Solved Examples

How to Use This Problem Bank

Eleven sections of theory, and now the part that actually earns marks. What follows is 55 worked problems, arranged easy first and hard last, covering everything from "which formula is this?" up to the multi-step arrangements that decide ranks.

Work them with a pen and paper. Cover the solution, try it, then compare — including the check at the end of each one, because the check is where the marks usually leak away.

The four questions to ask before you write anything

Nearly every mistake in fluids is made in the first ten seconds, before any arithmetic starts. Ask these four, in this order:

1. Is the fluid at rest or moving? At rest means $P = P_a + \rho g h$ and Archimedes. Moving means continuity and Bernoulli. Moving and sticky means Poiseuille and Stokes. Reaching for Bernoulli in a hydrostatics problem is the commonest wasted page in the chapter.
2. Gauge or absolute? Say it out loud before you write the first line. A tyre gauge, a manometer and a blood-pressure cuff all read gauge. A barometer, a boiling point and anything involving a gas law need absolute. Half the wrong answers in this chapter are a missing $P_a$.
3. Radius or diameter? Areas go as $r^{2}$ and Poiseuille goes as $r^{4}$, so one slip here is a factor of 4 or of 16. Circle the letter in the question before you start.
4. One surface or two? A liquid drop has one surface, a soap bubble and a soap film have two. That factor of 2 is the entire content of a great many one-mark questions.

Key Point — the master formulas, all in one place: $$P = \frac{F}{A}, \qquad P = P_a + \rho g h, \qquad \frac{F_1}{A_1} = \frac{F_2}{A_2}$$ $$F_B = \rho_{\text{fluid}} V_{\text{disp}}\, g, \qquad \frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}, \qquad A_1 v_1 = A_2 v_2$$ $$P + \frac{1}{2}\rho v^{2} + \rho g h = \text{constant}, \qquad v = \sqrt{2gh}, \qquad R = 2\sqrt{h(H-h)}$$ $$F = \eta A \frac{dv}{dx}, \qquad F = 6\pi\eta r v, \qquad v_t = \frac{2r^{2}\left(\rho_{\text{body}} - \rho_{\text{fluid}}\right)g}{9\eta}$$ $$Q = \frac{\pi P r^{4}}{8\eta L}, \qquad Re = \frac{\rho v d}{\eta}, \qquad S = \frac{F}{L}, \qquad W = S\,\Delta A$$ $$\Delta P_{\text{drop}} = \frac{2S}{r}, \qquad \Delta P_{\text{cavity}} = \frac{2S}{r}, \qquad \Delta P_{\text{bubble}} = \frac{4S}{r}, \qquad h = \frac{2S\cos\theta}{r\rho g}$$

A note on symbols, once

Throughout this chapter $\rho$ is density, $S$ is surface tension and $\eta$ is the coefficient of viscosity. Elsewhere $T$ or $\gamma$ often stands for surface tension and $\mu$ for viscosity. In the terminal-velocity formula the density of the surrounding fluid is written $\rho_{\text{fluid}}$ rather than $\sigma$, because $\sigma$ is doing other jobs elsewhere. $r$ is always a radius and $d$ always a diameter. And $h$ is a depth measured downwards from a free surface in every $\rho g h$ term, unless a solution says otherwise — each one states its reference level.

The data used below

Unless a problem states its own numbers, every solution here uses this table, and each problem also restates the constants it uses inside its own solution, so you never have to scroll back.

Quantity	Value used
Density of water	1000 kg/m$^3$
Density of sea water	1030 kg/m$^3$
Density of mercury	13600 kg/m$^3$
Density of air at 0°C, at sea level	1.29 kg/m$^3$
Atmospheric pressure $P_a$	$1.013 \times 10^{5}$ Pa
Surface tension of water	0.073 N/m
Surface tension of soap solution	0.025 N/m
Surface tension of mercury at 20°C	0.465 N/m
Viscosity of water at 20°C	$1.0 \times 10^{-3}$ Pa s
Viscosity of air	$1.8 \times 10^{-5}$ Pa s
Specific heat capacity of water	4200 J/(kg K)

On the value of $g$: every problem states whether it uses $9.8$ m/s$^2$ or $10$ m/s$^2$, and no problem mixes the two. Every problem here uses $9.8$ m/s$^2$ unless it says otherwise.

[Board Important] Every solution below writes the formula on its own line before any number goes into it, and every pressure answer says the word gauge or the word absolute. Do both in the exam. A correct formula with an arithmetic slip still earns most of the marks; a pressure quoted without saying which kind it is can lose them all.

Solved Examples

Part 1: Warm-Ups — One Formula Each

Nine problems and two sets of one-liners whose whole content is picking the right relation. Get fast at these and the rest of the chapter is arithmetic.

Example 1: The tractor and the bicycle

A tractor of total mass 1800 kg rests on four tyres, each with a contact patch of area 0.055 m$^2$ on the soil. A cyclist and bicycle together have a mass of 78 kg and two contact patches of area $6.0 \times 10^{-4}$ m$^2$ each. Taking $g = 9.8$ m/s$^2$, compare the pressure each puts on the ground.

Solution:

1. The load spreads over every patch. Pressure is the normal force per unit area: $$P = \frac{F}{A} = \frac{mg}{n\,A_{\text{patch}}}$$

2. The tractor. $$P_{\text{tractor}} = \frac{1800 \times 9.8}{4 \times 0.055} = \frac{17\,640}{0.220} = 8.02 \times 10^{4} \text{ Pa}$$

3. The bicycle. $$P_{\text{bicycle}} = \frac{78 \times 9.8}{2 \times 6.0 \times 10^{-4}} = \frac{764.4}{1.2 \times 10^{-3}} = 6.37 \times 10^{5} \text{ Pa}$$

4. The ratio. $$\frac{P_{\text{bicycle}}}{P_{\text{tractor}}} = \frac{6.37 \times 10^{5}}{8.02 \times 10^{4}} = 7.9$$

Both of these are gauge pressures — the extra push on top of the atmosphere, which is already pressing on the soil everywhere.

5. Read the answer. A machine twenty-three times heavier presses on the ground with about one eighth of the pressure, because its tyres give it about 180 times the contact area. That is the entire reason a tractor has balloon tyres: it must not sink into a field.

Final Answer: tractor $8.0 \times 10^{4}$ Pa gauge; bicycle $6.4 \times 10^{5}$ Pa gauge — the bicycle presses about 8 times harder.

Takeaway: Weight decides the force; area decides the pressure. Whenever a question compares two things pressing on a surface, count the areas first — the masses almost never win the argument.

Example 2: How deep before the water doubles the pressure

At what depth in a fresh-water lake is the absolute pressure twice atmospheric? And in the sea, where the density is 1030 kg/m$^3$? Take $P_a = 1.013 \times 10^{5}$ Pa and $g = 9.8$ m/s$^2$.

Solution:

1. Write the depth relation, with the free surface as the reference level. Measuring $h$ downwards from the surface, $$P_{\text{abs}} = P_a + \rho g h$$

2. The condition. "Twice atmospheric" means $P_{\text{abs}} = 2P_a$, so the water alone must supply one whole atmosphere: $$\rho g h = P_a \qquad \Longrightarrow \qquad h = \frac{P_a}{\rho g}$$

3. Fresh water. $$h = \frac{1.013 \times 10^{5}}{1000 \times 9.8} = \frac{1.013 \times 10^{5}}{9800} = 10.3 \text{ m}$$

4. Sea water. $$h = \frac{1.013 \times 10^{5}}{1030 \times 9.8} = \frac{1.013 \times 10^{5}}{10\,094} = 10.0 \text{ m}$$

5. Store the number. Roughly every 10 m of water is worth one atmosphere. A diver at 30 m is at about 4 atmospheres absolute, or 3 atmospheres gauge. That single fact answers a surprising number of questions without a calculator.

Final Answer: about 10.3 m in fresh water and 10.0 m in the sea, at which depth the absolute pressure is $2.026 \times 10^{5}$ Pa and the gauge pressure is $1.013 \times 10^{5}$ Pa.

Takeaway: Ten metres of water is one atmosphere. Say it once and you will never again be surprised that a suction pump cannot lift water from a well more than about 10 m deep.

Example 3: What a cylinder gauge is not telling you

The gauge on a compressed-air cylinder reads 12.0 bar, where 1 bar $= 1.0 \times 10^{5}$ Pa. (a) What is the absolute pressure of the air inside? (b) The cylinder is used until the gauge reads zero. Is it empty? Take $P_a = 1.013 \times 10^{5}$ Pa.

Solution:

1. What a gauge measures. Every ordinary pressure gauge compares the inside with the atmosphere outside it. It reads the difference: $$P_{\text{gauge}} = P_{\text{abs}} - P_a$$

2. (a) Add the atmosphere back on. $$P_{\text{abs}} = P_{\text{gauge}} + P_a = 12.0 \times 10^{5} + 1.013 \times 10^{5}$$ $$P_{\text{abs}} = 1.3013 \times 10^{6} \text{ Pa}$$ which is about 12.85 atmospheres absolute.

3. (b) A zero reading is not an empty cylinder. Gauge zero means $P_{\text{abs}} = P_a$. The cylinder is still full of air at ordinary atmospheric pressure — it simply will not push any more of it out, because there is no longer a pressure difference to drive the flow.

4. Why this matters. Anything that uses the gas laws — a diver's air supply, a gas cylinder's contents, a bubble changing size — must use the absolute pressure, because $PV =$ constant is a statement about absolute pressure. Anything that measures a difference — the strength a tank wall needs, the reading on a blood-pressure cuff — is naturally gauge.

Final Answer: (a) $1.30 \times 10^{6}$ Pa absolute; (b) no — the cylinder still holds air at $1.013 \times 10^{5}$ Pa absolute, it just cannot deliver any of it.

Takeaway: Gauge pressure can be zero, and it can even be negative; absolute pressure cannot go below zero. Decide which one a problem needs before you start, and write the word down.

