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

The Reynolds number is a dimensionless quantity used to predict flow patterns in fluid dynamics. It is given by the formula:

$$ \text{Re} = \frac{\rho v d}{\eta} $$

where:

• $$\rho$$ is the density of the fluid,
• $$v$$ is the flow velocity,
• $$d$$ is the diameter of the pipe,
• $$\eta$$ is the dynamic viscosity.

Given:

• Time to fill the bucket, $$t = 5$$ minutes = $$5 \times 60 = 300$$ seconds,
• Volume of bucket, $$V = 15$$ litres = $$15 \times 10^{-3}$$ m$$^3$$ (since 1 litre = $$10^{-3}$$ m$$^3$$),
• Diameter of tap, $$d = \frac{2}{\sqrt{\pi}}$$ cm,
• Density of water, $$\rho = 10^3$$ kg/m$$^3$$,
• Viscosity of water, $$\eta = 10^{-3}$$ Pa.s.

First, convert the diameter to meters because SI units are required. Since 1 cm = 0.01 m,

$$ d = \frac{2}{\sqrt{\pi}} \times 10^{-2} \text{ m} = \frac{0.02}{\sqrt{\pi}} \text{ m}. $$

Next, calculate the cross-sectional area $$A$$ of the tap. Since the tap is circular,

$$ A = \pi \left( \frac{d}{2} \right)^2. $$

Substitute $$d$$:

$$ \frac{d}{2} = \frac{0.02}{2\sqrt{\pi}} = \frac{0.01}{\sqrt{\pi}} \text{ m}, $$

so,

$$ A = \pi \left( \frac{0.01}{\sqrt{\pi}} \right)^2 = \pi \times \frac{(0.01)^2}{\pi} = \pi \times \frac{0.0001}{\pi} = 0.0001 \text{ m}^2 = 10^{-4} \text{ m}^2. $$

Now, find the volume flow rate $$Q$$, which is volume divided by time:

$$ Q = \frac{V}{t} = \frac{15 \times 10^{-3}}{300} = \frac{0.015}{300} = 0.00005 \text{ m}^3/\text{s} = 5 \times 10^{-5} \text{ m}^3/\text{s}. $$

The flow velocity $$v$$ is given by $$Q = A \times v$$, so

$$ v = \frac{Q}{A} = \frac{5 \times 10^{-5}}{10^{-4}} = \frac{5 \times 10^{-5}}{1 \times 10^{-4}} = 5 \times 10^{-1} = 0.5 \text{ m/s}. $$

Now, substitute all values into the Reynolds number formula:

$$ \text{Re} = \frac{\rho v d}{\eta} = \frac{(10^3) \times (0.5) \times \left( \frac{0.02}{\sqrt{\pi}} \right)}{10^{-3}}. $$

First, compute the numerator:

$$ \rho v d = 1000 \times 0.5 \times \frac{0.02}{\sqrt{\pi}} = 500 \times \frac{0.02}{\sqrt{\pi}} = \frac{10}{\sqrt{\pi}}. $$

Then, divide by $$\eta$$:

$$ \text{Re} = \frac{\frac{10}{\sqrt{\pi}}}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times \frac{1}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times 10^3 = \frac{10000}{\sqrt{\pi}}. $$

Now, compute the numerical value. Using $$\pi \approx 3.1416$$, $$\sqrt{\pi} \approx \sqrt{3.1416} \approx 1.77245$$, so

$$ \text{Re} = \frac{10000}{1.77245} \approx 5641.89. $$

Comparing with the options:

• A. 5500
• B. 550
• C. 1100
• D. 11000

The value 5641.89 is closest to 5500, with a difference of approximately 141.89, while the other options are farther away (550 is too small, 1100 is smaller, and 11000 is much larger). Therefore, the Reynolds number is close to 5500.

Hence, the correct answer is Option A.