A glass tube of uniform cross section is connected to a tap with a rubber tube. The tap is opened slowly. Initially the flow of water in the tube is streamline. What should be the speed of fl.ow of water to convert h into a turbulent flow?
[Given : radius of the tube=1cm, `eta=1xx10^(-3)` Pas and Reynold's number=2500]

Options

A. 0.15m/s
B. 0.2 m/s
C. 0.125 m/s
D. 0.3 m/s

Explanation

To solve the problem of determining the speed of water flow required to convert from laminar to turbulent flow in a glass tube, we can use the concept of Reynolds number (Re). The Reynolds number is given by the formula:

\[ Re = \frac{\rho V D}{\mu} \]

Where:
- \(Re\) is the Reynolds number (given as 2500),
- \(\rho\) is the density of the fluid (for water, approximately \(1000 \, \text{kg/m}^3\)),
- \(V\) is the velocity of the fluid,
- \(D\) is the diameter of the tube,
- \(\mu\) is the dynamic viscosity of the fluid (given as \(1 \times 10^{-3} \, \text{Pas}\)).

### Step 1: Calculate the diameter of the tube
The radius of the tube is given as \(1 \, \text{cm}\). To find the diameter \(D\):

\[ D = 2 \times \text{radius} = 2 \times 1 \, \text{cm} = 2 \, \text{cm} = 0.02 \, \text{m} \]

### Step 2: Rearrange the Reynolds number formula to solve for \(V\)
We can rearrange the formula to isolate \(V\):

\[ V = \frac{Re \cdot \mu}{\rho \cdot D} \]

### Step 3: Substitute the known values into the equation
Now we can substitute the known values into the equation:

- \(Re = 2500\)
- \(\mu = 1 \times 10^{-3} \, \text{Pas}\)
- \(\rho = 1000 \, \text{kg/m}^3\)
- \(D = 0.02 \, \text{m}\)

Substituting these values gives:

\[ V = \frac{2500 \cdot (1 \times 10^{-3})}{1000 \cdot 0.02} \]

### Step 4: Calculate the velocity \(V\)
Now we perform the calculations:

\[ V = \frac{2500 \cdot 1 \times 10^{-3}}{1000 \cdot 0.02} = \frac{2.5}{20} = 0.125 \, \text{m/s} \]

### Conclusion
The speed of flow of water required to convert from laminar to turbulent flow is:

\[ \boxed{0.125 \, \text{m/s}} \]