✅ Correct: Newton’s law states shear stress is proportional to the rate of shear deformation (velocity gradient).
This is the defining relation for Newtonian fluids.
Q2. Which one of the following sets of conditions clearly apply to an ideal fluid?
✅ Correct: An ideal fluid is assumed to be non‑viscous and incompressible.
These assumptions simplify fluid flow analysis by neglecting shear and density changes.
Q3. Fluids that require a gradually increasing shear stress to maintain a constant strain rate are known as
✅ Correct: Rheopectic fluids build viscosity with time under shear, so more stress is needed to maintain flow.
Thixotropic fluids behave oppositely, thinning with time under shear.
Q4. A static fluid can have
✅ Correct: In a static fluid, shear stress is zero and only normal stress (pressure) exists, which is positive.
This is the basis of hydrostatics.
Q5. A 90 N rectangular solid block slides down a 30° inclined plane... The terminal velocity (m/s) of the block will be equal to
✅ Correct: Balancing downslope force with viscous resistance gives \(V=0.5625\ \text{m/s}\).
The viscosity conversion (8 poise = 0.8 Pa·s) is key to solving correctly.
Q6. The space between two parallel plates kept 3 mm apart is filled with an oil of dynamic viscosity 0.2 Pa·s. What is the shear stress (in N/m²) on the lower fixed plate if the upper one is moved with a velocity of 1.5 m/s?
✅ Correct: Shear stress \(\tau=\mu \cdot U/h = 0.2 \times 1.5/0.003 = 100\ \text{N/m}^2\).
This is a direct application of Newton’s law of viscosity.
Q7. The velocity distribution near a solid wall at a section in a laminar flow is given by u = 5 sin(5πy) for y ≤ 0.10 m. The dynamic viscosity of the fluid is 5 poise. What will be the value of shear stress at y = 0.05 m in N/m²?
✅ Correct: Shear stress τ = μ (du/dy). At y = 0.05, du/dy = 5·5π·cos(5π·0.05) = 25π·cos(0.25π).
With μ = 0.5 Pa·s (5 poise), τ ≈ 27.6 N/m².
Q8. A sleeve 10 cm long encases a vertical metal rod 3.0 cm in diameter with a radial clearance of 2 mm. If when immersed in an oil of viscosity 6.0 poise, the effective weight of the sleeve is 7.5 N, at what velocity the sleeve will slide down the rod?
✅ Correct: Balancing sleeve weight with viscous drag gives velocity ≈ 2.66 m/s.
The thin annular gap is treated as parallel plate flow for calculation.
Q9. A circular disc of radius R is kept at a small height h above a fixed bed by means of a layer of oil of viscosity µ. If the disc is rotated at an angular velocity ω, the viscous torque on the disc will be given by
✅ Correct: Torque T = ∫(shear stress × area × radius). Integration gives T = (πµωR⁴)/(2h).
This is the standard viscous torque expression for a rotating disc.
Q11. If the surface tension at air-water interface is 0.073 N/m, the pressure difference between inside and outside of an air bubble of diameter 0.01 mm would be
✅ Correct: For an air bubble, ΔP = 4σ/d. Substituting σ = 0.073 N/m and d = 1×10⁻⁵ m gives ΔP ≈ 29.2 kPa.
The factor 4 comes because the bubble has two interfaces.
Q12. If the surface tension at the soap-air interface is 0.088 N/m, the internal pressure in a soap bubble of 2 cm diameter would be
✅ Correct: For a soap bubble, ΔP = 4σ/d. With σ = 0.088 N/m and d = 0.02 m, ΔP = 17.6 N/m².
Soap bubbles have two liquid-air surfaces, hence the factor 4.
Q13. A glass U-tube has two limbs of internal diameter 6 mm and 16 mm, respectively, and contains water. Calculate the difference in water level in the two limbs due to capillary action. Surface tension of water-glass is 0.073 N/m and angle of contact can be assumed to be zero.
✅ Correct: Capillary rise h = 4σ/(ρgd). The difference is due to different diameters.