Example 4: Upthrust on a granite block

A block of granite of mass 25.0 kg and density 2700 kg/m$^3$ is hung from a spring balance and lowered until it is completely submerged in water. Taking $\rho_{\text{water}} = 1000$ kg/m$^3$ and $g = 9.8$ m/s$^2$, find the upthrust on it and what the balance now reads.

Solution:

1. Find the volume, because the upthrust is about volume, not mass. $$V = \frac{m}{\rho_{\text{granite}}} = \frac{25.0}{2700} = 9.26 \times 10^{-3} \text{ m}^3$$

2. The upthrust is the weight of the water pushed aside. The block is fully under, so the displaced volume is the whole of $V$: $$F_B = \rho_{\text{water}} V g = 1000 \times 9.26 \times 10^{-3} \times 9.8 = 90.7 \text{ N}$$

3. The apparent weight. $$W_{\text{app}} = mg - F_B = 25.0 \times 9.8 - 90.7 = 245 - 90.7 = 154 \text{ N}$$

4. What the balance shows in kilograms. $$m_{\text{app}} = \frac{154}{9.8} = 15.7 \text{ kg}$$ The block appears to have lost 9.26 kg — which is exactly the mass of the water it displaced, as it must be.

5. A check that costs nothing. The fraction of weight lost is $$\frac{F_B}{mg} = \frac{1000}{2700} = 0.370$$ the ratio of the two densities. Any body loses the fraction $\frac{\rho_{\text{fluid}}}{\rho_{\text{body}}}$ of its weight when fully submerged.

Final Answer: upthrust $90.7$ N; the balance reads $154$ N, that is an apparent mass of $15.7$ kg.

Takeaway: Upthrust needs the volume of the body and the density of the fluid — never the density of the body. The body's density only enters when you have to work the volume out from a mass.

Example 5: From hose to nozzle

Water flows along a garden hose of internal diameter 2.0 cm at 1.2 m/s and leaves through a nozzle of diameter 0.60 cm. Find the speed of the jet and the volume flow rate in litres per minute.

Solution:

1. Continuity: what goes in must come out. $$A_1 v_1 = A_2 v_2 \qquad \Longrightarrow \qquad v_2 = v_1\left(\frac{d_1}{d_2}\right)^{2}$$ Notice the square — the diameters enter through the areas.

2. Substitute the diameters directly, since only their ratio matters: $$v_2 = 1.2 \times \left(\frac{2.0}{0.60}\right)^{2} = 1.2 \times (3.333)^{2} = 1.2 \times 11.11 = 13.3 \text{ m/s}$$

3. The flow rate, from the hose end. $$A_1 = \pi\left(1.0 \times 10^{-2}\right)^{2} = 3.142 \times 10^{-4} \text{ m}^2$$ $$Q = A_1 v_1 = 3.142 \times 10^{-4} \times 1.2 = 3.77 \times 10^{-4} \text{ m}^3\text{/s}$$

4. Convert. One cubic metre is 1000 litres and one minute is 60 s: $$Q = 3.77 \times 10^{-4} \times 1000 \times 60 = 22.6 \text{ litres per minute}$$

5. The flow rate is the same everywhere. Check it at the nozzle: $A_2 = \pi(0.30 \times 10^{-2})^{2} = 2.83 \times 10^{-5}$ m$^2$, and $2.83 \times 10^{-5} \times 13.3 = 3.77 \times 10^{-4}$ m$^3$/s. The same number, as continuity demands.

Final Answer: the jet leaves at $13.3$ m/s; the flow rate is $3.8 \times 10^{-4}$ m$^3$/s, about $23$ litres per minute.

Takeaway: Narrowing the bore by a factor of $k$ multiplies the speed by $k^{2}$, and leaves $Q$ untouched. $Q$ is the quantity that does not change along the pipe — always compute it once and re-use it.

Example 6: The pressure drop across a squeeze

Water flows steadily along a horizontal pipe. Where the cross-section is 6.0 cm$^2$ the speed is 1.5 m/s; further along the pipe narrows to 2.0 cm$^2$. Taking $\rho = 1000$ kg/m$^3$, find the speed in the narrow part and the pressure drop between the two sections.

Solution:

1. Continuity first, always. $$v_2 = v_1 \frac{A_1}{A_2} = 1.5 \times \frac{6.0}{2.0} = 4.5 \text{ m/s}$$

2. Bernoulli along the streamline. The pipe is horizontal, so the $\rho g h$ terms are equal on both sides and cancel: $$P_1 + \frac{1}{2}\rho v_1^{2} = P_2 + \frac{1}{2}\rho v_2^{2}$$

3. Rearrange for the drop. $$P_1 - P_2 = \frac{1}{2}\rho\left(v_2^{2} - v_1^{2}\right) = \frac{1}{2}\times 1000 \times \left(4.5^{2} - 1.5^{2}\right)$$ $$P_1 - P_2 = 500 \times (20.25 - 2.25) = 500 \times 18.0 = 9.0 \times 10^{3} \text{ Pa}$$

4. Gauge or absolute? It does not matter here, and that is worth understanding. Both sides of Bernoulli have a $P_a$ in them if you use absolute pressures, and it cancels in the difference. So a pressure difference is the same number either way. The moment a problem asks for a pressure value, you must choose and say which.

5. Feel the size. 9 kPa is about 9% of an atmosphere, or 92 cm of water. That is a real, easily measured drop — which is exactly why a constriction makes a usable flowmeter.

Final Answer: $v_2 = 4.5$ m/s and the pressure falls by $9.0 \times 10^{3}$ Pa — a difference, identical in gauge and in absolute terms.

Takeaway: Continuity gives you the speeds, then Bernoulli gives you the pressures. Never try to do it the other way round, and never write Bernoulli before you have both speeds.

Example 7: Drag on a bead in oil

A small glass bead of radius 0.50 mm moves through oil of viscosity 0.20 Pa s at a steady 4.0 cm/s. Find the viscous drag on it.

Solution:

1. The sphere is small and the speed low, so Stokes' law applies: $$F = 6\pi\eta r v$$

2. Everything into SI first. $r = 0.50$ mm $= 5.0 \times 10^{-4}$ m and $v = 4.0$ cm/s $= 4.0 \times 10^{-2}$ m/s.

3. Substitute. $$F = 6\pi \times 0.20 \times 5.0 \times 10^{-4} \times 4.0 \times 10^{-2}$$ $$F = 6\pi \times 4.0 \times 10^{-6} = 7.54 \times 10^{-5} \text{ N}$$

4. The direction. Viscous drag always opposes the relative motion of the sphere and the fluid. If the bead is falling, the drag points up; if it is being dragged sideways, the drag points backwards along its path.

5. A sanity check on Stokes' law itself. It is valid only while the flow around the sphere stays laminar, which needs $Re = \frac{\rho v d}{\eta}$ to be well below about 1. For an oil of density around 900 kg/m$^3$, $Re = \frac{900 \times 0.040 \times 1.0 \times 10^{-3}}{0.20} = 0.18$. Comfortably fine.

Final Answer: $F = 7.5 \times 10^{-5}$ N, directed opposite to the bead's motion.

Takeaway: Stokes' law is linear in every one of $\eta$, $r$ and $v$ — no squares anywhere. That is what makes it so different from the $v^{2}$ drag you meet on a cricket ball, and it is why it only works for small, slow spheres.

Example 8: A raindrop and a soap bubble, same size

A spherical water drop and a soap bubble both have radius 1.5 mm. Taking $S_{\text{water}} = 0.073$ N/m and $S_{\text{soap}} = 0.025$ N/m, find the excess pressure inside each, and say which is greater.

Solution:

1. Count the surfaces. A liquid drop is solid liquid all the way through, so it has one surface — the outside. A soap bubble is a thin film with air inside and air outside, so it has two surfaces.

2. The drop. $$\Delta P_{\text{drop}} = \frac{2S}{r} = \frac{2 \times 0.073}{1.5 \times 10^{-3}} = \frac{0.146}{1.5 \times 10^{-3}} = 97.3 \text{ Pa}$$

3. The soap bubble — the two surfaces double it: $$\Delta P_{\text{bubble}} = \frac{4S}{r} = \frac{4 \times 0.025}{1.5 \times 10^{-3}} = \frac{0.100}{1.5 \times 10^{-3}} = 66.7 \text{ Pa}$$

4. Which wins? The drop, at 97.3 Pa against 66.7 Pa — even though the bubble gets the factor of 4 and the drop only gets 2. Water's surface tension is nearly three times that of soap solution, and that beats the factor of two.

5. These are excess pressures, which are gauge quantities by construction — each is the amount by which the inside exceeds the immediate outside. To get an absolute pressure inside the drop you would add $P_a$: $1.013 \times 10^{5} + 97.3 = 1.0140 \times 10^{5}$ Pa absolute.

Final Answer: drop $97.3$ Pa excess; soap bubble $66.7$ Pa excess. The drop has the higher excess pressure.

Takeaway: The factor of 4 belongs to a soap bubble, the factor of 2 to a drop and to a cavity — but only after you have checked whose $S$ you are using. A bigger factor multiplying a smaller $S$ does not automatically win.

Example 9: Five explanations worth having ready

Give the physical reason for each of the following, in one or two sentences.

(a) Blood pressure in a standing person is higher at the feet than in the brain. (b) Hydrostatic pressure is a scalar even though it is a force divided by an area. (c) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. (d) Water spreads out on clean glass while mercury beads up on the same glass. (e) A free liquid drop, with gravity and air resistance negligible, is always spherical.

Solution:

1. (a) Blood is a fluid, so the pressure in a connected column rises with depth exactly as $\rho g h$ says. Between a brain about 1.7 m above the feet and blood of density around 1060 kg/m$^3$, the extra pressure at the feet is $$\rho g h = 1060 \times 9.8 \times 1.7 = 1.77 \times 10^{4} \text{ Pa}$$ which is about 133 mm of mercury — a large fraction of a normal reading. The heart has to work against exactly this, which is why standing up too fast makes you dizzy.

