The viscous drag force F acting on a spherical body moving through a fluid at speed v is g… — JAMB UTME

The viscous drag force $F$ acting on a spherical body moving through a fluid at speed $v$ is given by Stokes' law, $F = 6 \pi \eta r v$ , where $r$ is the radius of the sphere and $\eta$ is the coefficient of viscosity. What is the dimensional formula of the coefficient of viscosity $\eta$ ?

$\text{M L}^{-1} \text{T}^{-1}$ Answer
B
$\text{M L T}^{-2}$
C
$\text{M L}^2 \text{T}^{-1}$
D
$\text{M L}^{-1} \text{T}^{-2}$

Answer

The dimensional formula of the coefficient of viscosity is

\text{M L}^{-1} \text{T}^{-1}

Isolating the coefficient of viscosity from Stokes' law yields

\eta = \frac{F}{6\pi r v}

. Substituting the dimensions

[F] = \text{M L T}^{-2}

[r] = \text{L}

, and

[v] = \text{L T}^{-1}

gives

[\eta] = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{-1} \text{T}^{-1}

Step-by-Step Solution

Identify the dimensional formulas for force, radius, and velocity.

[F] = \text{M L T}^{-2}

[r] = \text{L}

, and

[v] = \text{L T}^{-1}

Base physical quantities must be represented by their fundamental dimensions.

Rearrange Stokes' law to express

\eta

in terms of the other variables, treating

6\pi

as a dimensionless constant.

[\eta] = \frac{[F]}{[r][v]}

Numerical constants do not carry physical dimensions.

Substitute the base dimensions into the formula and simplify using index rules.

[\eta] = \frac{\text{M L T}^{-2}}{\text{L} \cdot \text{L T}^{-1}} = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{-1} \text{T}^{-1}

Subtracting denominator indices from numerator indices yields the final dimensional expression.

Key Concept

Dimensional analysis of physical constants and equations

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic