Water is flowing in a pipe of diameter 6 cm with an average velocity `7.5 cm^(-1).s^(-1)` and its density is `10^(3) kg m^(-3)`. What is the nature of flow ? Given coefficient of viscosity of water is `10^(-3) kgm^(-1) s^(-1).`

Options

A. Streamline
B. Turbulent
C. Both a and b
D. None of these

Answer

B

Explanation

Reynolds number for the given situation is given as
`R_e=(rhovD)/(eta)`
Here `rho " density " = 10^3 kgm^(-3)`
Coefficient of viscosity, `eta = 10^(-3) kgm^(-1) s^(-1)`
Average velocity of water, `v=7.5cms^(-1) = 0.075ms^(-1)`
Diameter of pipe , D=6cm=0.06m
Hence , `R_e=(10^3xx0.075xx0.06)/(10^(-3))=10^6xx0.0045=4500`
`therefore R_e gt 2000`
therefore , the flow is turbulent