Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2018-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.
Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and manometry (Ch. 2), Bernoulli and the energy equation (Ch. 3), viscous flow in ducts and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), potential flow and the Magnus effect (Ch. 8), open-channel flow and the hydraulic jump (Ch. 10), turbomachinery and jet propulsion (Ch. 11).
Check — assumptions used across this paper:
Q1's manometer chain is read off the extraction as a two-stage water–mercury–glycerine–(air)–glycerine–mercury system. The enclosed air pocket between the two glycerine columns is treated as weightless (uniform pressure), so only the one described open end is needed to close the hydrostatic chain back to pipe P; the second "opening" is not load-bearing for this calculation.
Q5's wave/hydraulic-jump analysis takes the depth "in front of the wave" (0.15 m, undisturbed, at rest) as the upstream state and "behind the wave" (0.75 m) as the downstream state, per the question's own prose (the raw figure-label ordering in the extraction is a reconstruction and is not used to override the stated text). The classic hydraulic-jump head-loss formula is applied to the given depths, and the swept flow rate uses the measured wave celerity directly — a standard engineering estimate, not a fully momentum-self-consistent bore solution.
Q6's air properties are taken at 20°C (domestic ambient, ρ=1.19 kg/m³) since no duct-air temperature is stated.
Q7(c)'s "discharged at right angles to the initial direction (20° becomes 0°)" is read as: the reverser's exhaust jet is normally angled 20° forward of the fully-radial (right-angle) direction; part (c) removes that forward lean entirely, leaving a purely radial (90° to the engine axis) discharge with zero axial velocity component.
Q8's terminal velocity and Q9's cable drag coefficient are obtained from the Reynolds-number relations the attached charts themselves plot (Morrison's sphere-drag correlation for Q8; the flat subcritical Cd≈1.2 plateau of the smooth-cylinder curve for Q9, since Re≈3×104 falls solidly within it).
Warm cubicle air (30°C) is less dense than the bathroom air (15°C); the resulting pressure imbalance pushes the curtain inward through angle θ from vertical.
Find. The equilibrium deflection angle $\theta$.
Approach. Both air masses are equal in pressure at the top rod (where air can mix freely) and diverge hydrostatically below it because of the density difference; integrate the resulting triangular pressure-difference profile into a net horizontal force, then balance moments about the rod with the curtain's own weight, applying both resultants at the curtain's midpoint per the hint.
Air densities (ideal gas, $p=\rho RT$).
$$\rho_{in}=\frac{p}{RT_{in}}=\frac{100{,}000}{(287)(303.15)}=1.1494\ \text{kg/m}^3\qquad \rho_{out}=\frac{100{,}000}{(287)(288.15)}=1.2092\ \text{kg/m}^3$$
Net horizontal force on the curtain. The pressure difference grows linearly from zero at the rod to $(\rho_{out}-\rho_{in})gH$ at the floor; the resultant of this triangular load is
$$F=\tfrac12(\rho_{out}-\rho_{in})g H^2 b=\tfrac12(1.2092-1.1494)(9.81)(2)^2(1)=\boxed{1.174\ \text{N}}$$
Moment balance about the rod, force and weight both applied at the midpoint (hint). With the curtain tilted at $\theta$, the weight's restoring moment is $mg(L/2)\sin\theta$ and the pressure force's driving moment is $F(L/2)\cos\theta$; equating,
$$F\cos\theta=mg\sin\theta\ \Rightarrow\ \tan\theta=\frac{F}{mg}=\frac{1.174}{(0.400)(9.81)}=0.2992$$
$$\theta=\boxed{16.7^\circ\ \text{from vertical}}$$
