Water is flowing in a pipe of radius 1.5 cm with an average velocity of `15 cm s^(-1)` . What is the nature flow ? Given coefficient of viscosity of water is `10^(-3) kg m^(-1)s^(-1)` and its density is `10^3 kg m^(-3)`.

Explanation

`D = 2 xx 1.5 cm = 3 xx 10^(-2) m`
`v = 15 cm s^(-1) = 0.15 ms^(-1)`
`eta = 10^(-3) kg m^(-1)s^(-1)`
`R_c = (rho v D)/(eta) = (10^3 xx 0.15 xx 3 xx 10^(-2))/(10^(-3))`
`R_c = 4500`
As `R_c > 3000`, so the flow is turbulent.