StreakPeaked· Practice

ExamsJEE MainPhysics

The radius of a pipeline's cross-section decreases gradually as r = r0*e^(-alpha*x), where alpha = 0.50 m⁻¹ and x is the distance from the pipe inlet. For an incompressible fluid in steady flow, find the ratio of Reynolds numbers Re_final/Re_initial between two cross-sections separated by deltaₓ = 3.2 m.

  1. e^(-1.6)
  2. e^(1.6)
  3. e^(-3.2)
  4. e^(3.2)

Correct answer: e^(1.6)

Solution

Continuity gives v proportional to 1/r². Reynolds number Re = rho v (2r)/eta is proportional to v*r proportional to (1/r²)*r = 1/r. So Re_final/Re_initial = r_initial/r_final = e^(-alpha*x1)/e^(-alpha*x2) = e^(alpha(x2-x1)) = e^(alpha*deltaₓ) = e^(0.5*3.2) = e^(1.6).

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →