Question

The terminal velocity of a small sphere settling in a viscous fluid varies as the

• A. First power of its diameter
• B. Inverse square of its diameter
• C. Inverse of the fluid viscosity
• D. Square of the difference in specific weight of solid and fluid

Hint:

Under Stokes’ law, terminal velocity of a small sphere varies as \( d^2 \) (diameter squared) and \( \frac{1}{\mu} \) (inverse of viscosity), assuming laminar flow.

Answer 1

The Correct Option is C

Step 1: Understand terminal velocity and Stokes’ law. 
The terminal velocity of a small sphere settling in a viscous fluid occurs when the gravitational force (minus buoyancy) is balanced by the drag force, so the sphere falls at a constant velocity. For small spheres under laminar flow conditions (low Reynolds number, \( \text{Re}<1 \)), Stokes’ law applies. The terminal velocity (\( v_t \)) is given by: \[ v_t = \frac{g (\rho_s - \rho_f) d^2}{18 \mu}, \] where:
\( g \): gravitational acceleration (\( \text{m/s}^2 \)),
\( \rho_s \): density of the sphere (\( \text{kg/m}^3 \)),
\( \rho_f \): density of the fluid (\( \text{kg/m}^3 \)),
\( d \): diameter of the sphere (\( \text{m} \)),
\( \mu \): viscosity of the fluid (\( \text{Pa·s} \)).
Step 2: Analyze the dependency of terminal velocity. 
From Stokes’ law, \( v_t = \frac{g (\rho_s - \rho_f) d^2}{18 \mu} \), we can determine how \( v_t \) varies with each parameter:
Diameter (\( d \)): \( v_t \propto d^2 \), so terminal velocity varies as the square of the diameter, not the first power.
Viscosity (\( \mu \)): \( v_t \propto \frac{1}{\mu} \), so terminal velocity varies as the inverse of the fluid viscosity.
Difference in densities (\( \rho_s - \rho_f \)): \( v_t \propto (\rho_s - \rho_f) \), which is related to the difference in specific weights (since specific weight is \( \rho g \)), but it is a linear relationship, not squared. 
Step 3: Evaluate the options. 
(1) First power of its diameter: Incorrect, as \( v_t \propto d^2 \), not \( d \). Incorrect.
(2) Inverse square of its diameter: Incorrect, as \( v_t \propto d^2 \), not \( \frac{1}{d^2} \). Incorrect.
(3) Inverse of the fluid viscosity: Correct, as \( v_t \propto \frac{1}{\mu} \), matching the inverse relationship with viscosity. Correct.
(4) Square of the difference in specific weight of solid and fluid: Incorrect, as \( v_t \propto (\rho_s - \rho_f) \), a linear relationship, not squared. Incorrect. 
Step 4: Select the correct answer. 
The terminal velocity of a small sphere settling in a viscous fluid varies as the inverse of the fluid viscosity, matching option (3).