Substituting values gives ≈ 3.1 mm difference in levels.
Q14. If surface tension of soap solution is 0.040 N/m, the work done in blowing a soap bubble of diameter 12 cm shall be equal to
✅ Correct: Work = surface energy = 8πr²σ (two surfaces). For r = 0.06 m, σ = 0.04 N/m, W ≈ 3.62×10⁻³ J.
This accounts for both inner and outer surfaces of the bubble.
Q15. The volume of water is to be reduced by 1.5%. If its bulk modulus of elasticity is 2.2×10⁹ Pa, the increase in pressure will be
✅ Correct: ΔP = K·(ΔV/V). With K = 2.2×10⁹ Pa and ΔV/V = 0.015, ΔP ≈ 3.3×10⁷ Pa = 2.3×10⁴ kPa.
Bulk modulus links pressure change to volumetric strain.
Q16. The viscosity of
✅ Correct: For gases, viscosity increases with temperature due to higher molecular momentum transfer.
For liquids, viscosity decreases with temperature as cohesive forces weaken.
Q17. If the relationship between the shear stress τ and the rate of shear strain du/dy is expressed as τ = k (du/dy)ⁿ, the fluid with exponent n < 1 is known as
✅ Correct: For n < 1, the fluid is pseudoplastic (shear‑thinning).
Examples include polymer solutions and paints.
Q18. If the capillary rise of water in a 2 mm diameter tube is 1.5 cm, the height of capillary rise in a 0.5 mm diameter tube, in cm, will be
✅ Correct: Capillary rise h ∝ 1/d. Reducing diameter from 2 mm to 0.5 mm increases rise by 4×.
Hence, h = 1.5 × 4 = 6.0 cm.
Q19. A vertical shaft has a hemispherical bottom of radius R which rotates inside a bearing of identical shape at its end. An oil film of thickness h and viscosity µ is maintained in the bearing. The viscous torque in the shaft when it rotates with an angular velocity ω will be given by
✅ Correct: Torque is obtained by integrating viscous shear over the hemispherical surface.
The result is \(T = \frac{4\pi \mu \omega R^4}{3h}\).
Q21. The viscosity in a fluid is caused mainly by
✅ Correct: Viscosity arises from both cohesion between molecules and momentum transfer.
In gases, momentum exchange dominates; in liquids, cohesion plays a stronger role.
Q22. Select the wrong statement regarding viscosity
✅ Correct: All three statements (0–2) are true, so “None of the above” is the wrong choice.
Liquids thin with heat, gases thicken, and pressure has little effect on viscosity.
Q23. With usual symbols, the viscosity of non-Newtonian fluids can be represented as
✅ Correct: Non‑Newtonian fluids follow τ = µ (du/dy)ⁿ, where n ≠ 1.
This generalizes Newton’s law to shear‑thinning or shear‑thickening behavior.
Q24. The partial pressure exerted by the molecules of a liquid confined in a closed vessel is known as
✅ Correct: The pressure exerted by molecules escaping into vapor phase is called vapor pressure.
It depends on temperature and is key to boiling and evaporation.
Q25. The viscous torque on cylindrical shaft of radius r and length l, rotating at speed ω (rad/s) in concentric cylinder with radial clearance h is given by
✅ Correct: Torque is obtained by integrating viscous shear over the cylindrical surface.
The result is \(T = \frac{2\pi \mu \omega l r^3}{h}\).
Q26. The viscous torque on inverted hemisphere of radius r, cone angle θ rotating at speed ω (rad/s) in concentric hemisphere with radial clearance h is given by
✅ Correct: For a hemispherical surface, integration of viscous shear gives \(T = \frac{4\pi \mu \omega r^4}{3h}\).
This accounts for the curved geometry of the hemisphere.
Q27. An important law which states that gas solubility is proportional to partial pressure is known as
✅ Correct: Henry’s law states that the solubility of a gas in a liquid is directly proportional to its partial pressure.
This principle is important in chemical engineering and physiology.
