Calculate the viscous force on a liquid drop of radius 0.3 mm moving with a velocity of `5 cms^(-1)` through a medium of viscosity `2 xx 10^(-5) Nm^(-2)s.`

Answer

To calculate the viscous force on a liquid drop moving through a medium, we can use the formula for viscous force (Fs) given by Stokes' law:

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

where:
- \( F_s \) = viscous force (in Newtons)
- \( \eta \) = coefficient of viscosity (in \( \text{N} \cdot \text{m}^{-2} \cdot \text{s} \))
- \( r \) = radius of the drop (in meters)
- \( v \) = velocity of the drop (in meters per second)

### Step-by-Step Solution:

1. **Convert the given values to SI units:**
- Radius \( r = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \)
- Velocity \( v = 5 \, \text{cm/s} = 5 \times 10^{-2} \, \text{m/s} \)
- Viscosity \( \eta = 2 \times 10^{-5} \, \text{N} \cdot \text{m}^{-2} \cdot \text{s} \)

2. **Substitute the values into the formula:**
\[ F_s = 6 \pi (2 \times 10^{-5}) (0.3 \times 10^{-3}) (5 \times 10^{-2}) \]

3. **Calculate the numerical values:**
- First, calculate \( 6 \pi \):
\[ 6 \pi \approx 6 \times 3.14 = 18.84 \]
- Now, calculate \( 0.3 \times 5 = 1.5 \):
\[ 0.3 \times 5 = 1.5 \]
- Now, multiply all the components:
\[ F_s = 18.84 \times (2 \times 10^{-5}) \times (1.5 \times 10^{-3}) \]
- Combine the powers of ten:
\[ 2 \times 1.5 = 3 \quad \text{and} \quad 10^{-5} \times 10^{-3} = 10^{-8} \]
- Thus, we have:
\[ F_s = 18.84 \times 3 \times 10^{-8} \]
- Calculate \( 18.84 \times 3 \):
\[ 18.84 \times 3 = 56.52 \]
- Therefore:
\[ F_s = 56.52 \times 10^{-8} \, \text{N} \]

4. **Convert to standard scientific notation:**
\[ F_s = 5.652 \times 10^{-7} \, \text{N} \]

### Final Answer:
The viscous force on the liquid drop is approximately:
\[ F_s \approx 5.65 \times 10^{-7} \, \text{N} \]