A rectangular pontoon 6 m long and 3 m wide floats at a draft of 1 m. Its centre of gravity is 1.4 m above the keel. The pontoon is

unstable, with a metacentric height of −0.15 m

stable, with a metacentric height of 0.25 m

unstable, with a metacentric height of −0.15 m

stable, because the buoyant force always acts below the centre of gravity

neutrally stable, since the waterplane is rectangular

Which of the following statements about laminar boundary layers on a flat plate are correct?

The thickness grows in proportion to √x from the leading edge

The local skin-friction coefficient decreases with distance along the plate

The thickness grows in proportion to √x from the leading edge

The local skin-friction coefficient decreases with distance along the plate

The total drag on a plate of length L is proportional to L

The speed of sound in air at 300 K is about 347 m/s. At 600 K, with the same gas, it is about

347 m/s — it depends only on the gas

Bernoulli’s equation `p/ρg + V²/2g + z = constant` may not be applied between two points when

there is appreciable friction between them

the two points lie on the same horizontal level

there is appreciable friction between them

the flow accelerates between them

the pipe changes diameter between them

Fluid Mechanics: Statics, Control Volumes, Boundary Layers and Pipe Flow

Section 3 is the largest of the Mechanical paper’s three discipline sections, and its fluid-mechanics heading is the longest single sub-list in the whole syllabus. The good news is that it is held together by very few ideas. Fluid statics is one equation, dp/dz = −ρg, and everything about manometers and submerged gates is that equation plus geometry — with one consistent trap, which is that the resultant force acts at the centroid’s depth but is applied at the centre of pressure, which is always deeper. Control-volume analysis is the workhorse: pick a boundary, and continuity, momentum and energy each become a statement about what crosses it. Bernoulli is the energy statement under four conditions worth naming every time — steady, incompressible, inviscid, along a streamline — and a question that violates one of them is testing whether you noticed rather than whether you can rearrange it. Dimensional analysis turns an unmanageable experiment into a plot of two numbers: n variables in m dimensions give n − m dimensionless groups, and for drag on a sphere that is 5 − 3 = 2, which is why the whole subject of sphere drag is one curve of C_d against Re. Viscous flow splits at the Reynolds number, and laminar pipe flow is fully solved — the velocity profile is a parabola, u_max/u_avg = 2 exactly, and f = 64/Re is not a correlation but a derivation. Boundary layers grow as √x, which is why the drag on a plate is not proportional to its length. And compressible flow enters only through the Mach number and a = √(γRT), in which the speed of sound depends on temperature and on nothing else.

Statics: The Force Is at the Centroid, the Push Is Deeper

Fluid statics is one equation, dp/dz = −ρg, and the two quantities every question about it wants are the hydrostatic force and the centre of pressure. For a plane surface submerged in a static fluid, the resultant force is F = ρg h̄ A where h̄ is the depth of the centroid — that part is easy, and it is why an area-average is enough for the magnitude. Where the force acts is a different question, because pressure grows with depth, so the lower half of the surface carries more than the upper half and the resultant is pulled down. The exact location is h_cp = h̄ + I_G/(h̄A) for a vertical surface, and the second term is the whole content. Take a vertical rectangular gate 2 m wide and 3 m deep with its top edge at the free surface: h̄ = 1.5 m, A = 6 m², so F = 1000 × 9.81 × 1.5 × 6 = 88.29 kN. Then I_G = 2(3³)/12 = 4.5 m⁴, and h_cp = 1.5 + 4.5/9 = 2.0 m exactly — two thirds of the depth, which is the standard result for a rectangle breaking the surface, and it is worth knowing as a check rather than as a formula.

