d
Given: Diameter of water tap \( = \frac{2}{{\sqrt \pi  }}\,cm\)

\(\therefore \,\,Radius,\,r = \frac{1}{{\sqrt \pi  }} \times {10^{ - 2}}\,m\)

\(\frac{{dm}}{{dt}} = \rho AV\)

\(\frac{{15}}{{5 \times 60}} = {10^3} \times \pi {\left( {\frac{1}{{\sqrt \pi  }}} \right)^2} \times {10^{ - 4}}V\)

\( \Rightarrow V = 0.05\,m/s\)

Reynold's number, \({R_e} = \frac{{\rho Vr}}{n}\)

\( = \frac{{{{10}^3} \times 0.5 \times \frac{2}{{\sqrt \pi  }}{{10}^{ - 2}}}}{{{{10}^{ - 3}}}} \cong 5500\)