SSC JE Mechanical Engineering 2014 Set 1 Paper

Given:Kinetic viscosity $\mathit{\upsilon }=4\times {10}^{\mathit{–}2}{\mathrm{m}}^2$/s
dia of pipe d = 80 mm
discharge $\mathit{q}=3200\mathit{\pi }\times {10}^{\mathit{–}4}{\mathit{m}}^3$/s
Now $\mathit{q}=\mathit{a}\mathit{v}$
⇒ discharge = Area × velocity
$3200\times \mathit{\pi }\times {10}^{-4}=\frac{\mathit{\pi }}{4}{\mathit{d}}^2\times \mathit{v}$ ∵ d = 80 × 10${}^{-3}$m
v = 200 m/s
∵ Re (Reynold No. ) = $\frac{\mathit{\rho }\mathit{v}\mathit{d}}{\mathit{\mu }}=\frac{\mathit{v}\mathit{d}}{\mathit{\upsilon }}$
$=\frac{200\times 80\times {10}^{-3}}{4\times {10}^{-2}}$
Re = 400
Reynold number 400 means flow is laminar, because
Re < 2000 flow is laminar
2000 > Re < 4000 flow in transition
Re > 4000 flow is turbulant.