2. (b) Force is a vector, but pressure is not the ratio of a vector to a scalar. At a point in a fluid at rest, the fluid pushes with the same magnitude in every direction, and the direction of the resulting force is decided by the orientation of whatever surface you place there, not by the pressure. A quantity with a single value at each point and no direction of its own is a scalar.

3. (c) The angle of contact is set by the fight between adhesion — liquid to solid — and cohesion — liquid to liquid. For mercury on glass, cohesion is much the stronger, the liquid pulls itself away from the wall, and the angle measured inside the liquid comes out obtuse, about 140°. For water on clean glass, adhesion wins, the liquid climbs the wall, and the angle is acute, close to 0°.

4. (d) Same competition, seen from outside. Where adhesion wins the liquid maximises its contact with the solid and wets it, spreading into a film. Where cohesion wins the liquid minimises its contact with the solid and pulls itself into beads, whose shape is set by its own surface tension.

5. (e) Surface tension makes a liquid surface behave like a stretched skin that always tries to shrink, and shrinking the surface means minimising the area. Of all shapes enclosing a given volume, the sphere has the smallest surface area, so the drop settles there. Gravity spoils this for a drop resting on a table, which is why the statement carries the "no external forces" clause.

Final Answer: (a) the $\rho g h$ of a 1.7 m column of blood, about 17 kPa; (b) it has one value at a point with no direction of its own; (c) cohesion beats adhesion for mercury and loses for water; (d) the same competition decides wetting against beading; (e) a sphere minimises area for a given volume.

[Board Important] These five carry a mark each and take fifteen seconds if you have said them once before. Learn to name the mechanism — $\rho g h$, cohesion against adhesion, minimum area — rather than describing what happens.

Takeaway: A one-mark "explain why" wants a mechanism, not a description. "Because the pressure is greater at the feet" earns nothing; "because $\rho g h$ over 1.7 m of blood is about 17 kPa" earns the mark.

Example 10: Filling in five blanks

Complete each statement with the right word from the pair given.

(a) The surface tension of a liquid generally _ as the temperature rises. (increases / decreases) (b) The viscosity of a gas with temperature, whereas the viscosity of a liquid with temperature. (increases / decreases) (c) For a solid, the shearing force is proportional to , while for a fluid it is proportional to . (shear strain / rate of shear strain) (d) For a fluid in steady flow, the increase in speed at a constriction follows from . (conservation of mass / Bernoulli's principle) (e) In a wind tunnel, turbulence sets in at a _ speed for a scale model than for the real aircraft. (greater / smaller)

Solution:

1. (a) decreases. Surface tension comes from the net inward pull on the surface molecules. Heat them and they move faster and hold each other less tightly, so $S$ falls, reaching zero at the critical temperature.

2. (b) increases; decreases. This is the one people get backwards. In a gas, viscosity is momentum carried across by molecules wandering between layers — heat them and they wander faster, so $\eta$ rises. In a liquid, viscosity is the intermolecular grip between layers — heat it and the grip weakens, so $\eta$ falls.

3. (c) shear strain; rate of shear strain. A solid under a shear stress deforms by a fixed angle and stops: $\frac{F}{A} = G\theta$. A fluid under a shear stress never stops, it deforms at a steady rate: $\frac{F}{A} = \eta\frac{dv}{dx}$. That difference is the definition of a fluid.

4. (d) conservation of mass. The speeding-up at a constriction is the equation of continuity $A_1v_1 = A_2v_2$, which is nothing but mass conservation. Bernoulli then tells you what the pressure does as a consequence of the speed change — it does not cause the speed change.

5. (e) greater. Turbulence sets in at a critical Reynolds number, $Re = \frac{\rho v d}{\eta}$. A model has a smaller $d$, so in the same air it needs a larger $v$ to reach the same $Re$. That is exactly why a tunnel testing a one-tenth model in ordinary air would need ten times the real flight speed.

Final Answer: (a) decreases; (b) increases, decreases; (c) shear strain, rate of shear strain; (d) conservation of mass; (e) greater.

[NEET Important] (b) and (d) between them account for a remarkable share of the one-mark fluid questions asked each year. Both have a mechanism behind them; learn the mechanism and you can never write the wrong half.

Takeaway: Continuity is mass; Bernoulli is energy. Keep those two labels straight and part (d) — and a good deal of the rest of the chapter — answers itself.

Part 2: Pressure at Depth, Barometers, Manometers and Hydraulics

Everything here is $P = P_a + \rho g h$ and Pascal's law, and everything here needs the word gauge or the word absolute in the answer.

Key Point: $$P_{\text{abs}} = P_a + \rho g h, \qquad P_{\text{gauge}} = \rho g h$$ The depth $h$ is measured downwards from the free surface. The pressure depends on $h$, on $\rho$ and on $g$ — and on nothing else: not on the shape of the vessel, not on how much liquid it holds, not on the area of its base.

Example 11: Will the offshore platform hold?

A vertical offshore structure is built to withstand a maximum stress of $1.0 \times 10^{9}$ Pa. Is it suitable for standing on the seabed above an oil well, where the ocean is roughly 3.0 km deep? Take the density of sea water as 1030 kg/m$^3$, $g = 9.8$ m/s$^2$ and $P_a = 1.013 \times 10^{5}$ Pa, and ignore currents.

Solution:

1. What the structure actually has to resist is the pressure of the water column above the seabed, measured from the free surface downwards: $$P_{\text{gauge}} = \rho g h = 1030 \times 9.8 \times 3000$$ $$P_{\text{gauge}} = 3.03 \times 10^{7} \text{ Pa}$$

2. The absolute pressure down there adds the weight of the atmosphere sitting on the sea: $$P_{\text{abs}} = P_a + \rho g h = 1.013 \times 10^{5} + 3.03 \times 10^{7} = 3.04 \times 10^{7} \text{ Pa}$$ The atmosphere contributes only $0.33\%$ — at this depth the distinction barely matters numerically, but you should still say which one you are quoting.

3. Compare with the design limit. $$\frac{3.03 \times 10^{7}}{1.0 \times 10^{9}} = 0.0303$$

4. Verdict. The water loads the structure to about 3% of what it can take — a safety factor of roughly 33. It is comfortably suitable.

5. What the calculation leaves out. Ocean currents, wave slam, storm loading and the bending moments they produce are the things that actually govern an offshore design, and they are dynamic loads on a slender column. Standing pressure is the easy part.

Final Answer: the water exerts $3.0 \times 10^{7}$ Pa gauge ($3.0 \times 10^{7}$ Pa absolute to two figures), only about $3\%$ of the design stress. The structure is suitable.

Takeaway: Answer the "is it safe?" question by taking a ratio, not by comparing two big numbers by eye. And when the numbers are this large, say explicitly how much of the total the atmosphere contributed — here it was one third of one per cent.

Example 12: A barometer ten metres tall

Torricelli's barometer used mercury. Suppose the same experiment is done with a wine of density 984 kg/m$^3$ instead. How tall would the column be at normal atmospheric pressure? Take $P_a = 1.013 \times 10^{5}$ Pa and $g = 9.8$ m/s$^2$.

Solution:

1. What a barometer balances. The column stands until its own weight per unit area equals the atmospheric pressure pushing down on the open reservoir: $$P_a = \rho g h \qquad \Longrightarrow \qquad h = \frac{P_a}{\rho g}$$

2. Substitute. $$h = \frac{1.013 \times 10^{5}}{984 \times 9.8} = \frac{1.013 \times 10^{5}}{9643} = 10.5 \text{ m}$$

3. Cross-check against mercury. For mercury at 13600 kg/m$^3$, $$h_{\text{Hg}} = \frac{1.013 \times 10^{5}}{13600 \times 9.8} = 0.760 \text{ m}$$ and the ratio of the two heights is $$\frac{10.5}{0.760} = 13.8 = \frac{13600}{984}$$ exactly the inverse ratio of the densities, as $h \propto \frac{1}{\rho}$ requires.

4. Why mercury won. A wine barometer would need a glass tube over ten and a half metres tall, taller than a three-storey house, and would have to be read from a ladder. Mercury does the same job in 76 cm because it is 13.6 times denser.

5. The pressure being measured is absolute, and this is the one instrument where that is unavoidable: the space above the column is a vacuum, so the column is balancing the full atmospheric pressure with nothing pushing back from above.

Final Answer: about $10.5$ m of wine, measuring an absolute pressure of $1.013 \times 10^{5}$ Pa.

Takeaway: A barometer height is $\frac{P_a}{\rho g}$ and nothing else — not the tube's diameter, not its shape, not how much liquid is in the reservoir. Only the density of the liquid changes the number.

Example 13: What the small piston of a car lift must bear

A hydraulic automobile lift is designed to raise cars of mass up to 3000 kg. The piston that carries the load has a cross-sectional area of 425 cm$^2$. (a) What is the greatest pressure the smaller piston has to bear? (b) If the smaller piston has an area of 12 cm$^2$, what force must be applied to it, and what is the lift's mechanical advantage? Take $g = 9.8$ m/s$^2$.

Solution:

1. Pascal's law is the whole problem. A pressure applied to an enclosed fluid is transmitted undiminished, so the pressure at the small piston equals the pressure at the large one: $$P = \frac{F_1}{A_1} = \frac{F_2}{A_2}$$

2. (a) Find that common pressure from the load side. $$A_2 = 425 \text{ cm}^2 = 425 \times 10^{-4} \text{ m}^2 = 4.25 \times 10^{-2} \text{ m}^2$$ $$P = \frac{Mg}{A_2} = \frac{3000 \times 9.8}{4.25 \times 10^{-2}} = \frac{29\,400}{4.25 \times 10^{-2}}$$ $$P = 6.92 \times 10^{5} \text{ Pa}$$ This is the pressure in excess of atmospheric — that is, a gauge pressure, because the atmosphere is pushing down on the top of the load piston too and has already been accounted for. In absolute terms the oil is at $6.92 \times 10^{5} + 1.013 \times 10^{5} = 7.93 \times 10^{5}$ Pa.

