 9 N.
33. When ice melts in a glass of water, the water level:
Explanation: Floating ice displaces water equal to its weight. When it melts, it produces exactly that much water. Level stays the same.
34. Weight of object in liquid (W_apparent), in air (W). Relative density of object:
Explanation: RD = ρ_obj/ρ_water = W/Buoyancy = W/(W−W_apparent) = W/(W−W_app).
35. A ship can float in sea because:
Explanation: Ship floats because its average density (including air) .
36. Buoyant force on an object depends on:
Explanation: F_B = ρ_fluid × V_submerged × g. Depends on volume submerged and fluid density.
37. Object (m=100 g) fully submerged in liquid (ρ=0.8 g/cm³). Volume = 150 cm³. g=10 m/s². Net force:
Explanation: W = 0.1×10 = 1 N. F_B = 0.8×10³×150×10⁻⁶×10 = 1.2 N. Net = 1.2−1 = 0.2 N upward... hmm: Weight downward = 1 N, Buoyancy upward = 0.8×10³×1.5×10⁻⁴×10 = 1.2 N. Net = 0.2 N upward.
38. A body weighs 50 N in air and 35 N in water. Volume of body (g=10 m/s², ρ_water=1000):
Explanation: Buoyancy = 50−35 = 15 N = ρ_water×V×g. V = 15/(1000×10) = 0.0015 m³ = 1500 cm³.
39. Two identical objects: one in salt water (ρ=1030) and one in fresh water (ρ=1000). Which experiences more buoyant force?
Explanation: F_B = ρ_fluid × V × g. Higher ρ_fluid → more buoyancy. Salt water gives more buoyant force.
40. Metacentric height determines:
Explanation: Metacentric height determines the stability of a floating body. Positive metacentric height = stable.
41. An astronaut in a space station feels weightless because:
Explanation: Weightlessness in orbit = free fall. Both astronaut and station have same acceleration toward Earth.
42. A wooden block (density 600 kg/m³) of volume 10⁻³ m³ is tied by a string to pool bottom (water ρ=1000, g=10). Tension in string:
Explanation: F_B = 1000×10⁻³×10 = 10 N (up). Weight = 600×10⁻³×10 = 6 N (down). T = F_B − W = 10−6 = 4 N.
43. The line of action of buoyant force acts through:
Explanation: Buoyant force acts through centre of buoyancy = COM of displaced fluid.
44. Specific gravity of a substance is equal to:
Explanation: Specific gravity = density of substance / density of water at 4°C.
45. Submarine dives by flooding ballast tanks with seawater. This works by:
Explanation: Flooding ballast tanks increases average density above seawater → submarine sinks.
46. When an object is submerged deeper in a fluid (uniform density), buoyant force:
Explanation: F_B = ρ_f×V×g. Depends only on volume submerged and fluid density. At same submerged volume, F_B stays same.
47. A cube (side 10 cm, density 500 kg/m³) floats in water. Height above water:
Explanation: Fraction submerged = ρ/ρ_water = 500/1000 = 0.5. Submerged = 5 cm. Height above = 5 cm.
48. Reading of spring balance for object in liquid is the:
Explanation: Spring balance in liquid reads apparent weight = True weight − Buoyant force.
49. Plimsoll line on ships indicates:
Explanation: Plimsoll line shows the maximum safe loading level to ensure adequate buoyancy.
50. Object (W_air=100 N, W_water=80 N, W_liquid=85 N). Relative density of liquid:
Explanation: RD = (W_air−W_liquid)/(W_air−W_water) = (100−85)/(100−80) = 15/20 = 0.75.
51. Equation of continuity for fluid flow states:
Explanation: Continuity equation: A₁v₁ = A₂v₂ (conservation of mass for incompressible fluid).
52. Pipe narrows from 4 cm² to 1 cm² area. Fluid enters at 2 m/s. Exit velocity:
Explanation: A₁v₁ = A₂v₂. 4×2 = 1×v₂. v₂ = 8 m/s.
53. Bernoulli's equation is a statement of:
Explanation: Bernoulli's equation = conservation of energy applied to ideal (non-viscous, incompressible) fluid flow.
