The water flows form a tap of diameter `1.25` cm with a rate of `5xx10^(-5) m^(3) s^(-1).` The density and coefficient of viscosity of water are `10^(3)kg m ^(-3) and 10^(-3)` Pa. s respectively. The flow of water is

Options

A. Steady with Reynolds number 5100
B. turbulent with Reynolds number 5100
C. steady with Reynolds number 3900
D. turbulent with Reynolds number 3900

Answer

B

Explanation

Here, diameter, `D= 1.25 cm = 1.25 xx 10^(-2)m`
Density of water, `rho = 10^(3) kg m^(-3)`
Coefficient of viscosity `eta= 10^(-3)` Pa s
Rate of flow of water, `Q= 5 xx 10^(-5) m^(3) s^(-1)`
Reynolds number, `N_(R ) = (v rho D)/(eta)`....(i)
where v is the speed of flow.
Rate of flow of water Q= area of cross section `xx` speed of flow
`Q= (pi D^(2))/(4) xx v or v= (4Q)/(piD^(2))`
Substituting the value of v in eqn (i), we get
`N_(R ) = (4Q rho D)/(pi D^(2) eta) = (4Qrho)/(pi D eta)= (4 xx 5 xx 10^(-5) xx 10^(3))/(((22)/(7)) xx 1.25 xx 10^(-2) xx 10^(-3))~~ 5100`
For `N_(R ) gt 3000`, the flow is turbulent.
Hence, the flow of water is turbulent with Reynolds number 5100.