3. (b) The force on the small piston. $$F_1 = P A_1 = 6.92 \times 10^{5} \times 12 \times 10^{-4} = 830 \text{ N}$$ about the weight of an 85 kg person.

4. Mechanical advantage. $$\text{MA} = \frac{F_2}{F_1} = \frac{A_2}{A_1} = \frac{425}{12} = 35.4$$

5. And the price you pay. Energy is not created: to raise the car by 1.0 cm, the small piston must travel $35.4$ cm, because the same volume of oil has to move. Work in equals work out.

Final Answer: (a) $6.92 \times 10^{5}$ Pa gauge ($7.93 \times 10^{5}$ Pa absolute); (b) a force of $830$ N, with a mechanical advantage of $35.4$.

Takeaway: In a hydraulic machine the pressure is the shared quantity, not the force. Compute the pressure from whichever end you know, then use it on the other end.

Example 14: Oil floating on water in a road tanker

A tanker holds a 0.60 m layer of oil of density 820 kg/m$^3$ floating on a 1.20 m layer of water. Taking $g = 9.8$ m/s$^2$ and $P_a = 1.013 \times 10^{5}$ Pa, find the gauge pressure at the oil-water interface and at the floor of the tank, and the absolute pressure at the floor.

Solution:

1. Layered liquids just add their $\rho g h$ terms, taking each layer's own density over its own thickness. Measure depths downwards from the free oil surface.

2. At the interface, 0.60 m down, only the oil is above: $$P_{\text{gauge}} = \rho_{\text{oil}}\,g\,h_{\text{oil}} = 820 \times 9.8 \times 0.60 = 4.82 \times 10^{3} \text{ Pa}$$

3. At the floor, 1.80 m down, add the water layer's contribution: $$P_{\text{gauge}} = \rho_{\text{oil}}gh_{\text{oil}} + \rho_{\text{water}}gh_{\text{water}}$$ $$P_{\text{gauge}} = 4821.6 + 1000 \times 9.8 \times 1.20 = 4821.6 + 11\,760 = 1.66 \times 10^{4} \text{ Pa}$$

4. The absolute pressure at the floor. $$P_{\text{abs}} = P_a + P_{\text{gauge}} = 1.013 \times 10^{5} + 1.658 \times 10^{4} = 1.179 \times 10^{5} \text{ Pa}$$

5. The mistake to avoid. Do not average the densities and use 1.80 m of the average. That would give $\frac{820 + 1000}{2} = 910$ and $910 \times 9.8 \times 1.80 = 1.605 \times 10^{4}$ Pa — close, but wrong, and it is only close because the two layers happen to be of similar thickness. Add the layers, one at a time.

Final Answer: $4.82 \times 10^{3}$ Pa gauge at the interface, $1.66 \times 10^{4}$ Pa gauge at the floor, and $1.18 \times 10^{5}$ Pa absolute at the floor.

Takeaway: Walk down the column layer by layer, using each layer's own $\rho$ over its own thickness. Never average densities, and never use the total depth with one density unless there really is only one liquid.

Example 15: The mercury bridge — reading a spirit's density

A U-tube contains water in one arm and methylated spirit in the other, the two kept apart by mercury sitting in the bend. The mercury stands at the same level in both arms, with 10.0 cm of water above it on one side and 12.5 cm of spirit above it on the other. What is the relative density of the spirit?

Solution:

1. Use "same level, same pressure". Points at the same horizontal level in the same connected body of fluid at rest are at the same pressure. The mercury is one connected body, and its two surfaces are at the same level, so those two surfaces are at the same pressure.

2. Write the pressure at each mercury surface, coming down from the open air above each arm: $$P_a + \rho_{\text{water}}\,g\,h_{\text{water}} = P_a + \rho_{\text{spirit}}\,g\,h_{\text{spirit}}$$

3. The $P_a$ and the $g$ cancel — both arms are open to the same atmosphere: $$\rho_{\text{water}}\,h_{\text{water}} = \rho_{\text{spirit}}\,h_{\text{spirit}}$$

4. Solve. $$\rho_{\text{spirit}} = \rho_{\text{water}}\times\frac{h_{\text{water}}}{h_{\text{spirit}}} = 1000 \times \frac{10.0}{12.5} = 800 \text{ kg/m}^3$$ $$\text{relative density} = \frac{800}{1000} = 0.800$$

5. Read the answer physically. The lighter liquid needs the taller column to press down just as hard — 12.5 cm of spirit weighs the same, per unit area, as 10.0 cm of water. And notice that the density of the mercury never entered: it is only a bridge, and while its two surfaces are level it contributes nothing.

Final Answer: the spirit has density $800$ kg/m$^3$, a relative density of $0.800$.

Takeaway: When the mercury levels are equal, the mercury cancels out completely. The whole problem reduces to "equal weights per unit area of the two liquids above", which is one line.

Example 16: Fifteen centimetres more of each

Starting from the arrangement of the previous problem, a further 15.0 cm of water and 15.0 cm of spirit are poured into the respective arms. What is the new difference in the mercury levels? Take the relative density of mercury as 13.6 and $g = 9.8$ m/s$^2$.

Solution:

1. The columns are now unequal in weight. Water: $10.0 + 15.0 = 25.0$ cm. Spirit: $12.5 + 15.0 = 27.5$ cm. Their pressures at the old mercury level are $$\rho_{\text{water}}h_{\text{water}} = 1000 \times 0.250 = 250 \text{ kg/m}^2$$ $$\rho_{\text{spirit}}h_{\text{spirit}} = 800 \times 0.275 = 220 \text{ kg/m}^2$$ (These are $\frac{P}{g}$, kept in this form so nothing cancels by accident.) The water side now presses harder, so the mercury is pushed down under the water and up under the spirit.

2. Let $\Delta$ be the difference in the mercury levels, with the mercury standing higher in the spirit arm. Balance the pressures at the lower of the two mercury surfaces, working down each arm: $$P_a + \rho_{\text{water}}\,g\,h_{\text{water}} = P_a + \rho_{\text{spirit}}\,g\,h_{\text{spirit}} + \rho_{\text{Hg}}\,g\,\Delta$$

3. Cancel $P_a$ and $g$ and solve for $\Delta$. $$\Delta = \frac{\rho_{\text{water}}h_{\text{water}} - \rho_{\text{spirit}}h_{\text{spirit}}}{\rho_{\text{Hg}}} = \frac{250 - 220}{13\,600}$$ $$\Delta = \frac{30}{13\,600} = 2.21 \times 10^{-3} \text{ m} = 0.221 \text{ cm}$$

4. Sanity check the direction. The heavier column pushes its own mercury surface down, so the mercury must be higher in the spirit arm. It is, by about 2.2 mm.

5. Why so small? Because mercury is 13.6 times denser than water: a 3 cm imbalance of water column is worth only about 2 mm of mercury. That sensitivity ratio is exactly what makes mercury the right liquid for a manometer that must stay short.

Final Answer: the mercury levels differ by $0.221$ cm, about $2.2$ mm, standing higher in the spirit arm.

Takeaway: Set up a U-tube by balancing pressures at one chosen level, and choose the lowest interface as that level. Everything above it on each side contributes a $\rho g h$; everything below it is common and cancels.

Example 17: Reading a gas line off a mercury manometer

An open-tube mercury manometer is connected to a gas pipe. The mercury in the open arm stands 18.5 cm higher than in the arm connected to the gas. Taking $\rho_{\text{Hg}} = 13600$ kg/m$^3$, $g = 9.8$ m/s$^2$ and $P_a = 1.013 \times 10^{5}$ Pa, find the gauge and absolute pressures of the gas.

Solution:

1. Balance at a chosen level — take the lower mercury surface. The mercury is higher in the open arm, so the lower of the two mercury surfaces is the one in the gas arm. That surface is in direct contact with the gas, so the pressure there is simply $P_{\text{gas}}$.

2. Now reach the same level down the other arm. Starting at the open mercury surface, which sits at atmospheric pressure, you must descend 18.5 cm of mercury to get there: $$P_{\text{gas}} = P_a + \rho_{\text{Hg}}\,g\,h$$ So the gas is at the higher pressure — the taller mercury column in the open arm is being held up by the gas pushing on the other side. (Had the open arm been the lower one, the sign would flip and the gas would be below atmospheric.)

3. Compute the gauge pressure. $$P_{\text{gauge}} = \rho_{\text{Hg}}\,g\,h = 13\,600 \times 9.8 \times 0.185$$ $$P_{\text{gauge}} = 2.47 \times 10^{4} \text{ Pa}$$

4. And the absolute pressure. $$P_{\text{abs}} = P_a + P_{\text{gauge}} = 1.013 \times 10^{5} + 2.466 \times 10^{4} = 1.260 \times 10^{5} \text{ Pa}$$

5. Quote it as a height too, because that is how manometers are read in practice: the gas is at 18.5 cm of mercury gauge, or $76.0 + 18.5 = 94.5$ cm of mercury absolute.

Final Answer: $2.47 \times 10^{4}$ Pa gauge, which is $1.26 \times 10^{5}$ Pa absolute — equivalently 18.5 cm of mercury gauge.

Takeaway: Decide the direction by walking down each arm and adding a $\rho g h$ for every column you pass through, then set the two totals equal at a common level. Guessing from a picture is how signs get lost.

Example 18: Why the pressure halves after six kilometres

The atmosphere extends well over 100 km, yet atmospheric pressure at a height of about 6 km is only half its sea-level value. Explain, and estimate the height at which the pressure halves. Take $P_a = 1.013 \times 10^{5}$ Pa, the density of air at sea level as 1.29 kg/m$^3$, and $g = 9.8$ m/s$^2$.

Solution:

1. Why $\rho g h$ will not do here. That formula assumes a constant density. Air is a gas: as you climb, the pressure falls, so the air expands, so its density falls too. Less dense air contributes less pressure per metre, so the pressure falls off ever more slowly rather than running out at a definite ceiling.

