Question

The water flows from a tap of diameter $1.25cm$ with a rate of $5\times {10}^{-5}{m}^3{s}^{-1}$. The density and coefficient of viscosity of water are ${10}^{3}kg{m}^{-3}$ and ${10}^{-3}Pas$ respectively. The flow of water is

• A. Steady with Reynold's number $5100$
• B. Turbulent with Reynold's number $5100$
• C. Steady with Reynold's number $3900$
• D. Turbulent with Reynold's number $3900$

Answer 1

The correct option is B Turbulent with Reynold's number $5100$

Here, diameter D=1.25cm=1.25×10−2mD=1.25cm=1.25×10−2m
density of water ρ=103kgm−3ρ=103kgm−3
coefficient of viscosity η=10−3Pasη=10−3Pas
rate of flow of water Q=5×10−5m3s−1Q=5×10−5m3s−1
Reynold's number NR=vρDη....(i)NR=vρDη....(i)
where vv is the speed of flow
Rate of flow of water Q=Q= Area of cross section timestimes speed of flow
Q=πD24×v⇒v=4QπD2Q=πD24×v⇒v=4QπD2
Substituting the value of vv in eqn, (i), we get
NR=4QρDNRη=4QρπDηNR=4QρDNRη=4QρπDη
substituting the values, we get
NR=4×5×10−5×103(227)×1.25×10−2×10−3=5100NR=4×5×10−5×103(227)×1.25×10−2×10−3=5100
For NR>3000NR>3000, the flow is turbulent
hence, the flow of water is turbulent with Reynold's number 5100