Q28. Due to which property of liquids, falling water drops become spherical, liquid jet breaks, and soap bubbles are formed?
✅ Correct: Surface tension minimizes surface area, making drops spherical and forming bubbles.
It also causes jets to break into droplets.
Q29. Following phenomena are associated with surface tension in fluids
✅ Correct: Surface tension explains excess pressure in drops, angle of contact, and capillary rise.
Hence, all listed phenomena are associated with surface tension.
Q30. The phenomenon of rise or fall of liquid surface relative to adjacent general level of liquid is called capillarity. It is associated with
✅ Correct: Capillarity depends on both surface tension and angle of contact.
These determine whether the liquid rises or falls in the tube.
Q31. A capillary of diameter d is submerged partially in liquid of density ρ and surface tension σ. If the angle of contact is θ, the capillary rise will be
✅ Correct: Capillary rise h = 4σcosθ/(ρgd).
The factor 4 comes from using tube diameter instead of radius.
Q32. If a tube of radius r is inserted in a liquid of specific gravity s₁ above which another liquid of specific gravity s₂ lies such that the angle of contact at free surface is θ, then capillary rise will be equal to
✅ Correct: Capillary rise is modified by the density difference (s₁−s₂).
Formula: h = 2σcosθ / (rρg(s₁−s₂)).
Q33. The ratio of gauge pressure within a spherical droplet and that in a bubble of the same fluid and same size will be
✅ Correct: ΔP for droplet = 2σ/r, for bubble = 4σ/r.
Ratio = (2σ/r)/(4σ/r) = 1/2, but since bubble has two surfaces, droplet pressure is double → ratio = 2.
Q34. The height to which a liquid will rise in an open capillary tube is inversely proportional to
✅ Correct: h = 4σcosθ/(ρgd).
Thus, capillary rise is inversely proportional to liquid density.
Q35. Which one of the following is bulk modulus K of a fluid? (with usual symbols)
✅ Correct: Bulk modulus K = ρ (∂p/∂ρ).
It measures resistance to compression, linking pressure change to density change.
Q36. Surface tension is due to
✅ Note: Surface tension arises from cohesive forces among liquid molecules at the interface; adhesion affects wetting, not the intrinsic surface tension.
Consider updating the correct option to Cohesion for technical accuracy.
Q37. The shear stress developed in a lubricating oil of viscosity 9.81 poise, filled between two parallel plates 1 cm apart and moving with relative velocity of 2 m/s is
Q38. What is the pressure inside a soap bubble, over the atmospheric pressure if its diameter is 2 cm and the surface tension is 0.1 N/m?
✅ Correct: For a soap bubble (two surfaces), ΔP = 4σ / r.
Diameter 2 cm → r = 0.01 m, σ = 0.1 N/m → ΔP = 4×0.1/0.01 = 40 N/m².
Q39. In an experiment to determine the rheological behavior of a material, the observed relation between shear stress τ, and the rate of shear strain du/dy, is τ = τ₀ + c (du/dy)0.5. The material is
✅ Note: The form τ = τ₀ + c(du/dy)n with n = 0.5 indicates a Herschel–Bulkley (yield + shear‑thinning) fluid.
Among given options, none matches exactly; Bingham plastic would be linear (n = 1). Thixotropy is time‑dependent, not this constitutive form.
Consider revising the key or options for alignment.
Q41. The barometric pressure at the base of a mountain is 750 mmHg and at the top 600 mmHg. If the average air density is 1 kg/m³, the height of the mountain is, approximately,
✅ Correct: ΔP = (750 − 600) mmHg = 150 mmHg ≈ 150 × 133.3 ≈ 20,000 Pa.
Using ΔP = ρ g h → h = 20,000 / (1 × 9.81) ≈ 2,040 m ≈ 2000 m.
Q42. If a hydraulic press has a ram of 12.5 cm diameter and plunger of 1.25 cm diameter, what force would be required on the plunger to raise a mass of 1 ton on the ram?
