The flow rate of water from a tap of diameter 1.25 cm is  $0.48L/\min$. The coefficient of viscosity of water is ${10}^{-3}Pa s$. Then find the Raynold’s number

Let the speed of the flow be   and the diameter of the tap be $d=1.25cm$ . The volume of the water flowing out per second is 
$Q=v\times \pi d^2/4$
$v=4Q/d^2\pi$
We then estimate the Reynolds number to be 
$R_e=4\rho Q/\pi d\eta$
$=4\times {10}^{3}kg{m}^{-3}\times Q/(3.14\times 1.25\times {10}^{-2}m\times {10}^{-3}Pas)$
$=1.019\times {10}^8{m}^{-3}sQ$
Since initially (a) 
$Q=0.48L/\min =8c{m}^3/s=8\times {10}^{-6}{m}^3{s}^{-1},$ e obtain,
$R_e=815$