Water is conveyed through a uniform tube of 8 cm in diameter and 3140 m in length at the rate `2xx10^-3 m^3 s^-1`. The pressure required to maintain the flow is (Viscosity of water `10^-3` SI units)

Options

A. `6.25xx10^3 N m^-2`
B. `0.65N m^-2`
C. `0.0625Nm^-2`
D. `0.00625 N m^-2`

Answer

A

Explanation

Rate of flow, `Q=(piPr^(2))/(8etal) = 2xx10^(-3)m^(3)//s`
`therefore P =(Q(8etal))/(pir^(4))`
`=(2xx10^(-3)xx8xx10^(-3)xx3140)/(3.14xx(4xx10^(-2))^(4))` [ `because eta=10^(-3)` I S.I. units]
`= (3140xx10^(2))/(3.14xx16) = (100xx10^(3))/16 = 6.25xx10^(3) N/m^(2)`