Core Practical 2: Investigating Viscosity | Edexcel International A Level Physics Revision Notes 2019

Core Practical 2: Investigating Viscosity of a Liquid

By allowing small spherical objects of known weight to fall through a fluid until they reach terminal velocity, the viscosity of the fluid can be calculated

Variables

Weigh the balls, measure their radius using Vernier callipers and calculate their density
Place three rubber bands around the tube. The highest should be far enough below the surface of the liquid to ensure the ball is travelling at terminal velocity when it reaches this band. The remaining two bands should be 10 – 15 cm apart so that time can be measured accurately
Release the ball and wait until it reaches the first rubber band. Start the timer at the first band, then use the lap timer to find the time to fall d₁ and also d₂
1. If lap timing is not available, two stopwatches operated by different people should be used
2. If the ball is still accelerating as it passes the markers, they need to be moved downwards until the ball has reached terminal velocity before passing the first mark
Measure and record the distances d₁ (between the highest and middle rubber band) and d₂ between the highest and lowest bands.
Repeat at least three times for balls of this diameter and three times for each different diameter
Ball bearings are removed from the bottom of the tube using the magnet against the outside wall of the measuring cylinder

Terminal velocity is used in this investigation since at terminal velocity the forces in each direction are balanced

W_s = F_d + U (equation 1)

Where;
- W_s = weight of the sphere
- F_d = the drag force (N)
- U = upthrust (N)

W_s = v_sρ_sg

$W_{s} = \frac{4}{3} π r^{3} ρ_{s} g$ (equation 2)

Where
- v_s = volume of the sphere (m³)
- ρ_s = density of the sphere (kg m³)
- g = gravitational force (N kg⁻¹)

F_d = 6πηrv_term (equation 3)

Upthrust equals the weight of the displaced fluid
- The volume of displaced fluid is the same as the volume of the sphere
- The weight of the fluid is found from volume, density and gravitational force as above

$U = \frac{4}{3} π r^{3} ρ_{f} g$ (equation 4)

$\frac{4}{3} π r^{3} ρ_{s} g = 6 π η r v_{t e r m} + \frac{4}{3} π r^{3} ρ_{f} g$

$\frac{4}{3} π r^{3} ρ_{s} g - \frac{4}{3} π r^{3} ρ_{f} g = 6 π η r v_{t e r m}$

$\frac{4 π r^{3} g (ρ_{s} - ρ_{f})}{3 \times (6 π r v_{t e r m})} = η$

$η = \frac{2 r^{2} g (ρ_{s} - ρ_{f})}{9 v_{t e r m}}$

Systematic Errors:

Ruler must be clamped vertically and close to the tube to avoid parallax errors in measurement
Ball bearing must reach terminal velocity before the first marker

Random errors:

Cylinder must have a large diameter compared to the ball bearing to avoid the possibility of turbulent flow
Ball must fall in the centre of the tube to avoid pressure differences caused by being too close to the wall which will affect the velocity

Measuring cylinders are not stable and should be clamped into position at the top and bottom
Spillages will be slippery and must be cleaned up immediately
Avoid getting fluids in the eyes