Determine viscosity by measuring terminal velocity of a sphere in liquid

Edited By Vishal kumar | Updated on Jul 02, 2025 07:16 PM IST

Viscosity is a measure of a fluid's resistance to flow, and understanding it is crucial in various fields ranging from engineering to biology. By determining the coefficient of viscosity of a given liquid, we gain insights into its flow behaviour, which can be essential for applications such as lubricant formulation, blood flow analysis, and industrial processes involving fluid transport. One practical method to measure viscosity involves observing the terminal velocity of a spherical body as it moves through the liquid. For example, consider a small ball bearing falling through a thick syrup; the speed it eventually reaches, where the forces of gravity and viscous drag balance out, provides valuable data to calculate the liquid's viscosity. This principle not only aids in academic experiments but also helps design efficient fluid systems in real-world applications like oil pipelines and medical diagnostics.

This Story also Contains

Aim
Apparatus
Calculation
Precautions
Solved Examples Based on Determine The Coefficient Viscosity Of A Given Viscous Liquid By Measuring The Terminal Velocity Of Given Spherical Body
Summary

Aim

To determine the coefficient of viscosity of a given viscous liquid by measuring the terminal velocity of a given spherical body.

Apparatus

A half-metre high, 5 cm broad glass cylindrical jar with millimetre graduations along its height, transparent viscous liquid, one steel ball, screw gauge, stop clock/watch, thermometer, and clamp withstand.

Theory

$\eta=\frac{2}{9}(\rho-\sigma) g \frac{r^2}{v}$
$
\begin{aligned}
& \eta=\text { coefficient of viscosity } \\
& \rho=\text { density of ball } \\
& \sigma=\text { density of liquid } \\
& \mathrm{r}=\text { radius of ball } \\
& \mathrm{v}=\text { terminal velocity of ball }
\end{aligned}
$
Knowing $\mathrm{r}, \rho$ and $\sigma_t$ and measuring $\mathrm{v}, \eta$ can be calculated.

Procedure

1. Take some viscous liquid which must be homogeneous and transparent.

2. Clean the glass cylinder and fill it with this liquid.

3. Find the least count of the vertical scale (graduation) of the glass cylinder.

4. Find the least count and zero error of the stopwatch.

5. Find the least count and zero error of the screw gauge.

6. Mark the given balls as 1, 2, 3, 4, 5. Giving number 1 to the ball of smallest diameter and number 5 to the ball of maximum diameter.

7. Find the diameters of the balls as explained in Experiment 2. (Section A)

8. Make two convenient marks on the cylinder about 40 cm apart, one at a distance of 40 cm from the top and the other at a distance of 20 cm from the bottom.

9. Drop the ball 1 gently in the liquid. For about 1/3^rd of the liquid column's height, the ball falls with the accelerated velocity. After that, this ball falls down with the uniform velocity called terminal velocity.

10. When the ball covers about 40 cm from the top, start the stopwatch.

11. Stop the stopwatch when the ball reaches the lower mark which is at 20 cm from the bottom.

12. Note the distance covered by the ball and the time taken to cover this distance.

13. Repeat the experiment one more time with the steel ball of the same diameter.

14. Repeat steps 9 to 13 with the other four balls.

Calculation

Mean diameter

$
D=\frac{D_1+D_2}{2} \mathrm{~mm}
$
Mean radius

$
r=\frac{D}{2} \mathrm{~mm}=
$

$\qquad$ cm

Meantime

$
t=\frac{t_1+t_2+t_3}{3}=
$

$\qquad$
Mean terminal velocity, $v=\frac{S}{t}=\ldots \mathrm{cms}^{-1}$
From formula,
$\eta=\frac{2 r^2(\rho-\sigma) g}{9 v}=$ $\qquad$ C.G.S. units.

Precautions

1. Liquid should be transparent to watch the motion of the ball.
2. The ball should be perfectly spherical.
3. Velocity should be noted only when it becomes constant.

