Manometers, barometers: Exercises — Physics deep dive

Before we start, the symbols we lean on constantly:

Definition What

$P$ , $Δ P$ , and the three letters in $ρ g h$ mean

$P$ = pressure = the force the fluid pushes with on each square metre of surface. Units: pascals (Pa), where $1 Pa = 1 N/m^{2}$ .
$Δ P$ (read "delta P") = a change (difference) in pressure between two points — the $Δ$ symbol always means "the change in" whatever follows it. Here it is the pressure at the lower point minus the pressure at the higher point.
$ρ$ (Greek "rho") = density = how many kilograms of fluid fit in one cubic metre. Units $kg/m^{3}$ .
$g$ = strength of gravity = $9.8 m/s^{2}$ . It converts mass into weight (a downward pull).
$h$ = the vertical height difference between the two points in the fluid, in metres. Multiply the last three and you get $ρ g h$ = the extra pressure gained by descending a vertical distance $h$ . See the picture below for what "vertical only" means.

Figure 1 (description): a fluid tube slanted from lower-left to upper-right. The black hypotenuse is labelled "tube length $L$ "; a red vertical segment on the right is labelled "vertical $h$ (this is what counts)"; a dashed horizontal base is labelled "horizontal (adds no pressure)". An arc marks the incline angle $θ$ . The message: whether the tube is fat, thin, or slanted, only the vertical drop (the red line) enters $ρ g h$ .

Level 1 — Recognition

Can you pick the right formula and read a height off a diagram?

L1.1

A mercury barometer stands with a column height $h = 0.76 m$ . What atmospheric pressure does it report?

Recall Solution

WHAT device: a barometer has a vacuum on top, so the column balances the full atmospheric push. That means it reads absolute pressure directly. WHY this formula: with vacuum ( $P \approx 0$ ) above and $P_{a t m}$ pressing at the dish, matching pressures at the dish level gives $P_{a t m} = 0 + ρ_{H g} g h$ . $P_{a t m} = ρ_{H g} g h = 13600 \times 9.8 \times 0.76 = 1.013 \times 1 0^{5} Pa$ That is exactly $760$ mmHg $= 1$ atm. ✓

L1.2

In an open-tube mercury manometer connected to a gas, the open arm's mercury surface sits $h = 0.12 m$ higher than the gas arm's surface. Does the gas pressure exceed or fall short of atmospheric, and by how much (gauge)?

Recall Solution

WHAT the height tells us: the open arm being higher means the gas has pushed its side down and shoved the other side up. So the gas is pushing harder than the atmosphere. WHY add: gas exceeds atmosphere, so $P_{g a s} = P_{a t m} + ρ g h$ ; the gauge part (the excess over atmosphere) is the $ρ g h$ term. $P_{g a ug e} = ρ_{H g} g h = 13600 \times 9.8 \times 0.12 = 15993.6 Pa \approx 1.60 \times 1 0^{4} Pa$ The gas is above atmospheric by about $1.6 \times 1 0^{4}$ Pa.

Level 2 — Application

Plug numbers into $P = P_{a t m} \pm ρ g h$ with the right sign.

L2.1

An open-tube water manometer is attached to a gas vessel. The gas arm's water level is $0.35 m$ higher than the open arm's. Find the absolute gas pressure.

Recall Solution

WHAT the geometry says: the gas arm is higher, so the atmosphere has pushed the gas side down less than the gas pushes back — actually the gas is weaker than atmosphere, letting the open side fall lower. Gas below atmospheric ⇒ subtract. $P_{ab s} = P_{a t m} - ρ_{w a t er} g h = 1.013 \times 1 0^{5} - 1000 (9.8) (0.35)$ $= 1.013 \times 1 0^{5} - 3430 = 9.787 \times 1 0^{4} Pa$ Gauge pressure is $- 3430$ Pa (a slight vacuum).

L2.2

A barometer is taken up a mountain where the mercury column drops to $h = 0.65 m$ . What is the atmospheric pressure there, and what fraction of sea-level pressure is it?

Recall Solution

$P = ρ_{H g} g h = 13600 \times 9.8 \times 0.65 = 86632 Pa \approx 8.66 \times 1 0^{4} Pa$ Fraction of sea level: $\frac{86632}{1.013 \times 1 0 ^{5}} = 0.855 \approx 85.5%$ WHY it dropped: higher up, less air sits above you, so the atmosphere pushes with less force and can only hold up a shorter column.

L2.3

How tall would a water barometer need to be to read standard atmospheric pressure ( $1.013 \times 1 0^{5}$ Pa)?