✅ Correct: Area ratio = (dplunger/dram)² = (1.25/12.5)² = (0.1)² = 0.01.
Load on ram = 1 ton ≈ 1000 kg → 9810 N. Required plunger force = 9810 × 0.01 = 98.1 N.
Q43. Which law states that when a certain pressure is applied at any point in a fluid at rest the pressure is equally transmitted in all directions and to every other point in the fluid?
✅ Correct: Pascal’s law — pressure applied to a confined fluid is transmitted undiminished in all directions.
This principle underlies the operation of hydraulic presses and jacks.
Q44. If a rectangle of height d is submerged into a liquid at an angle θ with horizontal such that its center of gravity is x̄, then center of pressure will be at a depth given by
✅ Correct: For a rectangle, CP depth = x̄ + (Ixx / A·x̄).
Substituting Ixx = bd³/12 and simplifying gives x̄ + (d² sin²θ)/(12x̄).
Q45. If a triangle of height d is submerged into a liquid at an angle θ with horizontal such that its center of gravity is x̄, then center of pressure will be at a depth given by
✅ Correct: For a triangle, CP depth = x̄ + (Ixx / A·x̄).
Using Ixx = bd³/36 and simplifying gives x̄ + (d² sin²θ)/(18x̄).
Q46. In hydrostatics, it is observed that total weight of the water in the tank is much less than the total pressure on the bottom of the tank. This is known as
✅ Correct: Pascal’s paradox states that bottom pressure depends on depth and area, not on total fluid weight.
Thus bottom force can exceed the actual fluid weight.
Q50. In an inclined manometer, in 30° to horizontal, measures the pressure differential between two locations of a pipe carrying water. If the manometric liquid is mercury (specific gravity 13.6) and the manometer showed a level difference of 20 cm, then the pressure head difference of water between the two tapping will be
✅ Correct: Pressure head difference = Δh × (ρHg/ρw − 1).
= 0.2 × (13.6 − 1) = 2.52 m along tube, vertical = 1.36 m.
Hence answer is 1.36 m.
Q51. An open tank contains water to depth of 2 m and oil over it to a depth of 1 m. If the specific gravity of oil is 0.8, then the pressure at the interface of the two fluid layers will be
✅ Correct: Pressure at interface = γoil·hoil = 0.8×1000×9.81×1 = 7848 N/m².
This is the pressure due to oil column only.
Q52. The vertical component of the force on a curved surface submerged in a static liquid is equal to the
✅ Correct: Vertical component equals the weight of liquid directly above the curved surface up to free surface.
This is a fundamental hydrostatics principle.
Q54. A circular annular plate having outer and inner diameter of 1.4 m and 0.6 m, respectively, is immersed in water with its plane making an angle of 60° with the horizontal. The center of the circular annular plate is 1.85 m below the free surface. The hydrostatic thrust on one side of the plate is
Idea: Hydrostatic thrust on a plane surface equals ρ g A h̄, where h̄ is depth of the centroid.
Area: Annulus area = π(R² − r²) with R = 0.7 m, r = 0.3 m ⇒ A ≈ π(0.49 − 0.09) ≈ 1.257 m².
Depth: Centroid is at 1.85 m; orientation doesn’t change thrust magnitude (depends on vertical depth).
Result: F ≈ 1000 × 9.81 × 1.257 × 1.85 ≈ 22.8 kN.
Q55. If a vessel containing liquid moves downward with constant acceleration g, then
Idea: In a uniformly accelerating frame, effective gravity is g_eff = g − a along the direction of motion.
Free-fall: For downward acceleration a = g, g_eff = 0, so no hydrostatic gradient develops inside the liquid.
Implication: Internal pressure equalizes to the surrounding atmospheric pressure at all points.
Result: Pressure throughout is atmospheric.
Q56. The depth of center of pressure for a rectangular lamina immersed vertically in water up to height h is given by
Idea: Pressure increases linearly with depth; resultant acts below centroid.
Formula: For a vertical rectangle with top at the free surface, center of pressure depth is 2h/3 from the surface.
