This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 1)

Option 2 : 10^{4}

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

8346

90 Questions
360 Marks
180 Mins

**Concept: **

In Fluid Mechanics, **Reynolds number** is used to represent the flow of fluid. It is dimensionless. It is defined as the ratio of inertial force to the viscous force.

**Calculation:**

We know that, Reynolds number,

\(Re=\frac{\rho \text{vd}}{\eta }\) ----(1)

Where, v = Velocity of the fluid with respect to the object

d = Diameter of the pipe

ρ = Density of the fluid

η = Viscosity of the fluid

In the question, the velocity is given as volume velocity in litres per minute (L/m). Its unit is m^{3}/s.

Since, we need m/s (velocity), we need to find the velocity using below given formula:

Volume velocity = velocity × area = v × πr^{2}

\(\Rightarrow v=\frac{Volume~velocity}{\pi {{r}^{2}}}\)

\(\Rightarrow v=\frac{100\times {{10}^{-3}}}{60}\times \frac{1}{\pi \times {{\left( 5\times {{10}^{-2}} \right)}^{2}}}\)

∴ v = 0.21 m/s

On substituting the given values in equation (1), we get,

\(\text{Reynolds }\!\!~\!\!\text{ number}=\frac{1000\times 0.21\times \left( 2\times 5\times {{10}^{-2}} \right)}{1\times {{10}^{-3}}}\)

Note: 1 mPa s = 1 milli-pascal second

Reynolds number = 21000 = 2.1 × 10^{4}