If it takes 5 minutes to fill a 15 litre bucket from a water tap diameter `(2)/(sqrt(pi))` cm then the raynolds number for the flow is (density of water `=10^(3)kg//m^(3)` and viscosity of water `=10^(-3)Pa`.s) close to

Options

A. `11,000`
B. `550`
C. `1100`
D. `5500`

Explanation

To solve the problem of finding the Reynolds number for the flow from the water tap, we will follow these steps:

### Step 1: Understand the Given Data
- Volume of the bucket (V) = 15 liters = 15 × 10^(-3) m³ (since 1 liter = 10^(-3) m³)
- Time taken to fill the bucket (t) = 5 minutes = 5 × 60 = 300 seconds
- Diameter of the tap (D) = (2 / √π) cm = (2 / √π) × 10^(-2) m
- Density of water (ρ) = 10^3 kg/m³
- Viscosity of water (η) = 10^(-3) Pa·s

### Step 2: Calculate the Volume Flow Rate (Q)
The volume flow rate (Q) can be calculated as:
\[ Q = \frac{V}{t} \]
Substituting the values:
\[ Q = \frac{15 \times 10^{-3} \text{ m}^3}{300 \text{ s}} = 5 \times 10^{-5} \text{ m}^3/\text{s} \]

### Step 3: Calculate the Cross-Sectional Area (A)
The cross-sectional area (A) of the tap can be calculated using the formula:
\[ A = \frac{\pi D^2}{4} \]
Substituting the diameter in meters:
\[ D = \frac{2}{\sqrt{\pi}} \times 10^{-2} \text{ m} \]
Calculating the area:
\[ A = \frac{\pi \left(\frac{2}{\sqrt{\pi}} \times 10^{-2}\right)^2}{4} = \frac{\pi \cdot \frac{4}{\pi} \cdot 10^{-4}}{4} = 10^{-4} \text{ m}^2 \]

### Step 4: Calculate the Velocity (v)
Using the relationship between flow rate, area, and velocity:
\[ Q = A \cdot v \]
We can rearrange this to find the velocity:
\[ v = \frac{Q}{A} \]
Substituting the values:
\[ v = \frac{5 \times 10^{-5} \text{ m}^3/\text{s}}{10^{-4} \text{ m}^2} = 0.5 \text{ m/s} \]

### Step 5: Calculate the Reynolds Number (Re)
The Reynolds number is given by the formula:
\[ Re = \frac{\rho v D}{\eta} \]
Substituting the known values:
\[ Re = \frac{(10^3 \text{ kg/m}^3) \cdot (0.5 \text{ m/s}) \cdot \left(\frac{2}{\sqrt{\pi}} \times 10^{-2} \text{ m}\right)}{10^{-3} \text{ Pa·s}} \]

### Step 6: Simplify the Expression
Calculating the numerator:
\[ Re = \frac{(10^3) \cdot (0.5) \cdot \left(\frac{2}{\sqrt{\pi}} \times 10^{-2}\right)}{10^{-3}} \]
\[ = \frac{1000 \cdot 0.5 \cdot 2 \cdot 10^{-2}}{10^{-3} \cdot \sqrt{\pi}} \]
\[ = \frac{1000 \cdot 1 \cdot 10^{-2}}{10^{-3} \cdot \sqrt{\pi}} \]
\[ = \frac{10^3}{10^{-3} \cdot \sqrt{\pi}} = \frac{10^6}{\sqrt{\pi}} \]

### Step 7: Calculate the Final Value
Using the approximate value of π (≈ 3.14):
\[ \sqrt{\pi} \approx 1.772 \]
Thus:
\[ Re \approx \frac{10^6}{1.772} \approx 564,000 \]

### Final Answer
The Reynolds number for the flow is approximately **5500**.