Solved Examples Based on Determine The Coefficient Viscosity Of A Given Viscous Liquid By Measuring The Terminal Velocity Of Given Spherical Body

Example 1: A uniform rod of length L and density $\rho$ is being pulled along A smooth floor with a horizontal acceleration $\alpha$ what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod?

1) $\frac{1}{3} \mathrm{~L} \rho \alpha$
2) $\frac{2}{3} \mathrm{~L} \rho \alpha$
3) $\mathrm{L} \rho \alpha$
4) $\frac{1}{2} \mathrm{~L} \rho \alpha$

Solution:

Let the cross-sectional area of the rod be A.

So, the mass of the rod $=\rho(A L)$

Now, we have to find the tension at the midpoint of the rod.

$\begin{aligned} & T=\frac{\rho A L}{2}(\alpha) \\ & \text { stress }=\frac{T}{\text { area }}=\frac{\rho \alpha L}{2}\end{aligned}$

The correct option is (4).

Example 2: Which of the following is responsible for the viscous force acting on a spherical body?

1) The radius of the spherical body

2) Coefficient of viscosity of the liquid

3) Terminal velocity of the spherical body

4) All of the above

Solution:

wherein

$\begin{aligned} & \eta=\frac{2}{9}(\rho-\sigma) g \frac{r^2}{v} \\ & \eta=\text { coefficient of viscoscity } \\ & \rho=\text { density of ball } \\ & \sigma=\text { density of liquid } \\ & \mathrm{r}=\text { radius of ball } \\ & \mathrm{v}=\text { terminal velocity of ball } \\ & F=6 \pi \eta r v \\ & \Rightarrow F \alpha \eta \\ & \quad \alpha r \\ & \alpha v\end{aligned}$

Example 3: A cylindrical glass tube with an inner radius of 0.20 mm is dipped into a beaker of mercury. Given the surface tension of mercury is 0.075 N/m and the density is 1000 kg/m³, calculate the height to which the mercury rises in the tube. Assume standard gravity as(g=10 m/s² ).

1) 0.0160 Pa·s

2) 0.0075 Pa·s

3) 10.44 Pa·s

4) 0.2320 Pa·s

Solution:

Given:

$T=0.075 \mathrm{~N} / \mathrm{mtext}\left(\right.$ surface tension of mercury) $\theta$ is (contact angle) is not provided, typically assumed as ( 0 ) for mercury in glass. $r=0.20 \times 10^{-3} \mathrm{~m} \rho=1000 \mathrm{~kg} / \mathrm{m}^3 g=10 \mathrm{~m} / \mathrm{s}^2$
Substituting the given values into the formula:

$
\begin{aligned}
& h=\frac{2 T \cdot \cos (\theta)}{r \cdot \rho \cdot g} \\
& h=\frac{2 \times 0.075 \mathrm{~N} / \mathrm{m} \times 1}{0.20 \times 10^{-3} \mathrm{~m} \times 1000 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2} \\
& h=\frac{0.15}{200} \mathrm{~m} h=0.00075 \mathrm{~m}=0.0075 \text { Pa.s }
\end{aligned}
$

Hence, the answer is the option (2).

Example 4: A spherical metal ball has a diameter (d) of 0.04 meters, and its coefficient of viscosity in a liquid is 0.8 Ns/m². The ball experiences a terminal acceleration (v) of 0.1 m/s² while moving through the liquid. Use this information to measure the coefficient of viscosity in standard units.

1) 1.60Ns/m²

2) 8.70Ns/m²

3) 0.8Ns/m²

4) 20Ns/m²

Solution:

Given:

Diameter of the ball (d) = 0.04 meters

Coefficient of viscosity (η) = 0.8 Ns/m²

Terminal acceleration (v) = 0.1 m/s²

To measure the coefficient of viscosity, we can use Stokes’ law equation:

$F_{\text {drag }}=6 \pi \eta r v$ ....... (1)

Where:

F_drag is the drag force on the ball

η is the coefficient of viscosity

r is the radius of the ball

v is the terminal acceleration

The radius is half the diameter:
$
r=\frac{d}{2}=\frac{0.04}{2}=0.02 \mathrm{~m}
$
Using equation (1), we rearrange to solve for $\eta$ :

$
\eta=\frac{F_{\text {drag }}}{6 \pi r v}
$
Plugging in values:

$
\eta=\frac{F_{\mathrm{drag}}}{6 \pi \times 0.02 \mathrm{~m} \times 0.1 \mathrm{~m} / \mathrm{s}^2}=0.8 \mathrm{Ns} / \mathrm{m}^2
$

The coefficient of viscosity is found to be 0.8 Ns/m² based on the terminal acceleration of the spherical metal ball moving through the liquid.

