An air bubble of 1 cm radius is rising at a steady rate of `2.00ms^-1` through a liquid of density `1.5gcm^-3`. Neglect density of air. If `g=1000cms^-2`, then the coeffieciet of viscosity of the liquid is

Options

A. `0.166xx10^(3)` poise
B. `166xx10^(3)` poise
C. `1.66xx10^(3)` poise
D. `16.6xx10^(3)` poise

Answer

To solve the problem, we will use the formula for terminal velocity of a sphere rising in a viscous fluid, which is given by:

\[ v_t = \frac{2}{9} \frac{r^2 g (\rho_s - \rho_l)}{\eta} \]

Where:
- \( v_t \) = terminal velocity (in cm/s)
- \( r \) = radius of the bubble (in cm)
- \( g \) = acceleration due to gravity (in cm/s²)
- \( \rho_s \) = density of the sphere (air bubble, which we neglect)
- \( \rho_l \) = density of the liquid (in g/cm³)
- \( \eta \) = coefficient of viscosity (in poise)

### Step 1: Identify the given values
- Radius of the bubble, \( r = 1 \, \text{cm} \)
- Terminal velocity, \( v_t = 2 \, \text{m/s} = 200 \, \text{cm/s} \) (conversion from m/s to cm/s)
- Density of the liquid, \( \rho_l = 1.5 \, \text{g/cm}^3 \)
- Acceleration due to gravity, \( g = 1000 \, \text{cm/s}^2 \)

### Step 2: Substitute the values into the terminal velocity formula
Since we are neglecting the density of air, we can assume \( \rho_s \) to be negligible. Thus, the formula simplifies to:

\[ 200 = \frac{2}{9} \frac{(1)^2 (1000) (0 - 1.5)}{\eta} \]

### Step 3: Rearrange the equation to solve for \( \eta \)
Rearranging the equation gives:

\[ \eta = \frac{2}{9} \frac{(1)^2 (1000) (-1.5)}{200} \]

### Step 4: Calculate the coefficient of viscosity
Now, we simplify the expression:

1. Calculate the numerator:
\[ 2 \times 1^2 \times 1000 \times (-1.5) = -3000 \]

2. Now substitute this back into the equation:
\[ \eta = \frac{-3000}{9 \times 200} \]

3. Calculate \( 9 \times 200 = 1800 \):
\[ \eta = \frac{-3000}{1800} \]
\[ \eta = -\frac{3000}{1800} = -\frac{5}{3} \]

4. Since viscosity cannot be negative, we take the absolute value:
\[ \eta = \frac{5}{3} \, \text{poise} \]

### Final Answer
To express this in a more standard form:
\[ \eta = 1.6667 \, \text{poise} \approx 1.67 \, \text{poise} \]