The same 2 m × 3 m gate at three depths: the offset shrinks, the force does not
Top edge at	`h̄`	Force	`h_cp − h̄`
the free surface	1.5 m	88.29 kN	0.500 m — a third of the depth below centroid
1 m down	2.5 m	147.15 kN	0.300 m
5 m down	6.5 m	382.59 kN	0.115 m — nearly at the centroid

🧠 Why the offset shrinks, and when you may ignore it

The offset I_G/(h̄A) has h̄ in its denominator, so the deeper the surface, the smaller the offset — 0.500 m near the surface, 0.300 m at 1 m down, and only 0.115 m at 5 m down for the same gate. The physical reading is that pressure varies by ρg × 3 m across the gate either way, but at 5 m depth that variation is a small fraction of the mean pressure, so the loading is nearly uniform and the resultant sits nearly at the centroid. That is the licence for a real engineering habit: a deeply submerged small gate can be treated as uniformly loaded, and the error is bounded by I_G/(h̄A) which you can compute in one line before deciding. The trap is the opposite case, and it is the one exams use: for a surface breaking the free surface, the offset is at its largest and never negligible. One more warning worth attaching. h_cp = h̄ + I_G/(h̄A) is the vertical-surface form; for a plane inclined at θ the same derivation gives h_cp = h̄ + I_G sin²θ/(h̄A), and using the vertical form on an inclined gate overstates the offset by 1/sin²θ — at 30° that is a factor of four.

Floating stability is the one topic here where the answer is a sign, not a magnitude. A floating body is stable when its metacentric height GM = KB + BM − KG is positive, where BM = I/V with I the second moment of the waterplane area and V the displaced volume. Take a rectangular pontoon 6 m long, 3 m wide, floating at 1 m draft: BM = (6 × 3³/12)/(6 × 3 × 1) = 13.5/18 = 0.75 m and KB = 0.5 m at half the draft. With the cargo stowed so that KG = 1.0 m, GM = 0.5 + 0.75 − 1.0 = +0.25 m — stable. Raise the centre of gravity to KG = 1.4 m by stacking the same cargo higher and GM = −0.15 m — it capsizes. Nothing about the hull, the water or the weight changed; only the height at which the weight sits. That single comparison is why loading plans are regulatory documents and why BM growing as the cube of beam is the reason wide, shallow hulls are hard to overturn.

Control Volumes, Bernoulli’s Four Conditions, and Two Dimensionless Numbers

Draw a boundary and every conservation law becomes an accounting statement about what crosses it. Continuity for steady incompressible flow is A₁V₁ = A₂V₂ — the whole basis of a nozzle. Momentum is ΣF = ṁ(V_out − V_in), a vector equation, and its usual examination form is the force on a bend or a jet striking a plate; the reliable habit is to draw the control volume so that the unknown force crosses the boundary and every pressure you do not know does not. Energy becomes Bernoulli, p/ρg + V²/2g + z = constant, and the four conditions attached to it should be recited every time: steady, incompressible, inviscid, and along a streamline. Each one is a question type. Drop inviscid and you are in pipe flow and need a head-loss term. Drop along a streamline and you need the flow to be irrotational for the constant to be global. Drop incompressible and you are in gas dynamics. Drop steady and you need the unsteady term, which is where water-hammer lives.

🎯 Dimensional analysis is not bookkeeping — it is what makes the experiment possible

Suppose you want the drag force on a sphere. It depends on the diameter D, the velocity V, the density ρ and the viscosity μ — five variables including F itself, expressed in three dimensions (mass, length, time). Buckingham’s theorem says there are n − m = 5 − 3 = 2 independent dimensionless groups, and the conventional choice is the drag coefficient C_d = F/(½ρV²A) and the Reynolds number Re = ρVD/μ. That reduction is not tidiness. Testing four variables at ten values each would be 10⁴ = 10 000 experiments; testing one dimensionless group against another is one curve, and it is the same curve for every fluid, every sphere size and every speed. That is why C_d versus Re appears in every handbook as a single plot. It is also the whole justification of model testing: a 1/50 scale ship model tested at the right Froude number predicts the full-size hull, because matching the dimensionless groups matches the physics regardless of size. The examinable subtlety is that you usually cannot match every group at once — a scale model matching Reynolds number will not match Froude number, and choosing which one to honour is an engineering judgement about which force dominates. Ship hulls match Froude, because wave-making dominates; aerofoils match Reynolds, because viscosity does.