Hence, the answer is the option (3).

Example 5: A student wants to determine the coefficient of viscosity (η) of a given viscous liquid using the method of measuring the terminal velocity of a spherical body falling through the liquid. The sphere has a radius (r) of 0.02 m and a density (ρ) of 1000 kg/m³. The density of the liquid (ρ₀) is 1200 kg/m³, and the acceleration due to gravity (g) is 9.81 m/s². The measured terminal velocity of the sphere in the liquid is 0.1 m/s. Calculate the coefficient of viscosity (η) of the given viscous liquid.

1) 1.60Ns/m²

2) 8.70Ns/m²

3) 0.8Ns/m²

4) 4.36Ns/m²

Solution:

The terminal velocity (v_t) of a spherical body falling through a viscous liquid is given by the following equation:

$v_t=\frac{2 \cdot\left(\rho-\rho_0\right) \cdot g \cdot r^2}{9 \cdot \eta}$

Where:

v_t is the terminal velocity of the sphere

ρ is the density of the sphere

ρ₀ is the density of the liquid g is the acceleration due to gravity

r is the radius of the sphere

η is the coefficient of viscosity of the liquid

We are given:

Sphere radius (r) = 0.02 m

Sphere density (ρ) = 1000 kg/m³

Liquid density (ρ₀) = 1200 kg/m³

Acceleration due to gravity (g) = 9.81 m/s²

Terminal velocity (v_t) = 0.1 m/s

Let’s rearrange the equation to solve for the coefficient of viscosity (η):
$
\eta=\frac{2 \cdot\left(\rho-\rho_0\right) \cdot g \cdot r^2}{9 \cdot v_t}
$
Now, substitute the given values:

$
\begin{gathered}
\eta=\frac{2 \cdot\left(1000 \mathrm{~kg} / \mathrm{m}^3-1200 \mathrm{~kg} / \mathrm{m}^3\right) \cdot 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot(0.02 \mathrm{~m})^2}{9 \cdot 0.1 \mathrm{~m} / \mathrm{s}} \\
\eta=\frac{2 \cdot\left(-200 \mathrm{~kg} / \mathrm{m}^3\right) \cdot 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot 0.0004 \mathrm{~m}^2}{0.9 \mathrm{~m} / \mathrm{s}} \\
\eta=\frac{-3.9248 \mathrm{Ns} / \mathrm{m}^2}{0.9 \mathrm{~kg} /(\mathrm{ms})} \\
\eta \approx-4.36 \mathrm{Ns} / \mathrm{m}^2
\end{gathered}
$

Since viscosity cannot be negative, it seems there might be an error or discrepancy in the given values or measurements. Please double-check the data provided for accuracy.
Hence, the answer is the option (4).

Summary

Measuring the terminal velocity of a spherical object moving through the fluid is a direct way of calculating the coefficient of viscosity of the fluid. At terminal velocity, the effects of gravitational force on the body are balanced by viscosity drag force for an equilibrium state. Thus, from an analysis of this equilibrium condition in accordance with Stokes’ law, it becomes possible to determine the fluidity co-efficient involved.

Frequently Asked Questions (FAQs)

1. What is a Newtonian fluid and why is it important for this experiment?

A Newtonian fluid is one where viscosity remains constant regardless of the applied stress or strain rate. This experiment assumes a Newtonian fluid, as Stokes' Law is derived for such fluids. Non-Newtonian fluids would require different analysis methods.

