Question Detail
Question
The volume flow rate Q (in m³/s) of a liquid through pipe having diameter d is related to viscosity of water η (unit Pascal.s) and the pressure gradient along the pipe dP/dx [pressure gradient dP/dx is rate of change of pressure per unit length along the pipe], by a formula of the form; Q = Kηᵃ dᵇ (dP/dx)ᶜ. Where K is a dimensionless constant. Find a, b, and c.
Correct Answer
Option A is the correct answer.
Detailed Explanation
This is based on dimensional analysis (Poiseuille's law). Q has dimensions L³T⁻¹. η has dimensions ML⁻¹T⁻¹. d has dimension L. dP/dx has dimensions ML⁻²T⁻² (since pressure is ML⁻¹T⁻²). Equating dimensions: L³T⁻¹ = (ML⁻¹T⁻¹)ᵃ (L)ᵇ (ML⁻²T⁻²)ᶜ. For M: 0 = a + c. For T: -1 = -a - 2c. For L: 3 = -a + b - 2c. Solving a + c = 0 => c = -a. Sub into T: -1 = -a - 2(-a) = -a + 2a = a => a = -1, so c = 1. Then L: 3 = -(-1) + b - 2(1) = 1 + b - 2 = b - 1 => b = 4. So (a, b, c) = (-1, 4, 1).
Hint
Use dimensional analysis. Write down the dimensions of each quantity and equate powers of M, L, T.