54. Bernoulli's equation: P + ½ρv² + ρgh = constant. When height is same, higher velocity means:
Explanation: At same height: P + ½ρv² = const. Higher v → lower P (Bernoulli effect).
55. Speed of efflux from hole at depth h (Torricelli's theorem):
Explanation: v = √(2gh) — same as free-fall velocity from height h (Torricelli's theorem).
56. Venturimeter works on:
Explanation: Venturimeter uses the pressure difference created by velocity change — based on Bernoulli's equation.
57. Water (ρ=1000 kg/m³) flows in pipe. At section 1: A=10 cm², v=2 m/s, P=2×10⁵ Pa. At section 2: A=5 cm², same height. v₂:
Explanation: Continuity: A₁v₁ = A₂v₂. v₂ = 10×2/5 = 4 m/s.
58. In the pipe problem above, P₂ at section 2 (same height):
Explanation: Bernoulli: P₁+½ρv₁² = P₂+½ρv₂². P₂ = 2×10⁵+½×1000×(4−16) = 2×10⁵−6000 = 1.94×10⁵ Pa.
59. Tank (H=2 m full) with hole at depth h=1 m from top. Horizontal range on ground (floor at base of tank):
Explanation: v = √(2gh) = √(2×10×1) = √20. Height to fall = H−h = 1 m. t = √(2×1/10) = √0.2. Range = v×t = √20×√0.2 = √4 = 2 m.
60. Aeroplane wings are designed so that air moves faster over top. This creates:
Explanation: Faster air on top → lower pressure on top. Pressure difference → upward lift (Bernoulli effect).
61. Volume flow rate Q = Av. If pipe splits into 5 equal branches (each area A/4), speed in each branch (original speed v₀):
Explanation: Q = A×v₀ = 5×(A/4)×v_branch. v_branch = A×v₀/(5A/4) = 4v₀/5 = 4v₀/5.
62. Pitot tube measures:
Explanation: Pitot tube measures flow velocity by comparing static and dynamic pressures (Bernoulli).
63. Hole in tank at middle (h = H/2). Speed of efflux vs hole at bottom (h = H):
Explanation: v ∝ √h. v_mid/v_bottom = √(H/2)/√H = 1/√2.
64. Venturimeter: inlet area A₁, throat area A₂, pressure difference ΔP. Volume flow Q:
Explanation: Standard venturimeter formula: Q = A₁A₂√(2ΔP/ρ(A₁²−A₂²)).
65. Two buildings close together experience strong wind between them. The corridor is dangerous because:
Explanation: Faster wind between buildings → lower pressure (Bernoulli) → can create suction forces.
66. Ideal fluid is incompressible and:
Explanation: Ideal fluid is incompressible and non-viscous. Bernoulli's equation applies to ideal fluids.
67. Spray pumps and perfume atomizers work on:
Explanation: Spray pumps: fast air over nozzle → low pressure → liquid sucked up. Bernoulli's effect.
68. Roof of a house can blow off in storm because:
Explanation: High wind speed outside → low pressure outside (Bernoulli). Inside at atmospheric. Inside pressure > outside → roof lifts.
69. Volume flow rate through a pipe (area A, velocity v):
Explanation: Volume flow rate Q = Av m³/s.
70. In a venturimeter, at the throat (narrowest part):
Explanation: At throat: smallest area → maximum velocity (continuity) → minimum pressure (Bernoulli). P min, v max.
71. Magnus effect (spinning ball curves) is due to:
Explanation: Spinning ball: faster air on one side → lower pressure → net force (curve). Bernoulli's principle.
72. Maximum range of efflux from a tank (height H, hole at height y from bottom):
Explanation: Range R = 2√(y(H−y)). Maximized when y = H/2 → R_max = H. Hole at y = H/2.
73. Blood flows faster in capillaries OR arteries? (Capillaries have small radius but far more in number.)
Explanation: Total cross-section of capillaries >> arteries. By continuity, v is slower in capillaries, faster in arteries.
74. Dynamic pressure (½ρv²) represents:
Explanation: ½ρv² = kinetic energy per unit volume (dynamic pressure).