2. What role does gravity play in this experiment?

Gravity provides the driving force for the sphere's motion. The gravitational force must be balanced against the drag force to reach terminal velocity. Variations in local gravity can affect results, though this is usually negligible for most locations.

3. What assumptions are made when using Stokes' Law in this experiment?

Key assumptions include: the fluid is incompressible and Newtonian, flow is laminar, there are no wall effects, the sphere is perfectly spherical and smooth, and there's no slip at the sphere's surface.

4. What is the significance of the "coefficient" in coefficient of viscosity?

The term "coefficient" indicates that viscosity is a proportionality constant relating shear stress to velocity gradient in a fluid. It quantifies the fluid's resistance to flow under applied stress.

5. How does the viscosity of a liquid relate to its molecular structure?

Viscosity is influenced by intermolecular forces and molecular size. Stronger intermolecular attractions and larger molecules generally result in higher viscosity. Understanding this relationship helps in predicting and explaining viscosity trends among different liquids.

6. How does the shape of the container affect the experiment?

A tall, narrow container is preferred to minimize wall effects and ensure the sphere reaches terminal velocity. The container should be wide enough to prevent significant interaction between the sphere and container walls.

7. How do you determine the appropriate sphere size for your experiment?

The sphere should be small enough to ensure laminar flow (low Reynolds number) but large enough to be easily observable. Its size should also allow for a reasonable terminal velocity that can be accurately measured with available equipment.

8. How do you measure the terminal velocity accurately?

Measure the time taken for the sphere to travel a known distance in the fluid, ensuring this distance is in the region where terminal velocity has been reached. Repeat measurements several times and average the results for better accuracy.

9. How does the density difference between the sphere and fluid affect measurements?

The density difference influences the net force on the sphere. A larger difference results in faster terminal velocity. The density difference must be sufficient to overcome buoyancy and allow the sphere to sink, but not so large that terminal velocity is reached too quickly to measure accurately.

10. What are wall effects and why are they important to consider?

Wall effects occur when the container boundaries influence the fluid flow around the falling sphere. They can increase drag and affect terminal velocity measurements, especially if the container is too narrow relative to the sphere size.

11. What is Stokes' Law and how does it relate to this experiment?

Stokes' Law describes the drag force on a small sphere moving through a viscous fluid. It relates the drag force to the fluid's viscosity, the sphere's radius, and its velocity. In this experiment, we use Stokes' Law to calculate viscosity from the measured terminal velocity.

12. What is terminal velocity in the context of this experiment?

Terminal velocity is the constant speed reached by an object falling through a viscous fluid when the drag force equals the gravitational force. In this experiment, it's the steady speed of a sphere falling through the liquid being studied.

13. Why do we use terminal velocity to determine viscosity?

We use terminal velocity because it represents a balance between gravitational and viscous forces. At this point, the net force on the object is zero, allowing us to relate the object's speed to the fluid's viscosity using Stokes' Law.

14. How does the size of the sphere affect the terminal velocity?

Larger spheres generally reach higher terminal velocities. This is because the gravitational force increases with the cube of the radius, while the drag force only increases with the square of the radius. However, the sphere must still be small enough for Stokes' Law to apply.

15. What is the significance of Reynolds number in this experiment?

The Reynolds number indicates whether the flow around the falling sphere is laminar or turbulent. For Stokes' Law to apply, we need a low Reynolds number (typically less than 0.1) to ensure laminar flow.

16. What factors can affect the accuracy of this experiment?

Factors affecting accuracy include temperature variations, impurities in the liquid, non-spherical shape of the object, wall effects of the container, and human error in timing and measurement.

17. How does temperature affect viscosity measurements?

Temperature significantly affects viscosity. Generally, as temperature increases, viscosity decreases. It's crucial to maintain a constant temperature throughout the experiment and record it accurately for reliable results.

18. How do you ensure the sphere reaches terminal velocity before measurement?

Allow the sphere to fall for a short distance before starting measurements. This "pre-fall" distance allows the sphere to accelerate and reach terminal velocity. The exact distance depends on the fluid and sphere properties.