2. Set up the honest version. Over a thin slab of thickness $dz$, $$dP = -\rho g\,dz$$ and for an isothermal atmosphere the density is proportional to the pressure, $\rho = \frac{\rho_0}{P_0}P$. Substituting, $$\frac{dP}{P} = -\frac{\rho_0 g}{P_0}\,dz \qquad \Longrightarrow \qquad P = P_0\,e^{-z/H}$$

3. The scale height $H$. $$H = \frac{P_0}{\rho_0 g} = \frac{1.013 \times 10^{5}}{1.29 \times 9.8} = \frac{1.013 \times 10^{5}}{12.64} = 8.0 \times 10^{3} \text{ m}$$ About 8 km. This is the height of a hypothetical atmosphere of uniform sea-level density that would give the same surface pressure.

4. Where does the pressure halve? Set $e^{-z/H} = \frac{1}{2}$: $$z_{1/2} = H\ln 2 = 8012 \times 0.693 = 5.55 \times 10^{3} \text{ m}$$ about 5.6 km — which is the observed "roughly 6 km", allowing for the fact that the real atmosphere is not isothermal.

5. Now answer the question as asked. The atmosphere is 100 km deep but the pressure halves in under 6 km because the density is not uniform: most of the mass sits in the bottom few kilometres, and everything above 30 km contributes barely one per cent of the total. The exponential never quite reaches zero, which is why "the top of the atmosphere" is a matter of definition rather than of measurement.

Final Answer: the pressure falls exponentially with a scale height of about $8.0$ km, halving at about $5.6$ km. All the pressures here are absolute.

Takeaway: $\rho g h$ is for liquids; gases need the exponential. The moment a problem spans kilometres of air, the constant-density formula is not an approximation, it is simply wrong.

Example 19: The cone that presses three times its own weight

A conical vessel stands on its circular base of radius 10.0 cm and tapers to a point 30.0 cm above it. It is filled to the brim with water. Taking $\rho = 1000$ kg/m$^3$ and $g = 9.8$ m/s$^2$, find the weight of the water and the force the water exerts on the base, and explain the difference.

Solution:

1. The weight of the water. A cone of base radius $R$ and height $H$ has volume $\frac{1}{3}\pi R^{2}H$: $$V = \frac{1}{3}\pi (0.100)^{2}(0.300) = \frac{1}{3}\pi \times 3.00 \times 10^{-3} = 3.14 \times 10^{-3} \text{ m}^3$$ $$W = \rho V g = 1000 \times 3.1416 \times 10^{-3} \times 9.8 = 30.8 \text{ N}$$

2. The force on the base. The base is flat and horizontal at a depth of 0.300 m below the free surface, so the gauge pressure is the same everywhere on it: $$P_{\text{gauge}} = \rho g H = 1000 \times 9.8 \times 0.300 = 2940 \text{ Pa}$$ $$F = P_{\text{gauge}} \times \pi R^{2} = 2940 \times \pi (0.100)^{2} = 2940 \times 3.1416 \times 10^{-2}$$ $$F = 92.4 \text{ N}$$

3. Three times the weight. Exactly three, in fact: $$\frac{F}{W} = \frac{\rho g H \pi R^{2}}{\rho\left(\frac{1}{3}\pi R^{2}H\right)g} = 3$$

4. Where does the extra force come from? From the walls. The cone slopes inwards as it rises, so the wall's normal reaction on the water has a downward component. Add that downward push from the walls to the weight of the water and you get exactly the 92.4 N pressing on the base. Nothing is created; the force balance on the water as a whole is perfectly satisfied.

5. This is the hydrostatic paradox, and it works the other way too. A vessel that flares outwards as it rises would press on its base with less than the weight of the water it holds, because the walls then hold part of the water up. In both cases the base feels $\rho g H A$, and only that.

Final Answer: the water weighs $30.8$ N but presses on the base with $92.4$ N gauge — three times as much. Adding $P_a$ over the same area would give the absolute force, $92.4 + 1.013 \times 10^{5} \times 3.1416 \times 10^{-2} = 3.27 \times 10^{3}$ N, but that is balanced by the atmosphere pushing up underneath the base.

Takeaway: The force on a horizontal base is $\rho g H \times A$, whatever the vessel is shaped like. If that disagrees with the weight of the liquid, the walls are making up the difference — and they always do, exactly.

Part 3: Upthrust, Apparent Weight and Floating

Key Point: $$F_B = \rho_{\text{fluid}}\,V_{\text{disp}}\,g \qquad \text{acting upward through the centre of buoyancy}$$ A floating body displaces its own weight of fluid, so the fraction of its volume under the surface is $$\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$$ This fraction cannot exceed 1 — if the ratio comes out above 1, the body sinks and $V_{\text{sub}} = V$.

Example 20: The diver's lead belt

A diver's weight belt carries 6.0 kg of lead of density 11300 kg/m$^3$. Taking the density of sea water as 1030 kg/m$^3$ and $g = 9.8$ m/s$^2$, find the upthrust on the belt under water and the effective weight the diver actually has to carry.

Solution:

1. Volume of the lead. $$V = \frac{m}{\rho_{\text{lead}}} = \frac{6.0}{11\,300} = 5.31 \times 10^{-4} \text{ m}^3$$ about half a litre.

2. Upthrust: the weight of the sea water displaced. $$F_B = \rho_{\text{sea}}\,V\,g = 1030 \times 5.31 \times 10^{-4} \times 9.8 = 5.36 \text{ N}$$

3. Weight in air. $$W = mg = 6.0 \times 9.8 = 58.8 \text{ N}$$

4. Apparent weight under water. $$W_{\text{app}} = W - F_B = 58.8 - 5.36 = 53.4 \text{ N}$$ which is what a 5.45 kg mass would weigh in air.

5. The fraction lost. $$\frac{F_B}{W} = \frac{\rho_{\text{sea}}}{\rho_{\text{lead}}} = \frac{1030}{11\,300} = 9.1\%$$ Lead is chosen for diving weights precisely because it is dense: a material of half the density would lose twice the fraction of its weight and you would need far more of it strapped to your waist to sink the same amount.

Final Answer: upthrust $5.36$ N; the belt weighs $53.4$ N under water instead of $58.8$ N in air, a loss of about $9\%$.

Takeaway: A submerged body loses the fraction $\frac{\rho_{\text{fluid}}}{\rho_{\text{body}}}$ of its weight. Dense materials lose almost none of it, which is exactly why they are used for ballast.

Example 21: A pine log in a river and in the sea

A pine log of density 520 kg/m$^3$ floats first in a fresh-water river ($\rho = 1000$ kg/m$^3$) and then drifts out into the sea ($\rho = 1030$ kg/m$^3$). What percentage of its volume stands above the surface in each case?

Solution:

1. Write the equilibrium condition. Floating means the upthrust exactly balances the weight: $$\rho_{\text{fluid}}\,V_{\text{sub}}\,g = \rho_{\text{log}}\,V\,g$$

2. The $g$ and the $V$ tidy up into a pure ratio: $$\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{log}}}{\rho_{\text{fluid}}}$$

3. In the river. $$\frac{V_{\text{sub}}}{V} = \frac{520}{1000} = 0.520 \qquad \Longrightarrow \qquad \text{above the surface} = 48.0\%$$

4. In the sea. $$\frac{V_{\text{sub}}}{V} = \frac{520}{1030} = 0.505 \qquad \Longrightarrow \qquad \text{above the surface} = 49.5\%$$

5. Read the change. The log rides $1.5\%$ of its volume higher in salt water. Denser fluid means more upthrust per unit of submerged volume, so less volume needs to be submerged. It is the same reason a ship rides higher when it leaves a river estuary, and the reason a swimmer floats more easily in the sea.

Final Answer: $48.0\%$ of the log stands above the water in the river, and $49.5\%$ in the sea.

Takeaway: The floating fraction is a ratio of densities and nothing else — the size of the log, the shape of the log and the value of $g$ all cancel. That is why the same wood floats identically in a bathtub and in the ocean.

Example 22: A balloon that just floats

A weather balloon of volume 8.0 m$^3$ is filled with helium of density 0.18 kg/m$^3$ in air of density 1.29 kg/m$^3$. (a) What is the greatest total mass — envelope, instruments and all — that it can just support? (b) If the envelope itself has a mass of 3.2 kg, what payload is left?

Solution:

1. The balloon floats in air exactly as a log floats in water. The upthrust is the weight of the air displaced: $$F_B = \rho_{\text{air}}\,V\,g$$

2. What has to be lifted is the helium plus everything attached: $$W = \left(\rho_{\text{He}}V + m_{\text{extra}}\right)g$$

3. Set them equal for the "just floats" condition and cancel $g$ throughout: $$\rho_{\text{air}}V = \rho_{\text{He}}V + m_{\text{extra}}$$ $$m_{\text{extra}} = \left(\rho_{\text{air}} - \rho_{\text{He}}\right)V$$

4. (a) Substitute. $$m_{\text{extra}} = (1.29 - 0.18)\times 8.0 = 1.11 \times 8.0 = 8.88 \text{ kg}$$

5. (b) Subtract the envelope. $$m_{\text{payload}} = 8.88 - 3.2 = 5.7 \text{ kg}$$

6. What limits a real balloon. As it rises, the air thins out and $\rho_{\text{air}}$ falls, so the upthrust falls while the load does not. The balloon stops climbing at the height where $\rho_{\text{air}}V$ has dropped to match the total weight — which is why weather balloons expand as they rise and eventually burst rather than floating forever.

Final Answer: (a) $8.88$ kg total; (b) about $5.7$ kg of payload.

Takeaway: The lifting capacity of a balloon is $\left(\rho_{\text{fluid}} - \rho_{\text{gas}}\right)V$. The gas inside is not weightless — it is part of the load, and forgetting it is the standard slip in this problem.

Example 23: How many drums under the raft?

A raft is to be built from sealed empty drums, each of volume 0.21 m$^3$ and mass 18 kg, in fresh water of density 1000 kg/m$^3$. It must carry a platform of mass 120 kg and six people averaging 70 kg each. How many drums are needed? Take $g = 9.8$ m/s$^2$.

