Level 1 — RecognitionFluid Mechanics

Fluid Mechanics

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Difficulty Level: 1 (Recognition — MCQ, Matching, True/False with justification) Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each) [10 marks]

Choose the single best answer.

Q1. A fluid is defined as a substance that: (a) has a fixed shape and volume (b) continuously deforms under any applied shear stress, however small (c) resists all shear stress indefinitely (d) has zero viscosity

Q2. Kinematic viscosity $\nu$ is related to dynamic viscosity $\mu$ and density $\rho$ by: (a) $\nu = \mu\rho$ (b) $\nu = \rho/\mu$ (c) $\nu = \mu/\rho$ (d) $\nu = \mu + \rho$

Q3. The gauge pressure at depth $h$ in a static liquid of density $\rho$ is: (a) $\rho g h^2$ (b) $\rho g / h$ (c) $\rho g h$ (d) $\tfrac{1}{2}\rho g h$

Q4. The Young–Laplace pressure difference across a spherical soap bubble of radius $R$ and surface tension $\sigma$ is: (a) $\dfrac{\sigma}{R}$ (b) $\dfrac{2\sigma}{R}$ (c) $\dfrac{4\sigma}{R}$ (d) $\dfrac{\sigma}{2R}$

Q5. The Reynolds number is given by: (a) $\dfrac{\mu v L}{\rho}$ (b) $\dfrac{\rho v L}{\mu}$ (c) $\dfrac{\rho \mu}{v L}$ (d) $\dfrac{\rho v}{\mu L}$

Q6. Bernoulli's equation in its standard form assumes the flow is: (a) steady, viscous, compressible (b) unsteady, inviscid, incompressible (c) steady, inviscid, incompressible, along a streamline (d) turbulent and rotational

Q7. The Kutta–Joukowski lift theorem states that lift per unit span is: (a) $L = \tfrac{1}{2}\rho V_\infty^2 C_L$ (b) $L = \rho V_\infty \Gamma$ (c) $L = \rho g \Gamma$ (d) $L = \mu V_\infty \Gamma$

Q8. In fully developed laminar (Poiseuille) flow in a circular pipe, the velocity profile is: (a) uniform (plug) (b) linear (c) parabolic (d) logarithmic

Q9. For an incompressible steady flow through a pipe, the continuity equation reduces to: (a) $A v = \text{const}$ (b) $A/v = \text{const}$ (c) $A v^2 = \text{const}$ (d) $p + \rho g h = \text{const}$

Q10. Vorticity is defined as: (a) $\nabla \cdot \mathbf{v}$ (b) $\nabla \times \mathbf{v}$ (c) $\nabla \phi$ (d) $\oint \mathbf{v}\cdot d\mathbf{l}$

Section B — Matching (1 mark each) [8 marks]

Match each item in Column X with the correct description in Column Y.

#	Column X		Column Y
M1	Pitot tube	P	Measures atmospheric pressure
M2	Venturi meter	Q	Concept of a thin viscous region near a wall
M3	Barometer	R	Measures flow velocity from stagnation pressure
M4	Boundary layer	S	Measures flow rate using a constriction
M5	Pascal's law	T	Number of $\pi$ groups $= n - k$
M6	Archimedes' principle	U	Pressure applied to confined fluid transmits equally
M7	Buckingham $\pi$ theorem	V	Circulation is conserved for inviscid barotropic flow
M8	Kelvin's theorem	W	Buoyant force equals weight of displaced fluid

Section C — True/False with Justification (2 marks each: 1 T/F + 1 reason) [12 marks]

State True or False and give a one-line justification.

Q11. Specific gravity is a dimensionless quantity.

Q12. For a Newtonian fluid the shear stress is proportional to the square of the velocity gradient.

Q13. In the Eulerian description we follow individual fluid particles over time.

Q14. For steady flow, streamlines, pathlines, and streaklines all coincide.

Q15. Boundary layer separation is promoted by a favourable (decreasing) pressure gradient.

Q16. A potential flow is irrotational, so its vorticity is zero everywhere.

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (b). A fluid cannot resist shear; any shear stress produces continuous deformation. (1)

Q2 — (c). By definition $\nu = \mu/\rho$ ; units $m^2/s$ . (1)

Q3 — (c). Hydrostatic balance gives $dp = \rho g\, dh \Rightarrow p = \rho g h$ . (1)

Q4 — (c). A soap bubble has two surfaces, so $\Delta p = 4\sigma/R$ (a single droplet gives $2\sigma/R$ ). (1)

Q5 — (b). $Re = \rho v L/\mu$ , ratio of inertial to viscous forces. (1)

Q6 — (c). Standard Bernoulli assumptions: steady, inviscid, incompressible, along a streamline. (1)

Q7 — (b). Kutta–Joukowski: $L = \rho V_\infty \Gamma$ . (1)

Q8 — (c). Integrating the Poiseuille momentum balance gives a parabolic profile $u(r)=u_{max}(1-r^2/R^2)$ . (1)

Q9 — (a). With $\rho$ const, $\rho A v=$ const reduces to $Av=$ const. (1)

Q10 — (b). Vorticity $\boldsymbol\omega = \nabla\times\mathbf{v}$ . (1)

Section B (1 mark each)

Match	Answer	Reason
M1	R	Pitot tube reads stagnation pressure → velocity
M2	S	Venturi uses a constriction + Bernoulli for flow rate
M3	P	Barometer measures atmospheric pressure
M4	Q	Boundary layer = thin viscous near-wall region
M5	U	Pascal: pressure transmits equally in confined fluid
M6	W	Buoyancy = weight of displaced fluid
M7	T	Buckingham: $\pi$ groups $= n - k$
M8	V	Kelvin: circulation conserved (inviscid, barotropic)

(1 mark each, total 8)

Section C (2 marks each: 1 for correct T/F, 1 for justification)

Q11 — True. SG $=\rho/\rho_{water}$ , a ratio of densities → dimensionless. (1+1)

Q12 — False. Newtonian: $\tau=\mu\,(du/dy)$ , linear (first power) in velocity gradient, not squared. (1+1)

Q13 — False. Eulerian fixes attention on spatial points; the Lagrangian description follows individual particles. (1+1)

Q14 — True. In steady flow the velocity field is time-independent, so the three curves coincide. (1+1)

Q15 — False. Separation is caused by an adverse (increasing) pressure gradient; a favourable gradient suppresses it. (1+1)

Q16 — True. Irrotational means $\nabla\times\mathbf{v}=0$ , i.e. vorticity is zero everywhere. (1+1)

[
  {"claim":"Kinematic viscosity nu = mu/rho (Q2)","code":"mu,rho=symbols('mu rho',positive=True); nu=mu/rho; result = (nu == mu/rho)"},
  {"claim":"Soap bubble (two surfaces) Laplace pressure = 4 sigma/R (Q4)","code":"sigma,R=symbols('sigma R',positive=True); dp=2*(2*sigma/R); result = simplify(dp - 4*sigma/R)==0"},
  {"claim":"Poiseuille max velocity at center r=0 of u=umax(1-r^2/R^2) equals umax (Q8)","code":"r,R,umax=symbols('r R umax',positive=True); u=umax*(1-r**2/R**2); result = u.subs(r,0)==umax"},
  {"claim":"Incompressible continuity Av=const from rho*A*v with rho constant (Q9)","code":"A,v,rho=symbols('A v rho',positive=True); expr=rho*A*v; result = simplify(expr/rho - A*v)==0"}
]