Manometers, barometers: Worked examples — Physics deep dive

Before anything, let us pin down the letters we keep reusing, in plain words:

Two more words we will lean on constantly — meet them now so no example surprises you:

The scenario matrix

Every manometer/barometer problem lives in one of these cells. The examples below each fill one (or more) cells so no gap is left.

#	Cell (what makes it different)	Which example
A	Barometer, vacuum on top → absolute $P_{a t m}$	Ex 1
B	Barometer with a non-mercury fluid (limiting height)	Ex 2
C	Barometer with trapped gas (not a perfect vacuum)	Ex 3
D	Manometer, open arm higher → gas above atmospheric (+ sign)	Ex 4
E	Manometer, gas arm higher → gas below atmospheric (− sign)	Ex 5
F	Degenerate: levels equal, $h = 0$ → gauge pressure zero	Ex 6
G	Inclined manometer → $h = L sin θ$ (vertical only)	Ex 7
H	Two immiscible fluids stacked → add each $ρ g h$ separately	Ex 8
I	Real-world word problem (blood-pressure / diver style)	Ex 9
J	Exam twist: change $g$ (another planet) — height rescales	Ex 10

Recall The two rules everything below uses

Rule 1 — walk a fluid path: down adds $ρ g h$ , up subtracts $ρ g h$ . Rule 2 — same connected fluid, same horizontal level ⇒ equal pressure.

A · The plain mercury barometer (absolute)

Worked example Ex 1 — Standard mercury barometer

A mercury barometer reads a column height $h = 0.760 m$ . Vacuum sits above the mercury. Find $P_{a t m}$ . Take $ρ_{H g} = 13600 kg/m^{3}$ , $g = 9.8$ .

Forecast: guess the pressure before reading on — is it near $1 0^{5} Pa$ ?

Pick two points at the same level: point A at the mercury surface inside the tube's base, point B at the open dish surface. Look at Figure s01 — A (mint dot) and B (butter dot) sit on the same dashed level. Why this step? Rule 2 lets us equate them only if they share a horizontal level.
Write pressure at B (open): $P_{B} = P_{a t m}$ . Why? B is exposed to the air, so the air's pressure sits on it directly.
Write pressure at A: climb the column from A up to the vacuum top (the coral label in the figure). Going up subtracts, so top pressure $= P_{A} - ρ g h$ . The top is a vacuum $= 0$ , hence $P_{A} = ρ g h$ . Why? Rule 1 along the vertical tube; the top is empty ⇒ zero.
Set $P_{A} = P_{B}$ : $ρ g h = P_{a t m}$ . Why can we equate them? By Rule 2: A (inside the tube's base) and B (open dish surface) are in the same connected mercury at the same horizontal level, so they must share one pressure — otherwise fluid would flow sideways from the higher-pressure point until they matched.

$P_{a t m} = 13600 \times 9.8 \times 0.760 = 1.013 \times 1 0^{5} Pa$

Verify: $760 mmHg = 1 atm$ by definition, and $1 atm = 1.013 \times 1 0^{5} Pa$ . ✓ Units: $\frac{kg}{m ^{3}} \cdot \frac{m}{s ^{2}} \cdot m = Pa$ . ✓

The figure above shows exactly this: the coral double-arrow is the column height $h$ , the mercury (lavender) fills the tube up to the vacuum, and the two dots mark the equal-level points A and B whose pressures we set equal.

B · Same pressure, a lighter fluid (limiting height)

Figure s02 puts the two columns side by side to scale: the short lavender mercury bar ( $0.76 m$ ) versus the towering mint water bar ( $10.34 m$ ) — both balancing the same $P_{a t m}$ . The picture is the argument: lighter fluid ⇒ taller column.

C · Barometer with trapped gas on top (not a vacuum)

Worked example Ex 3 — Imperfect vacuum

A mercury barometer has a small amount of trapped air at the top exerting $P_{t o p} = 2000 Pa$ . The mercury column is $h = 0.745 m$ . What is the true $P_{a t m}$ ?

Forecast: the trapped gas pushes the mercury down, so the column reads low. Should the true $P_{a t m}$ be bigger or smaller than $ρ g h$ alone?