Laminar Flow Is Solved, Boundary Layers Grow as √x, and Pipes Lose Head

Laminar flow in a circular pipe is one of the few problems in the section with a closed-form exact solution, and knowing that it is exact changes how you treat it. The velocity profile is a parabola, so u_max/u_avg = 2 precisely, the kinetic-energy correction factor is exactly 2 and the momentum correction factor exactly 4/3 — figures worth having, because a question that gives you the mean velocity and asks for the centreline velocity is asking for a factor of 2 and nothing else. The Hagen–Poiseuille result ΔP = 128μLQ/(πd⁴) then follows, and its d⁴ is the striking part: halving a pipe diameter at fixed flow rate multiplies the pressure drop by sixteen. Verify the whole chain on numbers. Take d = 10 mm, L = 2 m, μ = 10⁻³ Pa·s, Q = 10⁻⁵ m³/s. Then V = 0.1273 m/s and Re = 1273 — laminar, so the solution applies. Hagen–Poiseuille gives ΔP = 81.487 Pa. Now go round the other way: f = 64/Re = 0.05027, and Darcy–Weisbach ΔP = f(L/d)(ρV²/2) gives 81.487 Pa — the same number to six figures, because in laminar flow f = 64/Re is not an empirical correlation but the Hagen–Poiseuille solution rewritten. Above Re ≈ 2300 that stops being true and f becomes genuinely empirical, read from a Moody chart against Re and relative roughness.

Laminar circular pipe flow: exact factors worth memorising
Quantity	Laminar (Re < 2300)	Turbulent, for contrast
Velocity profile	Parabolic — exact	Roughly 1/7-power, flat in the core
`u_max/u_avg`	2.000	≈ 1.22
KE correction `α`	2.000	≈ 1.06
Momentum correction `β`	4/3 = 1.333	≈ 1.02
Friction factor	`f = 64/Re`, derived	Moody chart — empirical
Roughness matters?	No — the wall layer is thick	Yes, and dominates at high Re

🎯 Boundary layers grow as √x, and that is why drag is not proportional to length

Prandtl’s idea was to split the flow: viscosity matters only inside a thin layer next to the surface, and outside it the fluid behaves as though it were inviscid. Blasius solved that layer for a flat plate and got δ = 5x/√Re_x, which because Re_x = Ux/ν means δ grows as √x, not as x. Put numbers to it: air at U = 10 m/s with ν = 1.5 × 10⁻⁵ m²/s. At x = 0.1 m, Re_x = 66 667 and δ = 1.937 mm. At x = 0.5 m, five times further along, Re_x = 333 333 and δ = 4.330 mm — only √5 = 2.236 times thicker, exactly as the square root demands. The consequence for drag is what makes this examinable rather than descriptive. The local skin-friction coefficient falls as 1/√x, so the downstream half of a plate contributes far less drag than the upstream half, and total drag scales as √L rather than L. Two related facts fall out of the same picture. First, the layer thickens until it trips into turbulence at around Re_x = 5 × 10⁵, after which it thickens faster and drags harder — which is why a plate can have a laminar front and a turbulent back with two different formulae in one calculation. Second, a turbulent boundary layer separates later than a laminar one, because it carries more momentum near the wall; that is the entire reason a golf ball has dimples, and the reason a sphere’s drag coefficient famously drops as Reynolds number rises through the critical value.

Compressible flow enters the syllabus only at the basics, and the one number that organises all of it is the Mach number M = V/a with a = √(γRT). Read that formula for what is absent: the speed of sound depends on temperature alone — not on pressure, not on density independently. Air at 300 K has a = 347.19 m/s; at 600 K it has a = 491.00 m/s, and the ratio is exactly √2 because a goes as √T. That single dependence explains why an aircraft at cruise altitude reaches a given Mach number at a lower true airspeed than at sea level: the air is colder, so the sound speed is lower. The other fact worth carrying is the threshold: below about M = 0.3 the density change across the flow is under 5 per cent and the incompressible equations are legitimate, which is the licence under which every earlier result in this chapter was derived. Above it, area and velocity stop being inversely related — in supersonic flow a diverging duct accelerates the gas, which is why rocket nozzles have the shape they do.

