---
title: "Liquid rise in rotating U-tube"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:u-tube-rotating-liquid-rise"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/u-tube-rotating-liquid-rise
site: BCT Engineering
---

# Liquid rise in rotating U-tube

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2078 Kartik, 6 marks
- Video: [Vertical Motion Example + Intro to Rotating Vessel](https://www.youtube.com/watch?v=9JkY0FlcDyM) — Abrenian Lecturer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/u-tube-rotating-liquid-rise
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

The figure shows U-tube of base length L in which a liquid of density 0.85 is filled such that it completely fills the base length only. If the tube is now rotated at angular speed of 10 rad/sec as shown, find the level rise of liquid in outer arm of tube. [Figure: U-tube with base length L, rotating about the left vertical arm with $\omega$]

## Short answer

For rigid-body rotation, the rise h satisfies h = ω²(L²−h²)/(2g), giving h = (√(1+103.9L²)−1)/10.19 m; for L=0.1 m, h ≈ 42 mm, independent of liquid density.

## Answer

**Data:** U-tube of uniform bore with base length $L$, rotating about the left vertical arm at $\omega=10$ rad/s. The relative density 0.85 of the liquid does not affect the result.

At rest the liquid fills only the base. When the tube spins, the liquid is thrown outwards: it rises a height $h$ in the outer arm, and the same volume leaves the inner end of the base. With a uniform bore, the liquid now fills the base only from $r=h$ to $r=L$, with a free surface (at atmospheric pressure, since air enters through the inner arm) at $r=h$ on the base level.

For rigid-body rotation the pressure satisfies

$$
p+\rho g z-\tfrac{1}{2}\rho\omega^{2}r^{2}=\text{constant}
$$

Apply this between the two free surfaces, both at $p=0$: the inner surface in the base ($r=h$, $z=0$) and the surface in the outer arm ($r=L$, $z=h$):

$$
\rho g h-\tfrac{1}{2}\rho\omega^{2}L^{2}=-\tfrac{1}{2}\rho\omega^{2}h^{2}
\quad\Rightarrow\quad
h=\frac{\omega^{2}\left(L^{2}-h^{2}\right)}{2g}
$$

With $k=\dfrac{\omega^{2}}{2g}=\dfrac{100}{19.62}=5.097\ \text{m}^{-1}$, this is $k h^{2}+h-kL^{2}=0$, so

$$
h=\frac{\sqrt{1+4k^{2}L^{2}}-1}{2k}=\frac{\sqrt{1+103.9\,L^{2}}-1}{10.19}\ \text{m}\quad(L\text{ in metres})
$$

For example, $L=0.1$ m gives $h=0.042$ m (42 mm), and $L=0.3$ m gives $h=0.218$ m. If the loss of liquid from the inner end of the base were ignored, we would get $h=\omega^{2}L^{2}/2g=5.1\,L^{2}$, which overestimates the rise (51 mm for $L=0.1$ m).

**Answer: the rise in the outer arm satisfies $h=\dfrac{\omega^{2}(L^{2}-h^{2})}{2g}$, i.e. $h=\dfrac{\sqrt{1+103.9L^{2}}-1}{10.19}$ m (about 42 mm for $L=0.1$ m), independent of the liquid density.**

## Check yourself

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

1. How does pressure vary along the rotating base with radius r?
   A. p = p0 + rho omega^2 r
   B. p = p0 + 1/2 rho omega^2 r^2
   C. p = p0 - 1/2 rho omega^2 r^2
   D. p is constant
   Answer: B. Rigid-body rotation gives pressure rising as 1/2 rho omega^2 r^2 away from the axis. (Key point: Pressure in rotating liquid)

2. What is the rise h of the outer arm over the inner arm?
   A. omega^2 L^2 / g
   B. omega L / 2g
   C. omega^2 L^2 / 2g
   D. omega^2 L / 2g
   Answer: C. rho g h = 1/2 rho omega^2 L^2 gives h = omega^2 L^2/(2g) = 5.1 L^2 m. (Key point: Rise in outer arm)

3. Does the relative density 0.85 affect the result?
   A. Yes, h is proportional to density
   B. Yes, h is inversely proportional to density
   C. No, density cancels
   D. Only if L is large
   Answer: C. Density appears on both sides of rho g h = 1/2 rho omega^2 L^2 and cancels. (Key point: Role of density)

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