19. Why is it important to use a sphere in this experiment?

A sphere is used because Stokes' Law specifically applies to spherical objects. Its symmetry ensures uniform drag forces and simplifies calculations. Using non-spherical objects would introduce complexities and inaccuracies.

20. Why might you get different results with spheres of different densities?

Spheres with different densities will reach different terminal velocities. Denser spheres fall faster due to greater gravitational force. This affects the balance between gravity and drag, potentially influencing viscosity calculations.

21. What would happen if you used a non-spherical object instead of a sphere?

Non-spherical objects would experience different drag forces and potentially rotate as they fall, complicating the analysis. Stokes' Law wouldn't directly apply, making it difficult to relate the terminal velocity to viscosity without more complex models.

22. What is the significance of the fluid's viscosity in real-world applications?

Viscosity is crucial in many applications, including lubrication, fluid transport, food processing, and blood flow in the body. Understanding viscosity helps in designing efficient systems and predicting fluid behavior in various contexts.

23. How does the viscosity of the liquid affect the terminal velocity?

Higher viscosity liquids exert more drag force on the falling sphere, resulting in lower terminal velocities. Conversely, less viscous liquids allow for higher terminal velocities.

24. How do you account for buoyancy in your calculations?

Buoyancy is accounted for by using the effective weight of the sphere in the liquid, which is its weight in air minus the buoyant force. This is typically included in the equation derived from Stokes' Law for this experiment.

25. What precautions should be taken when handling viscous liquids?

Wear appropriate personal protective equipment (gloves, goggles), avoid spills, ensure proper ventilation, and be aware of any specific safety considerations for the liquid being used. Some viscous liquids may be toxic, corrosive, or have other hazardous properties.

26. How do you clean the equipment between measurements?

Clean the sphere and container thoroughly with appropriate solvents to remove all traces of the previous liquid. Dry components completely to prevent contamination or dilution of subsequent samples. Proper cleaning ensures accurate and consistent results.

27. What are some common sources of error in this experiment?

Common errors include inaccurate timing, temperature fluctuations, impurities in the liquid, non-spherical balls, wall effects, air bubbles, and parallax errors in reading measurements. Systematic errors can also arise from equipment calibration issues.

28. Can this method be used for all liquids?

This method is most suitable for Newtonian fluids with moderate viscosities. It may not be appropriate for extremely viscous liquids, non-Newtonian fluids, or very low-viscosity liquids where terminal velocity is reached too quickly or other forces become significant.

29. How does viscosity change with pressure?

For most liquids, viscosity increases with pressure, though this effect is often negligible at normal atmospheric conditions. In this experiment, pressure effects are typically not considered unless working with specialized high-pressure equipment.

30. What is the difference between dynamic and kinematic viscosity?

Dynamic viscosity (measured in this experiment) is the ratio of shear stress to shear rate in a fluid. Kinematic viscosity is the ratio of dynamic viscosity to fluid density. Kinematic viscosity is often used in fluid flow calculations.

31. How do you convert between different units of viscosity?

Common units for dynamic viscosity include Pascal-seconds (Pa·s) and poise (P). To convert, use the relationships: 1 Pa·s = 10 P = 1000 cP (centipoise). Always check the units in your final answer and convert if necessary.

32. Why might the experimental value differ from the accepted value for a given liquid?

Differences can arise due to experimental errors, variations in liquid purity or composition, temperature differences, or limitations in the experimental setup. Careful analysis of potential error sources can help explain discrepancies.

33. How does the concept of laminar flow apply to this experiment?

Laminar flow occurs when fluid layers move smoothly past each other with minimal mixing. This experiment relies on laminar flow around the falling sphere for Stokes' Law to apply accurately. Turbulent flow would invalidate the assumptions of the experiment.

34. How does this method compare to other viscosity measurement techniques?

The falling sphere method is simple and suitable for moderate viscosities. Other methods include rotational viscometers for a wider range of viscosities, capillary viscometers for low viscosities, and vibrational viscometers for high-precision measurements. Each has its advantages and limitations.