Key takeaways

The hydrostatic force uses the centroid depth; the centre of pressure is always deeper. A 2 m × 3 m vertical gate breaking the surface: F = 88.29 kN with h_cp at exactly 2.0 m, two thirds of the depth.
The offset I_G/(h̄A) shrinks with depth — 0.500 m at the surface, 0.115 m at 5 m down for the same gate. On an inclined plane it carries sin²θ, and omitting that overstates the offset by 1/sin²θ.
Floating stability is a sign, not a size: GM = KB + BM − KG. The 6 × 3 m pontoon at 1 m draft is stable at KG = 1.0 m (GM = +0.25) and capsizes at KG = 1.4 m (GM = −0.15) with the identical hull and weight.
Recite Bernoulli’s four conditions every time: steady, incompressible, inviscid, along a streamline. Each dropped condition names a different topic — pipe flow, gas dynamics, water hammer.
n − m dimensionless groups turn 10 000 experiments into one curve. Sphere drag has 5 − 3 = 2 groups, C_d and Re. You usually cannot match every group at model scale — hulls match Froude, aerofoils match Reynolds.
Laminar pipe flow is exact: u_max/u_avg = 2, α = 2, β = 4/3, and f = 64/Re is a derivation, not a correlation. At d = 10 mm, Q = 10⁻⁵ m³/s, Re = 1273, Hagen–Poiseuille and Darcy–Weisbach both give 81.487 Pa.
Boundary layers grow as √x, so drag scales as √L. Air at 10 m/s: δ = 1.937 mm at x = 0.1 m and 4.330 mm at x = 0.5 m — five times the distance, √5 times the thickness.
The speed of sound depends on temperature alone: a = √(γRT) gives 347.19 m/s at 300 K and 491.00 m/s at 600 K, a ratio of exactly √2. Below M ≈ 0.3 the incompressible equations are legitimate.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A vertical rectangular gate 2 m wide and 3 m deep has its top edge at the free surface of water. The depth of the centre of pressure below the free surface, in metres, is ______
Numerical answer — type the value.
Show answer
Answer: 2
2.0 m. The centroid is at h̄ = 1.5 m and the area is 6 m², so the resultant force is ρg h̄ A = 1000 × 9.81 × 1.5 × 6 = 88.29 kN. The centre of pressure is deeper, at h_cp = h̄ + I_G/(h̄A) with I_G = 2(3³)/12 = 4.5 m⁴, so h_cp = 1.5 + 4.5/(1.5 × 6) = 1.5 + 0.5 = 2.0 m — exactly two thirds of the depth, the standard result for a rectangle whose top edge is on the surface, and the fastest sanity check available. Answering 1.5 m is the trap: that is where the magnitude of the force is computed from, and confusing the two is the single most common error in fluid statics, because ρg h̄ A does use the centroid. The physical reason they differ is that pressure grows linearly with depth, so the lower half of the gate carries more load than the upper half and the resultant is dragged downwards. Note also how the offset behaves: push the same gate 5 m down and h_cp − h̄ falls to only 0.115 m, because the h̄ in the denominator grows while I_G and A do not — a deeply submerged small gate really is nearly uniformly loaded.
A rectangular pontoon 6 m long and 3 m wide floats at a draft of 1 m. Its centre of gravity is 1.4 m above the keel. The pontoon is
1. stable, with a metacentric height of 0.25 m
2. unstable, with a metacentric height of −0.15 m
3. stable, because the buoyant force always acts below the centre of gravity
4. neutrally stable, since the waterplane is rectangular
Show answer
Answer: B — unstable, with a metacentric height of −0.15 m
Unstable, GM = −0.15 m. Compute the three heights. BM = I/V where I is the second moment of the waterplane area about the longitudinal axis, 6 × 3³/12 = 13.5 m⁴, and V is the displaced volume, 6 × 3 × 1 = 18 m³, so BM = 0.75 m. The centre of buoyancy sits at half the draft, KB = 0.5 m. Then GM = KB + BM − KG = 0.5 + 0.75 − 1.4 = −0.15 m, and a negative metacentric height means any small heel produces a moment that increases the heel — it capsizes. Option A is the answer for KG = 1.0 m, and the contrast is the point: the same hull, the same weight and the same water, stable or not depending only on the height at which the load sits. Option C states something false and important to be clear about: the buoyant force acting below the centre of gravity is the normal condition for a floating body and is perfectly compatible with stability, because for a floating body — unlike a submerged one — the metacentre moves as the hull heels, and BM can be large enough to overcome an unfavourable KG. Option D confuses shape with stability; the rectangular waterplane is exactly what let us compute I in one line. The design lesson from BM = I/V is that I grows as the cube of the beam, which is why widening a hull is by far the cheapest way to buy stability.