Recall Solution

WHY invert the formula: we know $P$ and want $h$ , so solve $P = ρ g h$ for $h$ . $h = \frac{P}{ρ _{w a t er} g} = \frac{1.013 \times 1 0 ^{5}}{1000 \times 9.8} = 10.34 m$ This is why we use mercury: a $13.6 \times$ denser fluid needs a $13.6 \times$ shorter column ( $10.34/13.6 \approx 0.76$ m). ✓

Level 3 — Analysis

Now the tube is tilted, or two fluids stack — vertical height must be extracted.

Figure 2 (description): a right triangle. The black hypotenuse is the fluid path labelled " $L = 0.20$ m", with a red dot marking the mercury slug on it. The red vertical side is labelled " $h = L sin θ = 0.10$ m". The horizontal side is black, and a small square marks the right angle at the base. An arc labels the incline " $θ = 3 0^{\circ}$ ". The message: the fluid travels the hypotenuse $L$ , but only the vertical side $h$ feeds into $ρ g h$ .

L3.1

A manometer tube is inclined at $θ = 3 0^{\circ}$ from the horizontal. The mercury slug moves a length $L = 0.20 m$ along the tube. What vertical height $h$ does this represent, and what gauge pressure (in Pa) does it correspond to?

Recall Solution

WHY $sin$ , not the full length: only the vertical rise fights gravity. Looking at the right triangle (figure above), the tube length $L$ is the hypotenuse, and the vertical side is the "opposite" side to the angle $θ$ . So $h = L sin θ = 0.20 \times sin 3 0^{\circ} = 0.20 \times 0.5 = 0.10 m$ WHY inclined tubes are used: they magnify a small vertical change into a large, easy-to-read length along the tube — better sensitivity. $P_{g a ug e} = ρ_{H g} g h = 13600 \times 9.8 \times 0.10 = 13328 Pa \approx 1.33 \times 1 0^{4} Pa$

L3.2

A U-tube contains water. Oil ( $ρ_{o i l} = 800 kg/m^{3}$ ) is poured into the left arm and forms a column $0.10 m$ tall floating on the water. By how much does the water level in the right (open) arm rise above the water level in the left arm at equilibrium?

Recall Solution

WHAT balances: pick the reference line at the oil–water interface (the lowest point where both sides are pure water/continuous). Same fluid, same level on the right side of this line ⇒ equal pressure. Left arm at the interface: the oil column presses down: $P_{a t m} + ρ_{o i l} g (0.10)$ . Right arm at the same level: water rises a height $h_{w}$ above that interface level, giving $P_{a t m} + ρ_{w a t er} g h_{w}$ . Set equal: $ρ_{o i l} g (0.10) = ρ_{w a t er} g h_{w}$ $h_{w} = \frac{ρ _{o i l}}{ρ _{w a t er}} \times 0.10 = \frac{800}{1000} \times 0.10 = 0.08 m$ WHY less than 0.10 m: oil is lighter than water, so a $0.10$ m oil column weighs the same as only an $0.08$ m water column. The water surfaces differ by $0.08$ m.

Level 4 — Synthesis

Combine hydrostatics with a second principle: sealed gas, or a two-fluid barometer.

L4.1

A closed-tube manometer (vacuum sealed above the mercury on one arm) reads a mercury height difference of $h = 0.68 m$ , with the sealed arm higher. What absolute pressure of the connected gas does this indicate?

Recall Solution

WHAT is different from the open tube: the top of the sealed arm is a vacuum, $P \approx 0$ — just like a barometer. So the mercury difference balances the full gas pressure, giving absolute directly. WHY set the bottom level equal: the two arms are joined by connected mercury, so at the lowest common horizontal level the pressure must be identical on both sides (otherwise fluid would flow sideways until it wasn't). Working up from that shared level to each surface lets us equate the two arms. On the gas side, tracing from the gas down to the reference contributes nothing extra beyond $P_{g a s}$ at that level; on the sealed side, the reference pressure is the vacuum ( $0$ ) plus the weight of the taller mercury column, $ρ_{H g} g h$ . Equating the two: $P_{g a s} = 0 + ρ_{H g} g h = 13600 \times 9.8 \times 0.68 = 90630.4 Pa \approx 9.06 \times 1 0^{4} Pa$ This is an absolute reading — no $P_{a t m}$ to add, because the reference arm is vacuum, not atmosphere.