Solution:

1. What one drum can hold up. At best a drum is fully submerged and displaces its whole volume: $$F_{B,\text{drum}} = \rho_{\text{water}}\,V\,g = 1000 \times 0.21 \times 9.8 = 2058 \text{ N}$$

2. Take off the drum's own weight. The drum has to float itself before it floats anything else: $$W_{\text{drum}} = 18 \times 9.8 = 176.4 \text{ N}$$ $$F_{\text{spare}} = 2058 - 176.4 = 1881.6 \text{ N}$$ which in mass terms is $$m_{\text{spare}} = \rho_{\text{water}}V - m_{\text{drum}} = 1000 \times 0.21 - 18 = 192 \text{ kg per drum}$$

3. The load to be carried. $$m_{\text{load}} = 120 + 6\times 70 = 120 + 420 = 540 \text{ kg}$$

4. Divide. $$n = \frac{540}{192} = 2.81$$

5. Round the right way. You cannot use 2.81 drums, and rounding down would sink the raft. Take 3 drums, which carry $3 \times 192 = 576$ kg — a margin of 36 kg, or about $6.7\%$. That is uncomfortably tight for a raft that will also meet waves, so a real builder would use four.

Final Answer: each drum supports 192 kg beyond its own weight, so 3 drums are the mathematical minimum; four would be sensible.

Takeaway: Always subtract the float's own weight before dividing, and always round a floatation count upward. A drum that can lift 210 kg of water can only lift 192 kg of passengers, because 18 kg of that capacity is spent on the drum itself.

Part 4: Continuity and Bernoulli Together

The pairing that runs the second half of the chapter. Continuity gives the speeds; Bernoulli then gives the pressures. In that order, every time.

Key Point: $$A_1v_1 = A_2v_2 \qquad \text{and} \qquad P_1 + \frac{1}{2}\rho v_1^{2} + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^{2} + \rho g h_2$$ Bernoulli holds along a single streamline, for steady, incompressible, non-viscous flow. All three terms have the units of pressure. You may use gauge pressures on both sides or absolute pressures on both sides, but never one of each.

Example 24: Forty fine holes in a spray pump

The cylindrical barrel of a spray pump has a cross-section of 8.0 cm$^2$, and one end of it carries 40 fine holes each of diameter 1.0 mm. If the liquid inside the barrel moves at 1.5 m per minute, what is the speed at which it is ejected through the holes?

Solution:

1. This is pure continuity. Whatever volume the plunger pushes along the barrel each second has to come out through the holes in the same second: $$A_{\text{barrel}}\,v_{\text{barrel}} = n\,a_{\text{hole}}\,v_{\text{hole}}$$

2. Convert everything to SI first. $$A_{\text{barrel}} = 8.0 \text{ cm}^2 = 8.0 \times 10^{-4} \text{ m}^2$$ $$v_{\text{barrel}} = \frac{1.5}{60} = 0.025 \text{ m/s}$$ The holes have diameter 1.0 mm, so their radius is $0.50$ mm $= 5.0 \times 10^{-4}$ m: $$a_{\text{hole}} = \pi r^{2} = \pi\left(5.0 \times 10^{-4}\right)^{2} = 7.854 \times 10^{-7} \text{ m}^2$$

3. Total hole area. $$n\,a_{\text{hole}} = 40 \times 7.854 \times 10^{-7} = 3.142 \times 10^{-5} \text{ m}^2$$

4. Solve for the ejection speed. $$v_{\text{hole}} = \frac{A_{\text{barrel}}v_{\text{barrel}}}{n\,a_{\text{hole}}} = \frac{8.0 \times 10^{-4} \times 0.025}{3.142 \times 10^{-5}} = \frac{2.0 \times 10^{-5}}{3.142 \times 10^{-5}}$$ $$v_{\text{hole}} = 0.64 \text{ m/s}$$

5. Notice how modest the speed-up is. The barrel is 25 times wider in area than all forty holes together, so the jet is 25 times faster than the plunger — but the plunger is crawling at 2.5 cm/s, so the jet only reaches 64 cm/s. To get a fine mist you would push far harder or use fewer holes.

Final Answer: the liquid leaves the holes at about $0.64$ m/s.

Takeaway: Add the hole areas up before you divide. Forty holes of 1 mm diameter are not a 1 mm problem — they are one $3.14 \times 10^{-5}$ m$^2$ opening, and the number 40 is as important as the diameter.

Example 25: Which streamline picture cannot be real?

Two diagrams both claim to show the steady flow of a non-viscous liquid through a pipe that narrows. In one, the streamlines crowd together where the bore is narrow. In the other, the streamlines keep exactly the same spacing throughout, even where the pipe has narrowed. Which is impossible, and why?

Solution:

1. What a streamline diagram means. These diagrams are drawn so that the same volume of fluid per second passes between each neighbouring pair of streamlines. The gaps between the lines are little tubes of flow, and no fluid ever crosses a streamline.

2. Apply continuity to one such gap. If the gap has cross-sectional area $A$ and the fluid there moves at speed $v$, then $Av$ is fixed for that gap all the way along. Narrower gap means faster flow — that is the entire content of the picture.

3. Now judge picture (b). It shows the same gap width in a narrower pipe. But the same gap width at the same spacing means the same $A$ — while the pipe has narrowed, so there is less total room. The only way to fit the same total flow through a smaller pipe with unchanged streamline spacing would be for some fluid to vanish, or for fluid to cross a streamline. Both are forbidden.

4. Say it as an equation. Between the two sections, $$A_{\text{wide}}v_{\text{wide}} = A_{\text{narrow}}v_{\text{narrow}}$$ so $A_{\text{narrow}} < A_{\text{wide}}$ forces $v_{\text{narrow}} > v_{\text{wide}}$, and faster flow is crowded streamlines.

5. Verdict. Picture (a), with the lines crowding into the constriction, is the correct one. Picture (b) violates the equation of continuity, which is to say it violates conservation of mass.

Final Answer: the picture with unchanged streamline spacing in the narrowed section is impossible; it breaks the equation of continuity, and so conservation of mass.

Takeaway: Crowded streamlines mean fast flow, and by Bernoulli, low pressure. You can read speed and pressure straight off a well-drawn streamline diagram without a single calculation.

Example 26: What the water pays to speed up and to climb

Water flows through a pipe that is 8.0 cm in diameter at ground level, where the speed is 2.0 m/s and the gauge pressure is $2.4 \times 10^{5}$ Pa. Further along, the pipe has narrowed to 4.0 cm in diameter and risen 6.0 m. Taking $\rho = 1000$ kg/m$^3$, $g = 9.8$ m/s$^2$ and $P_a = 1.013 \times 10^{5}$ Pa, find the speed and the pressure at the upper section.

Solution:

1. Continuity for the speed. Halving the diameter quarters the area: $$v_2 = v_1\left(\frac{d_1}{d_2}\right)^{2} = 2.0 \times \left(\frac{8.0}{4.0}\right)^{2} = 2.0 \times 4 = 8.0 \text{ m/s}$$

2. Bernoulli, with the ground as the datum for height. $$P_1 + \frac{1}{2}\rho v_1^{2} + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^{2} + \rho g h_2$$ with $h_1 = 0$ and $h_2 = 6.0$ m. Both pressures will be gauge, which is legitimate because $P_a$ would appear on both sides.

3. Evaluate the three terms at the lower section. $$P_1 = 2.4 \times 10^{5} \text{ Pa}, \qquad \frac{1}{2}\rho v_1^{2} = 500 \times 4.0 = 2.0 \times 10^{3} \text{ Pa}, \qquad \rho g h_1 = 0$$ $$\text{total} = 2.42 \times 10^{5} \text{ Pa}$$

4. Now the known terms at the upper section. $$\frac{1}{2}\rho v_2^{2} = 500 \times 64 = 3.2 \times 10^{4} \text{ Pa}, \qquad \rho g h_2 = 1000 \times 9.8 \times 6.0 = 5.88 \times 10^{4} \text{ Pa}$$

5. Subtract to get $P_2$. $$P_2 = 2.42 \times 10^{5} - 3.2 \times 10^{4} - 5.88 \times 10^{4} = 1.512 \times 10^{5} \text{ Pa gauge}$$ and the total at the top is $1.512 \times 10^{5} + 3.2 \times 10^{4} + 5.88 \times 10^{4} = 2.42 \times 10^{5}$ Pa, matching the bottom exactly — the check that the arithmetic worked.

6. In absolute terms, $P_2 = 1.512 \times 10^{5} + 1.013 \times 10^{5} = 2.525 \times 10^{5}$ Pa.

7. Where did the pressure go? $32$ kPa went into speeding the water up and $58.8$ kPa into lifting it. Climbing cost nearly twice as much as accelerating, which is typical: a few metres of height beats a few metres per second of speed almost every time.

Final Answer: $v_2 = 8.0$ m/s and $P_2 = 1.51 \times 10^{5}$ Pa gauge, that is $2.53 \times 10^{5}$ Pa absolute.

Takeaway: Write all three Bernoulli terms out separately at both ends and check the sums match. It costs one line and catches every sign error you are ever going to make in this chapter.

Example 27: The water tower and the tap

A water tower holds its free surface 18.0 m above a tap on the ground floor. The supply pipe is wide compared with the tap. Taking $\rho = 1000$ kg/m$^3$, $g = 9.8$ m/s$^2$ and $P_a = 1.013 \times 10^{5}$ Pa, find (a) the speed of the water leaving the fully open tap, and (b) the gauge pressure at the tap when it is closed.

Solution:

1. (a) Apply Bernoulli from the tower's free surface to the open tap. Both points are exposed to the atmosphere, so with gauge pressures both $P$ terms are zero. The tower's surface is enormous compared with the tap, so the water there is essentially at rest. $$0 + 0 + \rho g h = 0 + \frac{1}{2}\rho v^{2} + 0$$

2. Everything cancels except the speed. $$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 18.0} = \sqrt{352.8} = 18.8 \text{ m/s}$$ which is the same speed the water would have reached by falling freely through 18 m — Torricelli's result, and a straight statement of energy conservation.