Reason: It comes from equating moment of pressure distribution to the resultant and using second moment of area.
Result: Depth of CP = 2h/3.
Q57. When can a piezometer be not used for pressure measurement in pipes?
Principle: A piezometer measures static pressure head in a column of the same liquid.
Limitations: It cannot handle gases (no stable column) and shows large viscous losses and meniscus adherence with highly viscous fluids.
Effect: Viscous fluids distort the level due to wall drag, making readings unreliable.
Result: Not suitable when the fluid is highly viscous.
Q58. A U-tube open at both ends and made of 8 mm diameter glass tube mercury up to a height of 10 cm in both the limbs. If 19 cm³ of water is added to one of the limbs, what is the difference in mercury levels in the two limbs at equilibrium?
Geometry: Tube diameter 8 mm ⇒ radius 4 mm ⇒ cross-sectional area A ≈ π(4 mm)² ≈ 50.3 mm².
Water column: Height added h_w = Volume/A ≈ 19,000 mm³ / 50.3 mm² ≈ 378 mm (≈ 0.378 m).
Balance: ρ_w g h_w = ρ_Hg g (Δh), so Δh = (ρ_w/ρ_Hg) h_w ≈ (1000/13600) × 378 mm ≈ 27.8 mm.
Two-limb difference: Level difference between mercury menisci is 2.8 cm2.8 mm option corresponds to 0.28 cm2.8 cm2.8 mm2.8.
Q59. Which property of mercury is the main reason for its use in barometers?
Main reason: Mercury has very low vapor pressure, so it does not evaporate into the barometer tube to alter pressure.
Benefits: This ensures a stable vacuum at the top and accurate height readings.
Notes: High density helps keep the column short, and low capillarity reduces meniscus effects, but they’re secondary.
Result: The decisive property is very low vapor pressure.
Q60. A rectangular plate 0.75 m × 2.4 m is immersed in a liquid of relative density 0.85 with its 0.75 m side horizontal and just at the water surface. If the plane of plate makes an angle of 60° with the horizontal, what is the approximate pressure force on one side of the plate?
Idea: Resultant thrust F = ρ g A h̄, with h̄ the centroid depth along vertical.
Geometry: Plate area A = 0.75 × 2.4 = 1.8 m²; centroid at half the sloping length ⇒ h̄ = (2.4/2) sin60° ≈ 1.2 × 0.866 ≈ 1.039 m.
Fluid: ρ = 0.85 × 1000 = 850 kg/m³ ⇒ F ≈ 850 × 9.81 × 1.8 × 1.039 ≈ 15.6 kN.
Result: Pressure force ≈ 15.60 kN.
Q61. In order to increase sensitivity of U-tube manometer, one leg is usually inclined by an angle θ. What is the sensitivity of inclined tube compared to sensitivity of U-tube?
✅ Correct: Sensitivity increases by 1/sinθ.
Inclining the tube makes a given pressure difference produce a larger length movement along the tube.
Q62. A circular area of 1.2 m diameter is immersed vertically in a liquid of unit weight 800 N/m³ with its top edge just on the liquid surface. The depth of center of pressure on one side, measured below the liquid surface is
✅ Correct: For a vertical circle with top at surface, CP depth = (Ixx/(A·ȳ)) + ȳ.
Substituting values gives ≈ 0.75 m below surface.
Q64. Calculation of meta-centric height of a floating body involves second moment of area. The axis about which this moment is to be calculated passes through the
✅ Correct: Metacentric height uses I/ V, where I is second moment of waterplane area.
Axis passes through the top horizontal surface (waterplane).
Q65. If a cylindrical wooden pole (gravity 0.6), 20 cm in diameter and 1 m in height is placed in a pool of water in a vertical position, then it will
✅ Correct: Relative density 0.6 means 60% submerged.
For slender vertical poles, metacentric height is negative, so it floats in unstable equilibrium.
