---
title: "Brass cylinder floating in mercury"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:brass-cylinder-floating-mercury"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/brass-cylinder-floating-mercury
site: BCT Engineering
---

# Brass cylinder floating in mercury

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2072 Chaitra, 3+3 marks
- Video: [How To Calculate The Fractional Volume Submerged & The Density of an Object In Two Fluids](https://www.youtube.com/watch?v=sMkHKMsBEl8) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/brass-cylinder-floating-mercury
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

Given: Container of mercury with vertical tubes $d_1$ = 39.5 mm. Brass cylinder with D = 37.5 mm and H = 76.2 mm is introduced into larger tube, where it floats. Take $S_{brass}$ = 8.5. Find: (a) Pressure on bottom of cylinder (b) New equilibrium level, h, of mercury. [Figure: two vertical tubes of mercury (diameters $d_1$ and $d_2$); the brass cylinder floats in the larger tube, raising mercury level by h]

## Short answer

Pressure on the cylinder bottom equals its weight over base area: 6.35 kPa. Immersion depth is 47.6 mm; mercury rise h = 42.9 mm if the second tube is neglected.

## Answer

**Data:** brass cylinder $D=37.5$ mm, $H=76.2$ mm, $S_{brass}=8.5$, mercury $S=13.6$, larger tube $d_1=39.5$ mm. The second tube diameter $d_2$ is not legible in the figure, so its effect is kept as a symbol below.

#### (a) Pressure on the bottom of the cylinder
The floating cylinder's weight acts over its base area $A_c$, so the pressure on the bottom is the weight divided by the area:

$$
p=\frac{W}{A_c}=S_{brass}\,\gamma_w\,H=8.5(9810)(0.0762)=6354\ \text{Pa}
$$

(about 6.35 kPa gauge.) This equals the hydrostatic pressure of mercury at the immersion depth $d$.

#### (b) New equilibrium level
Depth of immersion (weight = buoyancy):

$$
S_{brass}H=S_{Hg}\,d\ \Rightarrow\ d=\frac{8.5}{13.6}(76.2)=47.6\ \text{mm}
$$

Volume of mercury displaced:

$$
V=A_c\,d=\frac{\pi}{4}(37.5)^{2}(47.6)=1104.5\times47.6=52\,600\ \text{mm}^3
$$

Let the mercury level rise by $h$ in both tubes. The cylinder bottom is then $(d-h)$ below the original level, and the mercury it pushes out from below the original level fills the annulus and the second tube above it:

$$
A_c(d-h)=\left(A_1-A_c\right)h+A_2h\ \Rightarrow\ h=\frac{A_c\,d}{A_1+A_2}=\frac{V}{A_1+A_2}
$$

where $A_1=\dfrac{\pi}{4}d_1^{2}=\dfrac{\pi}{4}(39.5)^{2}=1225.4\ \text{mm}^2$ is the full area of the larger tube and $A_2=\dfrac{\pi}{4}d_2^{2}$.

- With the second tube neglected: $h=\dfrac{52\,600}{1225.4}=42.9$ mm.
- For example, with $d_2=12.7$ mm ($A_2=126.7\ \text{mm}^2$): $h=\dfrac{52\,600}{1352.1}=38.9$ mm.

**Answer: (a) $p\approx6.35$ kPa; (b) immersion depth 47.6 mm and rise $h=\dfrac{52\,600}{1225.4+0.785\,d_2^{2}}$ mm (42.9 mm if the second tube is neglected; 38.9 mm for $d_2=12.7$ mm).**

## Check yourself

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

1. How is the pressure on the bottom of the floating cylinder found?
   A. S_Hg γw H
   B. Weight / base area = S_brass γw H
   C. γw x immersion depth
   D. Atmospheric pressure only
   Answer: B. A floating body's weight is carried over its base: p = 8.5 x 9810 x 0.0762 = 6.35 kPa. (Key point: Pressure on the cylinder bottom)

2. Which condition gives the immersion depth d?
   A. Volume of cylinder = displaced volume, d = H
   B. S_brass d = S_Hg H
   C. Weight = buoyancy: S_brass H = S_Hg d, so d = 47.6 mm
   D. d = H x S_Hg / S_brass
   Answer: C. Equating weight and buoyancy gives d = (8.5/13.6)(76.2) = 47.6 mm. (Key point: Depth of immersion)

3. How is the rise h calculated from the displaced volume?
   A. Displaced volume divided by the free mercury surface area
   B. Displaced volume divided by the cylinder area
   C. Immersion depth divided by 2
   D. Displaced volume times the free area
   Answer: A. The displaced volume spreads over the free mercury surface, the annulus around the cylinder (plus the other tube). (Key point: Rise of mercury)

4. What happens to h if the second tube is connected and its area is included?
   A. h increases because more mercury is displaced
   B. h is unchanged
   C. h becomes zero
   D. h decreases because the free surface area increases
   Answer: D. h = V / (A_ann + π d2^2 / 4); a larger denominator gives a smaller rise. (Key point: Effect of the second tube)

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