The rate of water fow from a tap of diameter `1.2cm` is `2.4xx10^(-5)m^(3)s^(-1)`. The density and coefficient of viscosity of water are `10^(3)kgm^(-3)` and `10^(-3)` Pa-s respectively. The flow of water has Reynold's number of `C/(pi)`.Find C.

Answer

8000

Explanation

Reynold.s number, $ \mathrm {N}_{\mathrm{R}}=\frac{/mathrm{v}\rho
\mathrm{D}}{\eta}$..............i
Rate of flow of water $ =$ area of cross section
$\times$ speed of flow
$\therefore Q=\frac{\piD^{2}}{4}\timesv$
$\therefore\quadv=\frac{4Q}{\piD^({2}}$............ii
From equation i and ii
$\mathrm{N}_{\mathrm{R}}=\frac
{\rho\mthrm{D}}{\eta}\times\frac{4\mathrm{Q}}{\pi\mathrm{D}^(2){2}}=\frac{4\rho\mathrm{Q}}{\pi\eta\mthrm{D}}$
$\RightarrowC=\frac{4\rhoQ|{nD}$
$\therefore$ $\quadC=\frac{4\times10^({3}\times2.4\times10^{-5}}{10^
{-3}\times 1.2\times10^(2){-2}}=8000$