A small glass sphere of radius `2 x 10^(-3) m `is moving through a liquid of viscosity 0.11 decapaise. Calculate the viscos force acting on it if the speed of the ball is `0.05 ms ^(-1)`

Explanation

To calculate the viscous force acting on a small glass sphere moving through a liquid, we can use the formula for viscous force, which is given by:

\[ F = 6 \pi \eta r v \]

Where:
- \( F \) is the viscous force,
- \( \eta \) is the viscosity of the liquid,
- \( r \) is the radius of the sphere,
- \( v \) is the speed of the sphere.

### Step 1: Convert viscosity from decapoise to Pascal-seconds
The viscosity is given as \( 0.11 \) decapoise. We need to convert this to Pascal-seconds (Pa·s) since \( 1 \) decapoise = \( 1 \) Pa·s.

\[ \eta = 0.11 \, \text{decapoise} = 0.11 \, \text{Pa·s} \]

### Step 2: Identify the values
- Radius \( r = 2 \times 10^{-3} \, \text{m} \)
- Speed \( v = 0.05 \, \text{m/s} \)
- Viscosity \( \eta = 0.11 \, \text{Pa·s} \)

### Step 3: Substitute the values into the formula
Now we can substitute these values into the viscous force formula:

\[ F = 6 \pi (0.11) (2 \times 10^{-3}) (0.05) \]

### Step 4: Calculate the force
Calculating the values step by step:

1. Calculate \( 6 \pi \):
\[ 6 \pi \approx 6 \times 3.14 \approx 18.84 \]

2. Multiply by the viscosity:
\[ 18.84 \times 0.11 \approx 2.0724 \]

3. Multiply by the radius:
\[ 2.0724 \times (2 \times 10^{-3}) \approx 0.0041448 \]

4. Finally, multiply by the speed:
\[ 0.0041448 \times 0.05 \approx 0.00020724 \, \text{N} \]

### Step 5: Express in scientific notation
\[ F \approx 0.21 \times 10^{-3} \, \text{N} = 2.1 \times 10^{-4} \, \text{N} \]

### Final Answer
The viscous force acting on the sphere is approximately \( 0.21 \times 10^{-3} \, \text{N} \).

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