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Question: The flow rate of water from a tap of diameter \(1.25\,cm\) is \(3\,L\,per\,\min \). The coefficient ...

The flow rate of water from a tap of diameter 1.25 cm1.25\,cm is 3 L per min⁡3\,L\,per\,\min . The coefficient of viscosity of water is 10−3 Pa s{10^{ - 3}}\,Pa\,s. The nature of flow is
A. Unsteady
B. Turbulent
C. Laminar
D. None of these

Explanation

Solution

Here, in the question, we are given a water of viscosity is 10−3 Pa s{10^{ - 3}}\,Pa\,s. Here, we will calculate the Reynolds number to find out the nature of the flow of the water. If the Reynolds number is less than 20002000, then the flow of water will be laminar. On the other hand, if the Reynolds number is greater than 20002000, then the flow of water will be turbulent.

Formula used:
The formula of Reynolds number used to find the nature of flow is given below
Re=4ρQπDη{R_e} = \dfrac{{4\rho Q}}{{\pi D\eta }}
Here, Re{R_e} is the Reynolds number, ρ\rho is the density of the fluid, QQ is the volume of the fluid flowing out per second, DD of the object through which the fluid will flow and η\eta is the coefficient of viscosity of the fluid.

Complete step by step answer:
The terms given below are given in the question
The diameter of the tap, D=1.25 cm=1.25×10−2mD = 1.25\,cm = 1.25 \times {10^{ - 2}}m
Density of the water, ρ=103kgm−3\rho = {10^3}kg{m^{ - 3}}
Coefficient of viscosity of the flow, η=10−3 Pa s\eta = {10^{ - 3}}\,Pa\,s
Now, the volume of the water flowing out per second is given below
Q=3L per min⁡Q = 3L\,per\,\min
⇒ Q=3×10−3 m360 s\Rightarrow \,Q = \dfrac{{3 \times {{10}^{ - 3}}\,{m^3}}}{{60\,s}}
⇒ Q=5×10−5 m3 s−1\Rightarrow \,Q = 5 \times {10^{ - 5}}\,{m^3}\,{s^{ - 1}}
Now, the nature of the flow can be calculated by using the formula of the Reynolds number which is given below
Re=4ρQπDη{R_e} = \dfrac{{4\rho Q}}{{\pi D\eta }}
⇒ Re=4×103kg m−3×5×10−5m3 s−13.14×1.25×10−2m×10−3 Pa s\Rightarrow \,{R_e} = \dfrac{{4 \times {{10}^3}kg\,{m^{ - 3}} \times 5 \times {{10}^{ - 5}}{m^3}\,{s^{ - 1}}}}{{3.14 \times 1.25 \times {{10}^{ - 2}}m \times {{10}^{ - 3}}\,Pa\,s}}
∴ Re=5095\therefore \,{R_e} = 5095
Now, we can see that the value of the Reynolds number is greater than 20002000. This is the condition of turbulence flow of liquid. Therefore, the flow of the water is turbulent in nature.

Hence, option B is the correct answer.

Note: For solving these types of questions, we must first convert the larger units into smaller units. Here, the volume of the liquid is given in liters that is why we have converted it in meters. Also, the flow will not be unsteady because the unsteady flow depends on time and we don’t have the value of time in the question.