An air bubble of radius `5mm` rises through a vat of syrup at a steady speed `2mm//s`. If the syrup has a density of `1.4xx10^(3)kg//m^(3)`, what is its viscosity ?

Options

A. ` 1.15 kg//m-s`
B. 15 kg/m-s
C. `2.25 kg//m-s `
D. `19.05 kg//m-s `

Answer

To find the viscosity of the syrup through which the air bubble rises, we can use the formula for the terminal velocity of a sphere in a viscous fluid:

\[ v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]

Where:
- \( v \) = velocity of the bubble (2 mm/s = \( 2 \times 10^{-3} \) m/s)
- \( r \) = radius of the bubble (5 mm = \( 5 \times 10^{-3} \) m)
- \( \rho \) = density of the syrup (1.4 x \( 10^3 \) kg/m³)
- \( \sigma \) = density of the air (approximately 0 kg/m³)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( \eta \) = viscosity of the syrup (what we want to find)

### Step 1: Rearranging the formula to solve for viscosity (\( \eta \))

We can rearrange the formula to isolate \( \eta \):

\[ \eta = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{v} \]

### Step 2: Substituting the known values

Now we can substitute the known values into the equation:

- \( r = 5 \times 10^{-3} \, \text{m} \)
- \( v = 2 \times 10^{-3} \, \text{m/s} \)
- \( \rho = 1.4 \times 10^3 \, \text{kg/m}^3 \)
- \( \sigma = 0 \, \text{kg/m}^3 \)
- \( g = 9.81 \, \text{m/s}^2 \)

Substituting these values into the equation gives:

\[ \eta = \frac{2}{9} \frac{(5 \times 10^{-3})^2 (1.4 \times 10^3 - 0) \cdot 9.81}{2 \times 10^{-3}} \]

### Step 3: Calculating \( r^2 \)

Calculate \( r^2 \):

\[ r^2 = (5 \times 10^{-3})^2 = 25 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Plugging in the values and simplifying

Now we can plug in the values:

\[ \eta = \frac{2}{9} \cdot \frac{25 \times 10^{-6} \cdot 1.4 \times 10^3 \cdot 9.81}{2 \times 10^{-3}} \]

### Step 5: Performing the calculations

1. Calculate the numerator:

\[ 25 \times 10^{-6} \cdot 1.4 \times 10^3 \cdot 9.81 = 25 \cdot 1.4 \cdot 9.81 \times 10^{-3} = 343.5 \times 10^{-3} \]

2. Now substitute this back into the equation for \( \eta \):

\[ \eta = \frac{2}{9} \cdot \frac{343.5 \times 10^{-3}}{2 \times 10^{-3}} = \frac{2 \cdot 343.5 \times 10^{-3}}{18 \times 10^{-3}} = \frac{687}{18} \approx 38.33 \, \text{kg/m·s} \]

### Final Answer

The viscosity of the syrup is approximately \( 38.33 \, \text{kg/m·s} \).

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