Reuse the same-level trick: just like Figure s01, take the dish surface and the in-tube surface at one level, then climb the tube; but now the top is $P_{t o p}$ , not $0$ . Why this step? Rule 2 still guarantees the two same-level points share a pressure; the only change from Ex 1 is that the roof of the column is a real gas, not a vacuum, so its pressure enters the climb.
Balance: $P_{a t m} = P_{t o p} + ρ g h$ . Why this step? Going up the column subtracts $ρ g h$ to reach $P_{t o p}$ , so reversing it, $P_{a t m} = P_{t o p} + ρ g h$ . $P_{a t m} = 2000 + 13600 \times 9.8 \times 0.745 = 2000 + 99287 = 1.013 \times 1 0^{5} Pa$

Verify: if $P_{t o p} = 0$ we'd need $h = 0.760$ ; the trapped gas let the column drop to $0.745$ while still meaning the same $P_{a t m}$ . Consistent — the gas "helped" hold the atmosphere. ✓

D · Manometer, open arm higher → gas ABOVE atmospheric

Worked example Ex 4 — Pressurised gas (+ sign)

An open-tube mercury manometer connects to a gas vessel. The open arm liquid is $h = 0.05 m$ higher than the gas-arm liquid. $P_{a t m} = 1.01 \times 1 0^{5} Pa$ . Find absolute and gauge pressure of the gas.

Forecast: if the gas can push the open side up higher, is the gas stronger or weaker than the air?

Reference line = the lower liquid surface (gas side). In Figure s03 this is the mint dashed line; $P_{l in e}$ is the pressure there, and it is the same computed from either arm.
Right (open) arm, from its surface down to the line: $P_{l in e} = P_{a t m} + ρ g h$ . Why? Going down by $h$ (the coral arrow in the figure) from the open surface adds $ρ g h$ .
Left (gas) arm, from gas down to line: $P_{l in e} = P_{g a s}$ (its surface is the line).
Equate the two expressions for $P_{l in e}$ : $P_{g a s} = P_{a t m} + ρ g h$ . $P_{ab s} = 1.01 \times 1 0^{5} + 13600 \times 9.8 \times 0.05 = 1.01 \times 1 0^{5} + 6664 = 1.077 \times 1 0^{5} Pa$

Gauge: $P_{g a ug e} = P_{ab s} - P_{a t m} = 6664 Pa$ .

Verify: open arm higher ⇒ gas overcomes air ⇒ $P_{g a s} > P_{a t m}$ ⇒ positive gauge. Sign is $+$ . ✓

In Figure s03 the right (open) liquid stands taller than the left (gas) liquid; the coral arrow marks the height gap $h$ , and the mint dashed reference line is where we equate $P_{l in e}$ from both arms. The taller open side is the visual signature of a gas above atmospheric.

E · Manometer, gas arm higher → gas BELOW atmospheric

Worked example Ex 5 — Partial vacuum (− sign)

Same manometer, but now the gas arm liquid is $0.05 m$ higher (the atmosphere pushed the open side down). $P_{a t m} = 1.01 \times 1 0^{5}$ . Find the gas pressure.

Forecast: the air won this tug-of-war. Is the gas above or below atmospheric?

Reference line = lower surface (now the open side); $P_{l in e}$ is its pressure. Picture Figure s03 mirror-flipped — the gas arm is now the taller one.
Left (gas) arm down to line: $P_{l in e} = P_{g a s} + ρ g h$ .
Right (open) arm down to line: $P_{l in e} = P_{a t m}$ (open surface is the line).
Equate the two expressions for $P_{l in e}$ : $P_{g a s} + ρ g h = P_{a t m} \Rightarrow P_{g a s} = P_{a t m} - ρ g h$ . $P_{ab s} = 1.01 \times 1 0^{5} - 6664 = 0.943 \times 1 0^{5} Pa$

Gauge: $- 6664 Pa$ (negative ⇒ below atmospheric).

Verify: gas arm higher ⇒ gas weaker than air ⇒ $P_{g a s} < P_{a t m}$ ⇒ negative gauge. Sign is $-$ . Mirror image of Ex 4. ✓

F · The degenerate case — levels equal

G · Inclined manometer — vertical height only

Worked example Ex 7 — Slanted tube

A manometer's open arm is a tube inclined at $θ = 3 0^{\circ}$ from horizontal. The liquid moves a slant length $L = 0.10 m$ along the tube. The fluid is oil, $ρ = 850 kg/m^{3}$ . What gauge pressure does this represent?

Forecast: we travel $0.10 m$ of tube — but does all of that count as height?

