A lead sphere of `1.0 mm` diameter and relative density `11.20` attains a terminal velocity of `0.7 cm s^(-1)` in a liquid of relative density `1.26`.
What is the value of the Reynolds number?

Options

A. `0.01`
B. `0.03`
C. `0.15`
D. `0.26`

Explanation

To find the Reynolds number for the lead sphere, we will follow these steps:

### Step 1: Identify the given data
- Diameter of the lead sphere, \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \)
- Relative density of lead, \( \text{RD}_{\text{lead}} = 11.20 \)
- Terminal velocity, \( V = 0.7 \, \text{cm/s} = 0.7 \times 10^{-2} \, \text{m/s} \)
- Relative density of the liquid, \( \text{RD}_{\text{liquid}} = 1.26 \)

### Step 2: Calculate the densities
- Density of lead, \( \rho_{\text{lead}} = \text{RD}_{\text{lead}} \times 1000 \, \text{kg/m}^3 = 11.20 \times 1000 = 11200 \, \text{kg/m}^3 \)
- Density of the liquid, \( \rho_{\text{liquid}} = \text{RD}_{\text{liquid}} \times 1000 \, \text{kg/m}^3 = 1.26 \times 1000 = 1260 \, \text{kg/m}^3 \)

### Step 3: Calculate the radius of the sphere
- Radius of the sphere, \( r = \frac{d}{2} = \frac{1.0 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \, \text{m} \)

### Step 4: Calculate the coefficient of viscosity, \( \mu \)
Using the formula for terminal velocity:
\[ V = \frac{2}{9} r^2 \frac{(\rho_{\text{lead}} - \rho_{\text{liquid}}) g}{\mu} \]
Rearranging gives:
\[ \mu = \frac{2}{9} r^2 \frac{(\rho_{\text{lead}} - \rho_{\text{liquid}}) g}{V} \]
Substituting the values:
- \( g \approx 10 \, \text{m/s}^2 \)

Calculating \( \mu \):
\[ \mu = \frac{2}{9} (0.5 \times 10^{-3})^2 \frac{(11200 - 1260) \times 10}{0.7 \times 10^{-2}} \]
Calculating the values:
\[ \mu = \frac{2}{9} (0.25 \times 10^{-6}) \frac{(9980) \times 10}{0.007} \]
\[ \mu = \frac{2}{9} (0.25 \times 10^{-6}) \frac{99800}{0.007} \]
\[ \mu = \frac{2 \times 0.25 \times 99800}{9 \times 0.007} \times 10^{-6} \]
\[ \mu \approx 0.77 \, \text{N s/m}^2 \]

### Step 5: Calculate the Reynolds number
Using the formula:
\[ Re = \frac{\rho_{\text{liquid}} V d}{\mu} \]
Substituting the values:
\[ Re = \frac{1260 \times 0.7 \times 10^{-2} \times 1.0 \times 10^{-3}}{0.77} \]
Calculating:
\[ Re = \frac{1260 \times 0.007 \times 0.001}{0.77} \]
\[ Re = \frac{0.00882}{0.77} \approx 0.0114 \]

Thus, the Reynolds number is approximately \( 0.01 \).

### Final Answer
The value of the Reynolds number is \( \approx 0.01 \).

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