---
title: "Determination of viscosity by viscometer"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_3/fluid_mechanics:viscometer-viscosity-determination"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/viscometer-viscosity-determination
site: BCT Engineering
---

# Determination of viscosity by viscometer

Fluid Mechanics · Chapter 1: Fluid and its physical properties

- Asked in: 2073 Shrawan, 2 marks; 2072 Chaitra, 3 marks (asked 2 times)
- Video: [Rotational Viscometer: Measurement of Dynamic Viscosity (Newtonian and Non-Newtonian Fluids)](https://www.youtube.com/watch?v=hhgadLpKUac) — Fluid Matters
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/viscometer-viscosity-determination
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-1

## Question

Explain the determination of viscosity by a viscometer.

## Short answer

A viscometer determines dynamic viscosity using Newton's law τ = μ du/dy. In a rotating-cylinder type, torque on a cylinder in a fluid-filled gap yields μ = T t / (2πωR³h). The capillary-tube type compares flow times of test and reference liquids: μ₁/μ₂ = ρ₁t₁/(ρ₂t₂).

## Answer

A **viscometer** measures the dynamic viscosity $\mu$ of a fluid from a measured relation between shear stress and velocity gradient, using Newton's law $\tau=\mu\,\dfrac{du}{dy}$.

#### Rotating-cylinder (Couette) viscometer
The fluid fills the thin gap $t$ between a fixed outer cylinder and an inner cylinder (radius $R$, immersed height $h$) turned at angular speed $\omega$. The torque $T$ on the inner cylinder is measured.

```text
   fixed outer cylinder
   |   |  fluid gap t |   |
   |   |<->|  inner   |   |
   |   |   | cylinder |   |
   |   |   | (omega)  |   |
```

For a thin gap the velocity profile is linear, so

$$
\tau=\mu\frac{\omega R}{t},\qquad T=\tau\,(2\pi R h)\,R
$$

$$
\mu=\frac{T\,t}{2\pi\omega R^{3}h}
$$

#### Capillary-tube (Ostwald) viscometer
The time $t_1$ for a fixed volume of the test liquid to flow through a capillary under its own head is compared with the time $t_2$ for a reference liquid (water) of known viscosity:

$$
\frac{\mu_1}{\mu_2}=\frac{\rho_1 t_1}{\rho_2 t_2}
$$

For laminar flow (Hagen-Poiseuille), $\mu=\dfrac{\pi\,\Delta p\,r^{4}}{8QL}$.

#### Falling-sphere viscometer
A small sphere of diameter $d$ and density $\rho_s$ falls at terminal velocity $v$ in the liquid (density $\rho$). Balancing weight, buoyancy and Stokes drag $3\pi\mu d v$:

$$
\mu=\frac{d^{2}(\rho_s-\rho)\,g}{18\,v}
$$

#### Other types
The Saybolt (efflux) viscometer gives kinematic viscosity from the time taken for 60 mL of oil to flow out through a standard orifice.

## Check yourself

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

1. On what law does a viscometer base its determination of dynamic viscosity?
   A. Newton's law of viscosity, tau = mu du/dy
   B. Pascal's law
   C. Archimedes' principle
   D. Bernoulli's equation
   Answer: A. A viscometer measures a shear stress and velocity gradient and uses tau = mu du/dy to get mu. (Key point: Principle of viscometers)

2. In the rotating-cylinder viscometer with thin gap t, which expression gives mu from the measured torque T?
   A. mu = T t / (2 pi omega R^3 h)
   B. mu = T R / (2 pi omega t h)
   C. mu = 2 pi omega R^3 h / (T t)
   D. mu = T / (omega R t h)
   Answer: A. tau = mu omega R / t and T = tau (2 pi R h) R, so mu = T t / (2 pi omega R^3 h). (Key point: Rotating-cylinder formula)

3. In the Ostwald capillary viscometer, how is the viscosity of the test liquid found?
   A. From the ratio mu1/mu2 = rho1 t1 / (rho2 t2) against a reference liquid
   B. From the terminal velocity of a sphere
   C. From the torque on an inner cylinder
   D. From the height of capillary rise only
   Answer: A. The flow times of equal volumes of test and reference liquid, with their densities, give the viscosity ratio. (Key point: Capillary viscometer comparison)

4. In the falling-sphere viscometer, which forces are balanced at terminal velocity?
   A. Weight = buoyancy + Stokes drag
   B. Weight = Stokes drag only
   C. Buoyancy = Stokes drag only
   D. Weight + buoyancy = Stokes drag
   Answer: A. At terminal velocity there is no acceleration, so the downward weight equals buoyancy plus the 3 pi mu d v drag, giving mu = d^2 (rho_s - rho) g / (18 v). (Key point: Falling-sphere method)

## Related in Fluid Mechanics

- [Newtonian or non-Newtonian fluid from data](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Anewtonian-or-not-table) — Fluid Mechanics, chapter 1