SSC JE Mechanical Engineering 2014 Set 2 Paper

Given value : Kinematic viscosity
$\left(\mathit{\upsilon }\right)=2.25$ stokes
$=2.25\times {10}^{-4}{\mathit{m}}^2$/s
diameter of pipe (d) = 20 cm.
= 0.20 m
Rate of flow (Q)= 1.5 ltr/s
= 1.5 × 10${}^{-3}$ m${}^3$/s
We know that ${\mathit{R}}_{\mathit{e}}=\frac{\mathrm{\rho }\mathrm{vd}}{\mathrm{\mu }}$ or
$=\frac{\mathit{v}\mathit{d}}{\mathit{\upsilon }}=\frac{0.047\times .20}{2.25\times {10}^{-4}}=41$
Hence reynold no. less than 2000, the flow is laminar.