75. Which is conserved in the continuity equation for incompressible fluid?
Explanation: Continuity equation: A₁v₁ = A₂v₂ expresses conservation of mass (volume flow rate for incompressible fluid).
76. Viscosity is the property of fluid that:
Explanation: Viscosity is the internal friction of a fluid — it resists relative motion between fluid layers.
77. Coefficient of viscosity η has SI units:
Explanation: η has units Pa·s (= N·s/m² = kg/(m·s)). 1 Poise = 0.1 Pa·s.
78. Stokes' law: drag force on sphere (radius r, velocity v, viscosity η):
Explanation: Stokes' Law: F = 6πηrv.
79. Terminal velocity is reached when:
Explanation: At terminal velocity, net force = 0: weight = drag + buoyancy.
80. Terminal velocity of sphere (r, ρ_s, ρ_f, η, g):
Explanation: v_t = 2r²(ρ_s−ρ_f)g/9η.
81. If radius of falling sphere doubles (same fluid, density), terminal velocity changes by factor:
Explanation: v_t ∝ r². r→2r: v_t → 4v_t.
82. Reynolds number Re < 2000 indicates:
Explanation: Re laminar flow. Re > ~4000: turbulent. 2000−4000: transitional.
83. Reynolds number Re = ρvd/η. It is:
Explanation: Re = inertial forces / viscous forces = ρvL/η.
84. Surface tension is defined as force per unit:
Explanation: Surface tension T = Force / Length. Unit: N/m.
85. Excess pressure inside a soap bubble of radius r (surface tension T):
Explanation: Soap bubble has two surfaces. Excess pressure = 4T/r.
86. Excess pressure inside a liquid drop of radius r:
Explanation: Liquid drop has one surface. Excess pressure = 2T/r.
87. Capillary rise is due to:
Explanation: Capillary rise/fall is caused by surface tension and adhesion between liquid and tube walls.
88. Capillary rise h = 2T cosθ/(ρgr). If radius halves, h:
Explanation: h ∝ 1/r. r→r/2: h → doubles.
89. Surface tension of soap solution is less than water because:
Explanation: Surfactants (soap) reduce intermolecular cohesive forces at the surface → lower surface tension.
90. Mercury in a glass capillary tube:
Explanation: Mercury is non-wetting (contact angle > 90°): cosθ depression (falls).
91. For liquids, viscosity generally _______ with temperature:
Explanation: For liquids, viscosity decreases with temperature (molecules move more freely). Opposite for gases.
92. Surface tension decreases when:
Explanation: Surface tension decreases when: temperature increases or surfactants (soap, detergents) are added.
93. Contact angle for water on clean glass is approximately:
Explanation: Water wets clean glass completely — contact angle ≈ 0° (complete wetting).
94. A ball (r=1 mm, ρ=2000 kg/m³) falls in oil (ρ_f=800, η=0.1 Pa·s, g=10). Terminal velocity:
Explanation: v_t = 2r²(ρ_s−ρ_f)g/9η = 2×10⁻⁶×1200×10/0.9 = 24000×10⁻⁶/0.9 ≈ 0.0267 m/s ≈ 0.0235 m/s (closest).
95. Small insects can walk on water surface due to:
Explanation: Water surface behaves like an elastic membrane due to surface tension, supporting light insects.
96. Newton's law of viscosity: shear stress = η × (dv/dy). Here dv/dy is called:
Explanation: dv/dy = velocity gradient (or shear rate). η is coefficient of viscosity.
97. Water rises 10 cm in capillary (T=0.07 N/m, θ=0°, r=?). If ρ=1000, g=10:
Explanation: h = 2Tcosθ/ρgr. r = 2×0.07/(1000×10×0.1) = 0.14/1000 = 0.00014 m = 0.14 mm.
98. Energy required to blow a soap bubble of radius R (surface tension T):
Explanation: Soap bubble has 2 surfaces. Energy = T × total area = T × 2 × 4πR² = 8πR²T.
99. Why does a parachutist reach terminal velocity at a lower speed when parachute opens?
Explanation: Larger parachute area → much greater drag force → equilibrium (terminal v) reached at much lower speed.
100. For a highly viscous fluid (like honey), flow is typically:
Explanation: High viscosity → low Reynolds number → laminar flow.