Dimensional analysis — units of pressure gradient in fluid flow (Determine the dimensions of dp/dx for a continuum fluid.)

Introduction / Context:Dimensional consistency is a cornerstone in fluid mechanics. Pressure gradients drive flow and appear in the Navier–Stokes momentum equation alongside inertial and viscous terms; getting their dimensions right helps verify derivations and unit conversions.

Given Data / Assumptions:

• Pressure p = force/area.
• Force has dimensions M L T^-2.
• Gradient dp/dx divides pressure by length.

Concept / Approach:Start from pressure dimensions: p = (M L T^-2) / L^2 = M L^-1 T^-2. Taking a spatial derivative divides by L, so dp/dx = (M L^-1 T^-2) / L = M L^-2 T^-2. This matches the body-force-per-volume units in momentum balance (e.g., -dp/dx as a driving term).

Step-by-Step Solution:1) Pressure: p = Force / Area = (M L T^-2) / L^2.2) Simplify → p = M L^-1 T^-2.3) Apply gradient: divide by L → dp/dx = M L^-2 T^-2.4) Choose the option with M L^-2 T^-2.

Verification / Alternative check:Compare with viscous term mu * d^2u/dx^2, where mu has M L^-1 T^-1 and second derivative yields L^-2 T^-1; the product gives M L^-3 T^-2 times dimension of velocity L T^-1—consistent at equation level.

Why Other Options Are Wrong:(a) sign/exponents incorrect; (b) corresponds to pressure/volume, not gradient; (d) has inverted mass dimension.

Common Pitfalls:Forgetting that taking a derivative with respect to length introduces an extra L^-1; mixing up dynamic and kinematic quantities.

Final Answer:M L^-2 T^-2