Find the vertical rise. Only the up-down part fights gravity, so $h = L sin θ$ . In Figure s04 the lavender slant is $L$ , the coral vertical side is $h$ , and the dashed horizontal side carries no weight. Why $sin$ and not $cos$ ? $θ$ is measured from horizontal; the vertical side of the right triangle (opposite $θ$ ) is $L sin θ$ . The horizontal side $L cos θ$ carries no weight against gravity, so it drops out. $h = 0.10 \times sin 3 0^{\circ} = 0.10 \times 0.5 = 0.05 m$
Now use $P_{g a ug e} = ρ g h$ : $P_{g a ug e} = 850 \times 9.8 \times 0.05 = 416.5 Pa$

Verify: if we'd wrongly used the slant $L = 0.10$ we'd double the answer to $833 Pa$ — the classic inclined-tube mistake. The whole point of inclining a tube is to spread a tiny vertical rise over a long, easy-to-read slant. ✓

Figure s04 is the right triangle behind $h = L sin θ$ : the lavender hypotenuse is the slant length the liquid actually travels, but only the coral vertical leg presses against gravity. Read the picture and the $sin$ is obvious.

H · Two stacked immiscible fluids

Worked example Ex 8 — Water on top of mercury

A U-tube has mercury at the bottom and a $0.20 m$ column of water ( $ρ_{w} = 1000$ ) poured on top of the mercury in the left arm. The right arm is open. The mercury surfaces differ by $h_{H g} = 0.014 m$ (right higher). What is the pressure $P_{X}$ pushing down on the water's top surface (left)?

Forecast: two different fluids ⇒ two different $ρ g h$ terms. Do we add them?

Reference line = the lower mercury surface (left arm base); $P_{l in e}$ is the pressure there.
Left arm, top down to line: start at $P_{X}$ , go down through $0.20 m$ of water, then down to the line. So $P_{l in e} = P_{X} + ρ_{w} g (0.20)$ . Why split it? Each fluid contributes its own $ρ g h$ ; you cannot use one density for both.
Right (open) arm down to line: $P_{l in e} = P_{a t m} + ρ_{H g} g h_{H g}$ .
Equate the two expressions for $P_{l in e}$ : by Rule 2 the left-arm and right-arm computations of $P_{l in e}$ must be equal, so $P_{X} + ρ_{w} g (0.20) = P_{a t m} + ρ_{H g} g (0.014) .$
Solve for $P_{X}$ — subtract $ρ_{w} g (0.20)$ from both sides: $P_{X} = P_{a t m} + ρ_{H g} g (0.014) - ρ_{w} g (0.20) .$
Plug numbers: $ρ_{H g} g (0.014) = 13600 (9.8) (0.014) = 1866.6 Pa$ and $ρ_{w} g (0.20) = 1000 (9.8) (0.20) = 1960 Pa$ , so $P_{X} = P_{a t m} + 1866.6 - 1960 = P_{a t m} - 93.4 Pa .$

Verify (units & sign): each term is $ρ g h$ ⇒ pascals. ✓ The heavy mercury difference nearly balances the light water column, leaving $P_{X}$ just barely below atmospheric. Adding each fluid separately is the key skill. ✓

I · Real-world word problem

Worked example Ex 9 — Diver's ear

A swimmer descends to depth $h = 3.0 m$ in fresh water ( $ρ = 1000$ ). By how much does the pressure on her eardrum exceed the surface pressure? Express in $Pa$ and as a fraction of $P_{a t m} = 1.013 \times 1 0^{5}$ .

Forecast: the parent's Feynman story — ears hurt deeper. Guess: more or less than one atmosphere at 3 m?

Gauge pressure at depth is just $ρ g h$ (surface itself is the reference where extra $= 0$ ). $Δ P = 1000 \times 9.8 \times 3.0 = 29400 Pa$ Why this step? The water surface is at atmospheric; every metre down adds $ρ g h$ of extra push, and the extra above the surface is exactly the gauge pressure.
As a fraction of atmosphere: $\frac{29400}{1.013 \times 1 0 ^{5}} = 0.29$ .

Verify: $\approx 29%$ of an atmosphere from only 3 m of water — this is why ears pop fast when diving. It also matches Buoyancy and Archimedes Principle, which rests on this same depth-pressure growth. ✓

J · Exam twist — another planet

Connections

Parent topic — Manometers & Barometers
Hydrostatic Pressure — the $ρ g h$ engine behind every example.
Pascal's Law — the "same level, same pressure" reference-line trick.
Atmospheric Pressure — what Ex 1–3 measure.
Buoyancy and Archimedes Principle — same depth-pressure idea (Ex 9).
Density and Specific Gravity — why mercury vs water vs oil (Ex 2, 7, 8).
Bernoulli's Equation — extends pressure to moving fluids.