Explanation

(b) Here, diameter $D = 1.25cm = 1.25 \times 10^{- 2}m$
Density of water $\rho = 10^{3}kgm^{- 3}$
Coefficient of viscosity $\eta = 10^{- 3}$ Pas
Rate of flow of water $Q = 5 \times 10^{- 5}m^{3}s^{- 1}$
Reynold’s number $N_{R} = \frac{v\rho D}{\eta}$ …(i)
Where $v$ is the speed of flow
Rate of flow of water $Q =$ Area of cross section $\times$ speed of flow
$Q = \frac{\pi D^{2}}{4} \times v \Rightarrow v = \frac{4Q}{\pi D^{2}}$
Substituting the value of $v$ in eqn. (i), we get
$N_{R} = \frac{4Q\rho D}{\pi D^{2}\eta} = \frac{4Q\rho}{\pi D\eta}$
Substituting the values, we get
$N_{R} = \frac{4 \times 5 \times 10^{- 5} \times 10^{3}}{\left( \frac{22}{7} \right) \times 1.25 \times 10^{- 2} \times 10^{- 3}} \approx 5100$
For $N_{R} > 3000$, the flow is turbulent
Hence, the flow of water is turbulent with Reynold’s number 5100