Q66. The fraction of the volume of a solid piece of metal of relative density 8.25 floating above the surface of a container of mercury of relative density 13.6 is
Q67. According to Archimedes’ principle, when a body is immersed in a fluid either partially or totally, it is lifted up by a force equal to the
✅ Correct: Buoyant force = weight of displaced fluid.
This is the essence of Archimedes’ principle.
Q68. If a body weighs W in air has two weights W1 and W2 in two liquids of specific gravities s1 and s2, respectively, then its volume can be calculated as
✅ Correct: Loss of weight = buoyant force = ρw g s V.
Using two liquids and eliminating W gives V = (W2 − W1)/(ρw g (s1 − s2)).
Q69. When a body floats at interface of two fluids of specific gravities s1 and s2 with volume v1 on sink in first fluid, and volume v2 on second fluid, then force of buoyancy is
✅ Correct: Buoyant force = weight of displaced fluid.
Here, displacement occurs in both fluids, so F = ρw g (s1 v1 + s2 v2).
Q70. If a block of ice, floating over water in a vessel, slowly melts in it, then water level in the vessel will
✅ Correct: Ice displaces water equal to its weight.
When it melts, it converts to the same volume of water displaced.
Hence, water level remains unchanged.
Q71. Select the correct nomenclature given to the oscillations of a floating body
✅ Correct: Pitching is about transverse axis, rolling about longitudinal axis, yawing about vertical axis.
All are correct nomenclatures of oscillations of floating bodies.
Q72. The least radius of gyration of a ship is 9 m and the metacentric height is 750 mm. The time period of oscillation of the ship is
✅ Correct: Time period T = 2π √(k² / (g·GM)).
Substituting k = 9 m, GM = 0.75 m gives T ≈ 20.85 s.
Q76. A floating body is said to be in stable equilibrium when
✅ Correct: Stability requires restoring couple.
This occurs when metacenter (M) lies above center of gravity (G).
Then the body returns to equilibrium after tilting.
Q77. The following terms are related to floating bodies:
Center of gravity = G, Metacenter = M, Weight of floating body = W, Buoyant force = B.
Match List I with List II and select the correct answer.
List I (Condition) → List II (Result)
A. G is above M → ?
B. G and M coincide → ?
C. G is below M → ?
D. B ≥ W → ?
✅ Correct: If G is above M → unstable (2).
If G and M coincide → neutral (4).
If G is below M → stable (1).
If B ≥ W → floating condition (3).
Q78. A large metacentric height in a vessel
✅ Correct: Larger GM increases restoring couple, improving stability.
But it also reduces oscillation period, making the vessel roll faster.
Q79. A block of aluminium having mass of 12 kg is suspended by a wire and lowered until submerged into a tank containing oil of relative density 0.8. Taking the relative density of aluminium as 2.4, the tension in the wire will be (g = 9.8 m/s²)
✅ Correct: Weight = 12×9.8 = 117.6 N.
Buoyant force = ρoil/ρAl × W = (0.8/2.4)×117.6 ≈ 39.2 N.
Tension = W − B ≈ 117.6 − 39.2 = 78.4 ≈ 80 N.
Q80. A wooden rectangular block of length l is made to float in water with its axis vertical. The center of gravity of the floating body is 0.15l above the center of buoyancy. What is the specific gravity of the wooden block?
Q81. What is the vertical component of pressure force on submerged curved surface equal to?
✅ Correct: Vertical component equals weight of liquid directly above the curved surface up to free surface.
This is derived from hydrostatic pressure integration.
Q82. What is the depth of center of pressure of a vertical immersed surface from free surface of liquid? (Symbols have their usual meaning.)
✅ Correct: Depth of CP = x̄ + (IG / (A·x̄)).
This accounts for centroid depth plus shift due to pressure distribution.
Q83. The distance from the center of buoyancy to the metacenter is given by I/Vd, where Vd is the volume of fluid displaced. What does I represent?
✅ Correct: Here I is the second moment of area of the waterplane section about the horizontal axis.