For fully developed laminar flow in a circular pipe, the ratio of maximum to mean velocity and the kinetic-energy correction factor are respectively
1. 2 and 2
2. 2 and 4/3
3. 1.22 and 1.06
4. 4/3 and 2
Show answer
Answer: A — 2 and 2
2 and 2. The laminar profile in a circular pipe is an exact parabola, u = u_max[1 − (r/R)²], and integrating it gives u_avg = u_max/2, so the ratio is exactly 2. The kinetic-energy correction factor α = (1/A)∫(u/u_avg)³ dA also comes out to exactly 2 for that parabola — the two happen to coincide, and it is worth noticing that they do so only for this profile. Option B pairs the right velocity ratio with the momentum correction factor β = 4/3, which is the other exact laminar constant; the two corrections differ because one integrates the cube of velocity and the other the square. Option D swaps them. Option C gives the turbulent values, roughly 1.22 and 1.06, and they are worth remembering alongside for contrast: a turbulent profile is much flatter in the core, so the corrections are close to 1 and are usually neglected — which is exactly why textbooks can drop α from Bernoulli in turbulent pipe problems without comment. The reason this set of numbers is worth knowing precisely is that laminar pipe flow is one of the few exactly solved problems in the section: f = 64/Re is not a correlation but the same solution rewritten, and a question that gives you Re < 2300 is telling you exact answers are available.
Water (μ = 10⁻³ Pa·s, ρ = 1000 kg/m³) flows at 10⁻⁵ m³/s through a 10 mm diameter pipe of length 2 m. The pressure drop, in pascals, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 81.49
81.49 Pa. Check the regime first, because the whole method depends on it. V = Q/A = 10⁻⁵/(π × 0.01²/4) = 0.1273 m/s, so Re = ρVd/μ = 1000 × 0.1273 × 0.01/10⁻³ = 1273 — comfortably laminar, so the exact solution applies. Hagen–Poiseuille: ΔP = 128μLQ/(πd⁴) = 128 × 10⁻³ × 2 × 10⁻⁵/(π × 10⁻⁸) = 2.56 × 10⁻⁶/3.1416 × 10⁻⁸ = 81.487 Pa. Now the cross-check that is worth doing every time: f = 64/Re = 0.05027, and Darcy–Weisbach ΔP = f(L/d)(ρV²/2) = 0.05027 × 200 × (1000 × 0.1273²/2) = 81.487 Pa — the same to six figures, because in laminar flow f = 64/Re is the Hagen–Poiseuille solution rewritten and not an empirical fit. The d⁴ is the fact worth taking away: halve the diameter at the same flow rate and the pressure drop rises sixteen-fold, which is why a partly blocked artery or a scaled pipe is so much worse than the area reduction suggests. Two conditions to state honestly. Above Re ≈ 2300 this is all wrong — f becomes empirical and roughness enters, neither of which appears in the laminar solution. And the result assumes fully developed flow, so it under-predicts the drop across a short pipe where the entrance length still matters.
Which of the following statements about laminar boundary layers on a flat plate are correct?
1. The thickness grows in proportion to √x from the leading edge
2. The local skin-friction coefficient decreases with distance along the plate
3. The total drag on a plate of length L is proportional to L
4. A turbulent boundary layer separates later than a laminar one
Show answer
Answer: A — The thickness grows in proportion to √x from the leading edge; B — The local skin-friction coefficient decreases with distance along the plate; D — A turbulent boundary layer separates later than a laminar one
A, B and D. Blasius gives δ = 5x/√Re_x and Re_x = Ux/ν, so δ ∝ √x ✓. Check it on numbers: air at 10 m/s with ν = 1.5 × 10⁻⁵ gives δ = 1.937 mm at x = 0.1 m and 4.330 mm at x = 0.5 m — five times the distance for √5 = 2.236 times the thickness, exactly. B follows from the same solution: C_f,x = 0.664/√Re_x falls as 1/√x, from 2.57 × 10⁻³ at 0.1 m to 1.15 × 10⁻³ at 0.5 m. C is the false one, and it is false because of B: integrating a friction coefficient that decays as 1/√x gives total drag proportional to √L, not L, so the downstream half of a plate contributes much less drag than the upstream half. Doubling the length of a plate does not double its drag; it multiplies it by √2. D is true and is one of the more useful counter-intuitive facts in the subject: a turbulent layer carries more momentum near the wall, so it resists an adverse pressure gradient longer and separates later — which is why golf balls have dimples and why a sphere’s drag coefficient falls as Reynolds number crosses its critical value, a rare case where making the flow more turbulent reduces drag rather than increasing it.