3. (b) The tap closed. Nothing is moving now, so every $\frac{1}{2}\rho v^{2}$ term vanishes and Bernoulli collapses back into hydrostatics: $$P_{\text{gauge}} = \rho g h = 1000 \times 9.8 \times 18.0 = 1.764 \times 10^{5} \text{ Pa}$$ $$P_{\text{abs}} = 1.764 \times 10^{5} + 1.013 \times 10^{5} = 2.777 \times 10^{5} \text{ Pa}$$

4. Compare the two situations. Closed, the tap feels 176 kPa gauge. Open, it feels zero gauge and the whole of that pressure has turned into $\frac{1}{2}\rho v^{2} = 500 \times 352.8 = 1.764 \times 10^{5}$ Pa of dynamic pressure. The energy did not go anywhere — it changed form.

5. Real taps are slower than this, typically by a factor of two or three, because a real pipe has viscous resistance which Bernoulli ignores. The 18.8 m/s is the ideal ceiling, not a prediction.

Final Answer: (a) $18.8$ m/s; (b) $1.76 \times 10^{5}$ Pa gauge, that is $2.78 \times 10^{5}$ Pa absolute.

Takeaway: Static head and dynamic pressure are the same $\rho g h$ wearing different clothes. Close the tap and it is all pressure; open it and it is all speed.

Example 28: A Venturi meter read in mercury

A Venturi meter in a horizontal water main has a bore of cross-section 20 cm$^2$ narrowing to a throat of 8.0 cm$^2$. A mercury manometer connected across it shows a difference of 3.0 cm. Taking $\rho_{\text{water}} = 1000$ kg/m$^3$, $\rho_{\text{Hg}} = 13600$ kg/m$^3$ and $g = 9.8$ m/s$^2$, find the speed in the main and the volume flow rate.

Solution:

1. First convert the manometer reading into a pressure difference. The mercury column is pushed up on one side and down on the other, and the water sitting above it partly cancels the effect, so the effective density is the difference: $$P_1 - P_2 = \left(\rho_{\text{Hg}} - \rho_{\text{water}}\right)g\,h_m$$ $$P_1 - P_2 = (13\,600 - 1000)\times 9.8 \times 0.030 = 12\,600 \times 0.294 = 3.70 \times 10^{3} \text{ Pa}$$

2. Continuity relates the two speeds. $$v_2 = v_1\frac{A_1}{A_2} = v_1 \times \frac{20}{8.0} = 2.5\,v_1$$

3. Bernoulli, horizontal, so the height terms cancel. $$P_1 - P_2 = \frac{1}{2}\rho\left(v_2^{2} - v_1^{2}\right) = \frac{1}{2}\rho v_1^{2}\left(\left(\frac{A_1}{A_2}\right)^{2} - 1\right)$$

4. Solve for $v_1$. $$v_1 = \sqrt{\frac{2\left(P_1 - P_2\right)}{\rho\left(\left(\frac{A_1}{A_2}\right)^{2} - 1\right)}} = \sqrt{\frac{2 \times 3704}{1000\left(6.25 - 1\right)}} = \sqrt{\frac{7408}{5250}}$$ $$v_1 = \sqrt{1.411} = 1.19 \text{ m/s}$$ and hence $v_2 = 2.5 \times 1.188 = 2.97$ m/s.

5. The flow rate. $$Q = A_1v_1 = 20 \times 10^{-4} \times 1.188 = 2.38 \times 10^{-3} \text{ m}^3\text{/s}$$ that is about $2.4$ litres per second, or 143 litres a minute.

6. Note what was measured and what was not. The manometer gives only the difference $P_1 - P_2$, which is the same number in gauge or in absolute terms. A Venturi meter never tells you the actual pressure in the main — only the drop across the throat, which is all it needs.

Final Answer: $v_1 = 1.19$ m/s in the main and $v_2 = 2.97$ m/s at the throat, giving $Q = 2.4 \times 10^{-3}$ m$^3$/s. The pressure drop is $3.70 \times 10^{3}$ Pa, a difference and so identical in gauge and absolute.

Takeaway: Use $\left(\rho_{\text{manometer}} - \rho_{\text{pipe fluid}}\right)$, not $\rho_{\text{manometer}}$ alone. Forgetting to subtract the water overstates the pressure drop by about $8\%$ here, and by far more with a lighter manometer liquid.

Part 5: Efflux, Dynamic Lift and Reaction Thrust

Bernoulli put to work. Everything here follows from the single line $P + \frac{1}{2}\rho v^{2} + \rho g h =$ constant.

Key Point — Torricelli's law: $$v = \sqrt{2gh}$$ where $h$ is the depth of the hole below the free surface, not the height of the hole above the ground. The speed of efflux is the speed a body would reach falling freely through $h$ — it does not depend on the density of the liquid at all.

Example 29: Where the jet from the drum lands

An open drum stands on level ground and is filled with water to a depth of 1.25 m. A small hole is opened in its side at a point 0.45 m below the water surface. Taking $g = 9.8$ m/s$^2$, find the speed of the jet, how long it is in the air, and how far from the drum it lands.

Solution:

1. The speed, from Torricelli. The depth below the free surface is $h = 0.45$ m: $$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.45} = \sqrt{8.82} = 2.97 \text{ m/s}$$

2. The height of the hole above the ground is what is left of the depth: $$y = H - h = 1.25 - 0.45 = 0.80 \text{ m}$$

3. Now it is projectile motion. The jet leaves horizontally, so its vertical motion starts from rest: $$y = \frac{1}{2}gt^{2} \qquad \Longrightarrow \qquad t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 0.80}{9.8}} = \sqrt{0.1633} = 0.404 \text{ s}$$

4. The horizontal range. $$R = v\,t = 2.9698 \times 0.40406 = 1.20 \text{ m}$$

5. The compact form, worth memorising. Substituting $v = \sqrt{2gh}$ and $t = \sqrt{\frac{2(H-h)}{g}}$, $$R = \sqrt{2gh}\times\sqrt{\frac{2(H-h)}{g}} = 2\sqrt{h(H-h)}$$ Check: $2\sqrt{0.45 \times 0.80} = 2\sqrt{0.36} = 2 \times 0.60 = 1.20$ m. The same answer, and notice that $g$ has vanished — the range of the jet is the same on the Moon.

6. A note on pressure. Just outside the hole the jet is a free stream open to the air, so its pressure is atmospheric — that is, zero gauge. Inside the drum at the same level the gauge pressure is $\rho g h = 4410$ Pa. That difference is exactly what accelerated the water.

Final Answer: the jet leaves at $2.97$ m/s, stays in the air for $0.404$ s and lands $1.20$ m from the base of the drum.

Takeaway: Two different heights appear in this problem and you must keep them apart. The depth below the surface sets the speed; the height above the ground sets the time of flight. Mixing them is the standard error.

Example 30: Two holes with the same range

A tank is filled with water to a depth of 1.80 m. A small hole is made 0.40 m below the surface. (a) Where must a second hole be made so that its jet has the same horizontal range? (b) What is that common range? (c) Where should a hole be to get the greatest possible range, and what is it?

Solution:

1. Start from the range formula derived above: $$R = 2\sqrt{h(H-h)}$$ with $H = 1.80$ m.

2. (a) The symmetry is immediate. The expression $h(H-h)$ is unchanged if you swap $h$ for $H-h$. So the companion hole is at $$h^{\prime} = H - h = 1.80 - 0.40 = 1.40 \text{ m below the surface}$$ The first hole is 0.40 m down and 1.40 m up from the base; the second is 1.40 m down and 0.40 m up. They are mirror images about the mid-depth.

3. (b) The common range. $$R = 2\sqrt{0.40 \times 1.40} = 2\sqrt{0.560} = 2 \times 0.7483 = 1.50 \text{ m}$$

4. (c) The best hole. Maximise $h(H-h)$. For a fixed sum $h + (H-h) = H$, the product is greatest when the two are equal, so $$h = \frac{H}{2} = 0.90 \text{ m}$$ $$R_{\max} = 2\sqrt{\frac{H}{2}\times\frac{H}{2}} = 2 \times \frac{H}{2} = H = 1.80 \text{ m}$$

5. A result worth carrying. The greatest range a tank can throw equals its own depth of water, and it comes from a hole exactly halfway down. Neither the tank's width nor the liquid's density enters.

Final Answer: (a) 1.40 m below the surface; (b) a common range of $1.50$ m; (c) the mid-depth hole at 0.90 m gives $R_{\max} = 1.80$ m, equal to $H$.

Takeaway: $h$ and $H - h$ always give the same range, and the mid-point always gives the maximum $R_{\max} = H$. Two facts, no calculation, and they answer a great many objective questions on sight.

Example 31: How long the drum takes to empty

A cylindrical drum of cross-sectional area 0.45 m$^2$ has a hole of area 3.0 cm$^2$ in its base and is filled with water to a depth of 1.6 m. Taking $g = 9.8$ m/s$^2$, find how long it takes to empty, and what fraction of that time is spent draining the upper half of the water.

Solution:

1. The level falls slowly, so treat the flow as quasi-steady. When the water stands at depth $y$, the efflux speed is $\sqrt{2gy}$, so the volume leaving per second is $a\sqrt{2gy}$, and that must equal the rate at which the drum's contents shrink: $$-A\frac{dy}{dt} = a\sqrt{2gy}$$

2. Separate and integrate from $H$ down to 0. $$t = \frac{A}{a}\int_{0}^{H}\frac{dy}{\sqrt{2gy}} = \frac{A}{a}\cdot\frac{1}{\sqrt{2g}}\left[2\sqrt{y}\right]_{0}^{H} = \frac{A}{a}\sqrt{\frac{2H}{g}}$$

3. Substitute. $$\frac{A}{a} = \frac{0.45}{3.0 \times 10^{-4}} = 1500$$ $$\sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 1.6}{9.8}} = \sqrt{0.3265} = 0.5714 \text{ s}$$ $$t = 1500 \times 0.5714 = 857 \text{ s} \approx 14.3 \text{ minutes}$$

4. How good is the quasi-steady assumption? It ignores the fact that the surface itself is falling, which adds a $\frac{1}{2}\rho v_{\text{surface}}^{2}$ term to Bernoulli. That term is smaller than the efflux term by the factor $\left(\frac{a}{A}\right)^{2} = \left(6.7 \times 10^{-4}\right)^{2}$, about four parts in ten million. Integrating the falling level numerically, without the approximation, changes the answer by about one part in a million — so here the simple formula is not merely adequate, it is exact for every practical purpose. It would fail badly if the hole were comparable in size to the tank.