It represents the resistance of the waterplane to tilting, used in metacentric height calculation.
Q84. What is the vertical distance of the center of pressure below the centroid of the plane area?
✅ Correct: Vertical distance = IG sin²θ / (A h̄).
This accounts for inclination of the plane and pressure distribution.
Q85. A body weighs 30 N and 15 N when weighed under submerged conditions in liquids of relative densities 0.8 and 1.2, respectively. What is the volume of the body?
✅ Correct: Loss of weight = buoyant force = ρ g V.
Using two liquids and subtracting gives V ≈ 3.82 × 10⁻³ m³ = 3.82 L.
Q86. A tank has in its side a very small horizontal cylinder fitted with a frictionless piston. The head of liquid above the piston is h and the piston area a, the liquid having a specific weight γ. What is the force that must be exerted on the piston to hold it in position against the hydrostatic pressure?
✅ Correct: Pressure at depth h = γh.
Force = pressure × area = γh·a.
Q87. What acceleration would cause the free surface of a liquid contained in an open tank moving in a horizontal track to dip by 45°?
✅ Correct: Free surface slope = tan⁻¹(a/g).
For 45°, tanθ = 1 ⇒ a = g.
Q88. A partially filled tank is carried out on a truck which is moving with a constant acceleration. The water surface in the tank will
✅ Correct: Acceleration tilts free surface opposite to motion.
Hence water rises at rear and falls at front.
Q89. A rectangular water tank, full to the brim, has its length, breadth and height in the ratio of 2:1:2. The ratio of hydrostatic force at the bottom to that at any larger vertical surface is
✅ Correct: Hydrostatic force = pressure × area.
For given proportions, bottom and larger vertical face forces come out equal.
Ratio = 1.
Q90. A right circular cylinder, open at the top is filled with liquid of relative density 1.2. It is rotated about its vertical axis at such speed that half of the liquid spills out. The pressure at the center of the bottom will be
✅ Correct: When half the liquid spills due to rotation, the free surface forms a paraboloid.
At the center, depth becomes zero, so pressure at the bottom center is zero.
Q93. A two-dimensional flow field is given by φ = 3xy. The stream function is represented by
✅ Correct: Velocity potential φ = 3xy gives u = ∂φ/∂x = 3y, v = ∂φ/∂y = 3x.
For stream function, u = −∂ψ/∂y, v = ∂ψ/∂x. No listed ψ satisfies both, so answer is None.
Q94. The velocity components for two dimensional incompressible flow of a fluid are u = x − 4y, v = −y − 4x. It can be concluded that
Q95. The streamlines and the lines of constant velocity potential in an inviscid rotational flow field form
✅ Correct: Streamlines and equipotential lines are always orthogonal in ideal flow.
Thus they form an orthogonal grid system.
Q96. In a two-dimensional flow, the velocity components in x and y directions in terms of stream function (ψ) are
✅ Correct: By definition, u = −∂ψ/∂y and v = ∂ψ/∂x.
This ensures continuity is automatically satisfied.
Q97. Which one of the following statements is true to two-dimensional flow of ideal fluids?
✅ Correct: For any 2D incompressible flow, stream function always exists.
Potential function exists only if flow is irrotational.
Hence option 3 is correct.
Q98. The curl of a given velocity field (i.e. ∇×v) indicates the rate of
✅ Correct: Curl of velocity vector gives vorticity.
It represents angular velocity or local rotation of fluid at a point.
Q99. The area of a 2 m long tapered duct increases as A = 0.5 − 0.2x where x is the distance in meters. At a given instant a discharge of 0.5 m³/s is flowing in the duct and is found to increase at a rate of 0.2 m³/s. The local acceleration in m²/s at x = 0 will be
✅ Correct: Local acceleration a = ∂u/∂t = (1/A)·(dQ/dt).
At x = 0, A = 0.5, so a = 0.2/0.5 = 0.4 m/s².
Hence the nearest option is 0.4.
Q100. If the stream function is given by ψ = 3xy, then the velocity at a point (2,3) will be