Question

A form of momentum equation for an incompressible fluid is
\[ \rho \frac{D \mathbf{V}}{Dt} = -\nabla p + \mu \nabla^2 \mathbf{V} + \mathbf{B} \] where \( \rho \) is density, \( \mathbf{V} \) is velocity, \( t \) is time, \( p \) is pressure, \( \mu \) is viscosity, and \( \mathbf{B} \) represents body force per unit volume. The dimension of term (iii) is:

• A. ( [L]^1 [T]^{-2} \)
• B. ( [M]^1 [L]^{-2} [T]^{-2} \)
• C. ( [M]^1 [L]^1 [T]^{-2} \)
• D. ( [M]^1 [L]^1 [T]^{-1} \)

Hint:

In dimensional analysis, always break down complex terms like viscosity and derivatives to their fundamental units (mass, length, time) to find their dimensional expressions.

Answer 1

The Correct Option is B

The dimension of term (iii) in the momentum equation is given by \( \mu \nabla^2 \mathbf{V} \). To find the dimensions, we first calculate the dimensions of \( \mu \) (viscosity) and \( \nabla^2 \mathbf{V} \) (the second derivative of velocity with respect to distance). 1. The dimension of \( \mu \) is \( [M]^1 [L]^{-1} [T]^{-1} \), as it has the dimensions of force per unit area per unit velocity. 2. The dimension of \( \nabla^2 \mathbf{V} \) is \( [L]^{-2} [T]^{-1} \), as it involves the second derivative of velocity with respect to length. Multiplying these gives the dimension of term (iii): \[ \left[ [M]^1 [L]^{-1} [T]^{-1} \right] \times \left[ [L]^{-2} [T]^{-1} \right] = [M]^1 [L]^{-2} [T]^{-2}. \] Thus, the correct answer is option (B).