35. Can you use liquids of known viscosity to calibrate your experimental setup?

Yes, using standard liquids with known viscosities can help calibrate the setup and verify its accuracy. This process can identify systematic errors and improve the reliability of measurements for unknown liquids.

36. How do you account for non-ideal behavior of real fluids in this experiment?

While the experiment assumes ideal fluid behavior, real fluids may deviate slightly. Accounting for non-ideal behavior might involve using correction factors, limiting the experiment to conditions where deviations are minimal, or using more advanced models for data analysis.

37. What is the relationship between viscosity and fluid flow rate?

Higher viscosity fluids flow more slowly under a given pressure difference. The relationship is often described by Poiseuille's law for laminar flow in pipes, where flow rate is inversely proportional to viscosity.

38. How does the concept of shear stress relate to viscosity in this experiment?

Shear stress is the force per unit area exerted by the fluid on the falling sphere. Viscosity is a measure of the fluid's resistance to this shear stress. The relationship between shear stress and the resulting deformation rate (velocity gradient) defines the fluid's viscosity.

39. Can this experiment be adapted for use with gases?

While the principle is similar, measuring gas viscosity with falling spheres is challenging due to the much lower viscosities and densities of gases. Modified techniques, such as falling-ball viscometers in pressurized chambers, are sometimes used for gases.

40. How do you ensure the sphere falls straight without wobbling?

Use a perfectly spherical object and ensure it's released without initial rotation. A guide tube at the top of the liquid column can help initiate a straight fall. Proper experimental technique and practice can minimize wobbling and improve measurement accuracy.

41. What is the effect of surface tension on this experiment?

Surface tension can affect the entry of the sphere into the liquid and potentially influence its motion near the surface. To minimize these effects, ensure the sphere is fully submerged before starting measurements and use a sufficiently long column of liquid.

42. How do you determine if Stokes' Law is applicable for your specific experimental conditions?

Calculate the Reynolds number for your setup. If it's less than about 0.1, Stokes' Law should apply. Also, ensure the sphere diameter is much smaller than the container width (typically less than 10% of the container diameter) to minimize wall effects.

43. What are some practical applications of viscosity measurements in industry?

Viscosity measurements are crucial in quality control for products like oils, paints, and food items. They're also important in designing fluid handling systems, optimizing combustion processes, and developing new materials with specific flow properties.

44. How does the viscosity of a mixture relate to the viscosities of its components?

The viscosity of a mixture is not always a simple average of its components' viscosities. It can depend on the nature of the components, their proportions, and how they interact. Some empirical or semi-empirical models exist for predicting mixture viscosities.

45. What is the relationship between a liquid's viscosity and its boiling point?

Generally, liquids with higher boiling points tend to have higher viscosities. This is because both properties are influenced by the strength of intermolecular forces. However, this relationship is not always straightforward and can have exceptions.

46. How do you account for air resistance when the sphere is initially dropped into the liquid?

Air resistance is typically negligible compared to the resistance in the liquid. However, to minimize any potential effects, start timing measurements after the sphere has entered the liquid and traveled a short distance to reach terminal velocity.

47. What is the importance of using a stopwatch with sufficient precision in this experiment?

A precise stopwatch (preferably digital) is crucial for accurate time measurements, especially for less viscous liquids where the fall time may be short. The timing precision directly affects the accuracy of the calculated viscosity.

48. How does the concept of streamlines relate to this experiment?

Streamlines are theoretical lines in a fluid that are tangent to the velocity vector of the fluid at every point. In laminar flow around the falling sphere, these streamlines should be smooth and symmetrical. Visualizing streamlines can help understand the flow behavior in the experiment.

49. How do you determine the uncertainty in your viscosity measurement?

Calculate the uncertainty by considering errors in each measured quantity (sphere diameter, density, fall time, distance). Use error propagation techniques to combine these individual uncertainties and determine the overall uncertainty in the viscosity value.

50. Can this experiment be automated for more precise measurements?

Yes, automation can improve precision. Possibilities include using photoelectric sensors to detect the sphere's passage, computer-controlled release mechanisms, and automated data collection and analysis. This can reduce human error and allow for more repeated measurements.