---
title: "New mercury positions after pressure rise"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:mercury-inclined-manometer-pressure-rise"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/mercury-inclined-manometer-pressure-rise
site: BCT Engineering
---

# New mercury positions after pressure rise

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2074 Asoj, 6 marks
- Video: [Fluid Mechanics - U-Tube Manometer Connected to a Closed Tank](https://www.youtube.com/watch?v=UarMG_4Woe0) — R12 Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/mercury-inclined-manometer-pressure-rise
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

In the figure below the pressures at A and B are the same, 100 kPa. If water is introduced at A to increase $P_A$ to 130 kPa, find the new positions of the mercury. The connecting tube is an uniform 1-cm in diameter. Assume no change in the liquid densities. [Figure: bulb A (water) on the left and bulb B (air) on the right joined by a tube containing mercury, the right part inclined at 15° to the horizontal; $\Delta h$ is the drop on the water side and $\Delta L$ the movement along the incline]

## Short answer

The 30 kPa increase balances as Δh[γ_Hg(1+sin15°)−γ_w]; mercury drops 19.0 cm on the water side and moves 19.0 cm along the 15° incline.

## Answer

**Assumptions from the figure:** the left limb (water side) is vertical, the right limb is inclined at $15^\circ$, the bore is uniform, and the pressure at B (air, large bulb) stays at 100 kPa. Initially the mercury is in balance at $p_A=p_B=100$ kPa. $\gamma_{Hg}=13.6\times9.81=133.4$ kN/m³, $\gamma_w=9.81$ kN/m³.

```text
  A (water)                B (air)
     |                      /
     |  drop dh            /  dL along 15 deg
     |_____ mercury ______/
```

**Geometry.** The tube has uniform bore, so the mercury length moved in each limb is equal: the vertical drop on the water side is $\Delta h$, so the movement along the incline is $\Delta L=\Delta h$. The vertical rise on the inclined side is $\Delta L\sin15^\circ$.

**Pressure balance.** At the water-mercury interface the water column above has lengthened by $\Delta h$:

$$
p_A+\gamma_w\Delta h=p_B+\gamma_{Hg}\left(\Delta h+\Delta L\sin15^\circ\right)
$$

$$
130-100=\Delta h\left[133.4(1+\sin15^\circ)-9.81\right]
$$

$$
30=\Delta h\,(167.95-9.81)=158.14\,\Delta h
$$

$$
\Delta h=0.1897\ \text{m}
$$

$$
\Delta L=\Delta h=0.1897\ \text{m},\qquad \text{vertical rise on right}=0.1897\sin15^\circ=0.0491\ \text{m}
$$

**Answer: mercury on the water side drops about 19.0 cm; on the inclined side it moves about 19.0 cm along the tube (a rise of 4.9 cm).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The tube has a uniform bore. If the mercury drops by dh in the vertical water-side limb, how far does it move along the 15° incline?
   A. dL = dh / sin15°
   B. The same length, dL = dh
   C. dL = dh sin15°
   D. dL = dh cos15°
   Answer: B. A uniform bore means equal volume displaced equals equal length moved in each limb, so dL = dh. (Key point: Uniform-bore geometry)

2. How much does the mercury surface rise vertically in the inclined limb?
   A. dL cos15°
   B. dL
   C. dL sin15°
   D. dL / sin15°
   Answer: C. Only the vertical component of the movement along the incline adds to the pressure head: dL sin15°. (Key point: Vertical rise on inclined limb)

3. Which pressure balance is used at the water-mercury interface?
   A. pA + γw dh = pB + γHg (dh + dL sin15°)
   B. pA = pB + γHg dL sin15° only
   C. pA + γHg dh = pB + γw (dh + dL sin15°)
   D. pA - γw dh = pB + γHg dL
   Answer: A. The water column on the A side lengthens by dh, and the mercury level difference is dh plus the vertical rise dL sin15°. (Key point: Pressure balance)

4. With a 30 kPa rise in pA, 30 = dh[133.4(1 + sin15°) - 9.81]. What is dh?
   A. About 0.22 m
   B. About 0.05 m
   C. About 0.30 m
   D. About 0.19 m
   Answer: D. The bracket equals 158.14 kN/m^3, so dh = 30/158.14 = 0.19 m (dL equals this, vertical rise 0.049 m). (Key point: Result of the calculation)

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