L4.2

A barometer tube accidentally has a tiny air pocket trapped above the mercury, exerting a pressure of $2000$ Pa. The true atmospheric pressure is $1.013 \times 1 0^{5}$ Pa. What mercury column height will this faulty barometer show, and what percent error in $P_{a t m}$ does the reading imply if you (wrongly) assume vacuum on top?

Recall Solution

WHAT the trapped air does: it pushes down on the mercury from above, so the column can't rise as high. Balance at the dish level: $P_{a t m} = P_{ai r} + ρ_{H g} g h$ $h = \frac{P _{a t m} - P _{ai r}}{ρ _{H g} g} = \frac{1.013 \times 1 0 ^{5} - 2000}{13600 \times 9.8} = \frac{99300}{133280} = 0.74504 m$ If you assume vacuum and read $P_{a t m} = ρ_{H g} g h = 133280 \times 0.74504 = 99300 Pa$ , you undershoot the true value. $error = \frac{1.013 \times 1 0 ^{5} - 99300}{1.013 \times 1 0 ^{5}} = \frac{2000}{101300} = 0.01974 = 1.97%$ WHY: the reading is short by exactly the trapped-air pressure ( $2000$ Pa), which is $\approx 1.97%$ of true $P_{a t m}$ .

Level 5 — Mastery

Everything at once — two immiscible fluids, sign bookkeeping, and a sanity limit.

L5.1

A U-tube open at both ends contains mercury. Water is poured into the left arm to a height of $0.40 m$ (measured from the mercury–water interface). Oil ( $ρ_{o i l} = 850 kg/m^{3}$ ) is poured into the right arm. If the mercury levels in the two arms end up equal, what height of oil column stands in the right arm?

Recall Solution

WHY equal mercury levels simplify everything: if the mercury surfaces are at the same height, then at that common level the mercury contributes equally on both sides and cancels. The pressures pushing down onto the mercury from each arm must therefore be equal. Left arm (water) pushing down at the mercury surface: $P_{a t m} + ρ_{w a t er} g (0.40)$ . Right arm (oil) pushing down at the mercury surface: $P_{a t m} + ρ_{o i l} g h_{o i l}$ . Set equal (the $P_{a t m}$ and $g$ cancel): $ρ_{w a t er} (0.40) = ρ_{o i l} h_{o i l}$ $h_{o i l} = \frac{ρ _{w a t er}}{ρ _{o i l}} \times 0.40 = \frac{1000}{850} \times 0.40 = 0.4706 m$ Sanity limit: oil is lighter than water, so we need a taller oil column ( $0.47 > 0.40$ m) to press with the same weight. ✓ If oil were as dense as water, we'd get exactly $0.40$ m — the formula agrees in that limit.

L5.2

A single manometer measures the pressure of a gas, but the connecting fluid is water while the U-tube's measuring fluid is mercury, stacked without mixing. The gas pushes a water column of height $0.30 m$ down onto the mercury, and the mercury shows a height difference of $0.05 m$ (open arm higher). Find the absolute gas pressure. (Take $ρ_{w a t er} = 1000$ , $ρ_{H g} = 13600$ .)

Recall Solution

WHAT to track: walk from the gas, down through the water column (pressure increases going down), across the bottom, then up the mercury difference on the open side to reach atmosphere. Start at gas: $P_{g a s}$ . Go down the $0.30$ m water: add $ρ_{w a t er} g (0.30)$ . Now go up the $0.05$ m mercury on the open side to reach the atmosphere: subtract $ρ_{H g} g (0.05)$ . End at $P_{a t m}$ : $P_{g a s} + ρ_{w a t er} g (0.30) - ρ_{H g} g (0.05) = P_{a t m}$ $P_{g a s} = P_{a t m} - ρ_{w a t er} g (0.30) + ρ_{H g} g (0.05)$ $= 1.013 \times 1 0^{5} - 1000 (9.8) (0.30) + 13600 (9.8) (0.05)$ $= 101300 - 2940 + 6664 = 105024 Pa \approx 1.050 \times 1 0^{5} Pa$ WHY signs flip: going down toward the reference adds pressure; going up away from it subtracts. Bookkeeping the direction of each leg is the whole skill.

Connections

Hydrostatic Pressure — the $ρ g h$ engine behind every problem here.
Pascal's Law — justifies "same fluid, same level ⇒ same pressure".
Atmospheric Pressure — what the barometer problems (L1.1, L2.2) measure.
Density and Specific Gravity — the two-fluid problems (L3.2, L5.1) hinge on comparing $ρ$ .
Buoyancy and Archimedes Principle — the floating oil column rests on the same depth-pressure idea.
Bernoulli's Equation — where these static results generalise to moving fluids.