5. The upper half. Time to fall from $H$ to $\frac{H}{2}$ is $t$ evaluated between those limits: $$t_{\text{upper}} = \frac{A}{a}\sqrt{\frac{2}{g}}\left(\sqrt{H} - \sqrt{H/2}\right) = t\left(1 - \frac{1}{\sqrt{2}}\right) = 0.293\,t$$

6. So the top half drains in $29.3\%$ of the time and the bottom half takes $70.7\%$. The drum starts fast and finishes slowly, because the head driving the flow is shrinking. In seconds: 251 s for the upper half, 606 s for the lower.

Final Answer: about $857$ s, or $14.3$ minutes, of which the upper half takes $29.3\%$ — about $251$ s.

Takeaway: $t = \frac{A}{a}\sqrt{\frac{2H}{g}}$ — note the square root, so quadrupling the depth only doubles the time. And the emptying is front-loaded: the last half of the water takes more than twice as long as the first.

Example 32: Lift on a model wing in a tunnel

In a wind-tunnel test on a model aeroplane, the air flows over the upper surface of a wing at 70 m/s and along the lower surface at 63 m/s. The wing area is 2.5 m$^2$. Taking the density of the air as 1.3 kg/m$^3$, find the lift on the wing.

Solution:

1. Apply Bernoulli across the wing. The two surfaces are at essentially the same height, so the $\rho g h$ terms cancel and $$P_{\text{lower}} - P_{\text{upper}} = \frac{1}{2}\rho\left(v_{\text{upper}}^{2} - v_{\text{lower}}^{2}\right)$$

2. Do the squares first, and note the difference of squares: $$v_{\text{upper}}^{2} - v_{\text{lower}}^{2} = 70^{2} - 63^{2} = 4900 - 3969 = 931 \text{ m}^2/\text{s}^2$$ (Or as $(70+63)(70-63) = 133 \times 7 = 931$, which is faster and less error-prone.)

3. The pressure difference. $$\Delta P = \frac{1}{2}\times 1.3 \times 931 = 0.65 \times 931 = 605 \text{ Pa}$$

4. The lift. $$F = \Delta P \times A = 605.15 \times 2.5 = 1.51 \times 10^{3} \text{ N}$$

5. What that lift can hold up. $$m = \frac{F}{g} = \frac{1512.9}{9.8} = 154 \text{ kg}$$ comfortably enough for a model, and a useful sanity check that the answer is the right size.

6. Note the pressures involved. $\Delta P = 605$ Pa is a difference, so gauge and absolute give the same number. It is only $0.6\%$ of an atmosphere — the whole of flight rests on a pressure difference smaller than the change you feel going up in a lift.

Final Answer: the pressure difference is $605$ Pa and the lift is $1.5 \times 10^{3}$ N, enough to support about $154$ kg.

Takeaway: Lift is $\frac{1}{2}\rho\left(v_1^{2}-v_2^{2}\right)A$, and the difference of squares is where the arithmetic goes wrong. Factorise it as $(v_1+v_2)(v_1-v_2)$ and the slip cannot happen.

Example 33: The kick a leaking tank feels

A tank of water sitting on a frictionless trolley springs a leak through a hole of area 2.5 cm$^2$ in its vertical wall, 1.20 m below the water surface. Taking $\rho = 1000$ kg/m$^3$ and $g = 9.8$ m/s$^2$, find the backward thrust on the tank.

Solution:

1. Why there is a thrust at all. The tank is throwing water out one side. By Newton's third law, the water throws the tank the other way — exactly as a rocket works. The thrust is the rate at which the escaping water carries momentum away.

2. The efflux speed. $$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.20} = \sqrt{23.52} = 4.85 \text{ m/s}$$

3. The mass leaving per second. $$\frac{dm}{dt} = \rho\,a\,v = 1000 \times 2.5 \times 10^{-4} \times 4.8497 = 1.21 \text{ kg/s}$$

4. The thrust is the momentum carried off per second. $$F = \frac{dm}{dt}\times v = \rho a v^{2} = 1.2124 \times 4.8497 = 5.88 \text{ N}$$

5. The elegant form. Since $v^{2} = 2gh$, $$F = \rho a v^{2} = 2\rho g h\,a = 2 \times 1000 \times 9.8 \times 1.20 \times 2.5 \times 10^{-4} = 5.88 \text{ N}$$

6. Read that factor of 2, because it is the point of the problem. The gauge pressure at the hole is $\rho g h = 11\,760$ Pa, and $\rho g h \times a$ would be only $2.94$ N. The thrust is twice that. The reason is that the pressure is not $\rho g h$ everywhere on the wall opposite the hole — the missing patch of wall unbalances the pressure distribution and the water accelerating towards the hole contributes as well. The momentum argument gets it right without needing any of that detail.

Final Answer: a backward thrust of $5.88$ N, which is $2\rho g h a$ — twice the naive $\rho g h a$.

Takeaway: Use momentum, not pressure, for a reaction thrust: $F = \rho a v^{2}$. The pressure route needs the whole pressure distribution on the tank and is far harder to get right.

Example 34: Fingers over the tap

Water leaves a tap of cross-sectional area 1.8 cm$^2$ at 0.90 m/s. You clamp your fingers over the mouth so that only $15\%$ of the opening is left clear, without touching the tap handle. Find the speed of the water now, and the dynamic pressure of the jet. Take $\rho = 1000$ kg/m$^3$.

Solution:

1. The tap setting has not changed, so to a good approximation the volume delivered per second is unchanged: $$Q = A_1v_1 = 1.8 \times 10^{-4} \times 0.90 = 1.62 \times 10^{-4} \text{ m}^3\text{/s}$$

2. The remaining opening. $$A_2 = 0.15 \times 1.8 \times 10^{-4} = 2.7 \times 10^{-5} \text{ m}^2$$

3. Continuity gives the new speed. $$v_2 = \frac{Q}{A_2} = \frac{1.62 \times 10^{-4}}{2.7 \times 10^{-5}} = 6.0 \text{ m/s}$$ That is $\frac{1}{0.15} = 6.67$ times the original speed.

4. The dynamic pressure of the jet. $$\frac{1}{2}\rho v_2^{2} = 500 \times 36 = 1.8 \times 10^{4} \text{ Pa}$$ about $18\%$ of an atmosphere, which is why the jet stings when it hits your hand.

5. Why this is the answer to "why does water gush between my fingers?" Continuity, not pressure. The same volume per second forced through a smaller opening must move faster, and the speed goes up as the reciprocal of the area. Bernoulli then tells you the pressure in the jet has fallen to atmospheric — the pressure energy has become kinetic energy.

6. The caveat worth stating. In reality the tap does not deliver quite the same $Q$ once you block it, because the extra resistance slows the whole supply a little. The continuity answer is the ideal upper bound.

Final Answer: the jet speeds up to $6.0$ m/s, with a dynamic pressure of $1.8 \times 10^{4}$ Pa.

Takeaway: The finger-over-the-hose effect is conservation of mass, not a pressure trick. Squeeze the area to a fraction $f$ and the speed multiplies by $\frac{1}{f}$ — nothing else changes.

Example 35: Three more explanations, all from one line of Bernoulli

Give the physical reason for each of the following.

(a) To hold a sheet of paper horizontal, you should blow over it, not under it. (b) A doctor controls the flow rate of an injection far better by choosing the needle's bore than by pushing harder on the plunger. (c) A spinning cricket ball does not follow a parabola.

Solution:

1. (a) Blowing over the top makes the air there move fast. Along a streamline, faster air means lower pressure: $$P + \frac{1}{2}\rho v^{2} = \text{constant}$$ The still air underneath stays at atmospheric, so there is now a net upward pressure difference holding the sheet up. Blow underneath instead and you lower the pressure below it, and the atmosphere above pushes the paper down. Numerically, air blown at 12 m/s gives $\frac{1}{2}\times 1.2 \times 144 = 86$ Pa — over a strip of area 0.015 m$^2$ that is $1.3$ N, about a hundred times the strip's own weight of roughly 0.012 N.

2. (b) The flow through a fine tube is governed by Poiseuille's law: $$Q = \frac{\pi P r^{4}}{8\eta L}$$ $Q$ is only linear in the driving pressure $P$ but goes as the fourth power of the radius. Doubling the thumb pressure doubles the flow; halving the bore cuts it to a sixteenth. The needle gauge is a vastly stronger lever than the thumb, and it is also far more repeatable than a person's push.

3. (c) A spinning ball drags a thin layer of air around with it. On the side where the surface is moving with the airstream the relative speed is high; on the other side the surface moves against the stream and the relative speed is low. Bernoulli then puts a lower pressure on the fast side, so there is a sideways force perpendicular to the ball's velocity — the Magnus effect. A parabola requires a constant force (gravity alone); adding a lateral force that keeps turning with the ball's motion gives a curved, swerving path instead.

Final Answer: (a) fast air above means low pressure above, so the sheet is pushed up; (b) $Q \propto r^{4}$ but only $Q \propto P$; (c) the Magnus effect adds a lateral force, so the net force is no longer constant and the path is not parabolic.

[JEE Tip] Every one of these three is the same sentence in disguise: where a fluid moves faster, its pressure is lower. If you can say that sentence and name what is moving fast, you have the mark.

Takeaway: Name the mechanism and quote the relation. "Bernoulli" alone is worth little; "faster air above the paper, so lower pressure above, so a net upward push" is worth full marks.