The drag force on a sphere depends on its diameter, the free-stream velocity, the fluid density and the fluid viscosity. The number of independent dimensionless groups needed to describe it is ______
Numerical answer — type the value.
Show answer
Answer: 2
2. Count the variables including the one being predicted: F, D, V, ρ, μ — that is n = 5. Count the dimensions they are built from: mass, length and time, so m = 3. Buckingham’s π theorem gives n − m = 5 − 3 = 2 independent dimensionless groups, and the conventional pair is the drag coefficient C_d = F/(½ρV²A) and the Reynolds number Re = ρVD/μ. The value of that reduction is worth stating in numbers rather than in principle: testing four independent variables at ten values each would need 10⁴ = 10 000 experiments, while plotting one group against the other needs one curve — and it is the same curve for every fluid, every sphere size and every speed, which is exactly why every handbook prints C_d versus Re as a single plot. The theorem is also the entire justification of model testing, since matching the dimensionless groups matches the physics at any scale. The subtlety a 2-mark version of this question would probe is that you usually cannot match every group at once: a scale model matched on Reynolds number will not be matched on Froude number, so ship hulls are tested at the correct Froude number because wave-making dominates, and aerofoils at the correct Reynolds number because viscosity does.
The speed of sound in air at 300 K is about 347 m/s. At 600 K, with the same gas, it is about
1. 694 m/s
2. 491 m/s
3. 347 m/s — it depends only on the gas
4. 245 m/s
Show answer
Answer: B — 491 m/s
491 m/s. a = √(γRT), so the speed of sound goes as √T: doubling the absolute temperature multiplies a by √2 = 1.414, giving 347.19 × 1.414 = 491.00 m/s. Option A doubles it, which would be the answer if a went as T. Option C states something almost true and importantly wrong: γ and R are properties of the gas, so at a fixed temperature the speed of sound is a property of the gas alone — but temperature is in the formula and this question changed it. Option D divides by √2. Read the formula for what it does not contain, because that is the examinable content: pressure does not appear, and neither does density independently. Compressing air isothermally raises both pressure and density in proportion and the sound speed does not budge; only the temperature moves it. That single dependence has a practical consequence worth knowing: an aircraft at cruise altitude reaches a given Mach number at a lower true airspeed than at sea level, because the air up there is colder and the sound speed is lower — which is why Mach number, not airspeed, is the number that matters for compressibility effects.
Bernoulli’s equation p/ρg + V²/2g + z = constant may not be applied between two points when
1. the two points lie on the same horizontal level
2. there is appreciable friction between them
3. the flow accelerates between them
4. the pipe changes diameter between them
Show answer
Answer: B — there is appreciable friction between them
When there is appreciable friction. Bernoulli is the energy equation with four conditions attached — steady, incompressible, inviscid, along a streamline — and friction violates the third directly: viscous work converts mechanical energy to heat, so the sum is no longer constant and the equation needs a head-loss term, p/ρg + V²/2g + z + h_f = constant. That extended form is exactly what pipe-flow problems use, with h_f from Darcy–Weisbach. Options A, C and D are all things Bernoulli handles perfectly well and are there to test whether the four conditions are actually known rather than half-remembered. Same level (A) simply makes z cancel and is the easiest case, not a violation. Acceleration (C) is Bernoulli’s whole subject: the equation exists to trade velocity head against pressure head, which is what a nozzle does. Changing diameter (D) is what causes that acceleration through continuity, and a venturi meter is nothing but Bernoulli plus continuity across a diameter change. The useful way to hold this is to attach each dropped condition to a topic: drop inviscid and you are in pipe flow with head losses; drop incompressible and you are in gas dynamics; drop steady and you are in water hammer; drop along a streamline and you need irrotationality before the constant can be global.