Estimate the speed of vertically falling rain drops from the following data. Radius of the drops=0.02cm, viscosity of air `=1.8xx10^-4 poise, g=9.9ms^-2 and ` density of water `=1000 kg m^-3`.

Explanation

To estimate the speed of vertically falling raindrops, we can use Stokes' law, which relates the viscous force acting on a sphere moving through a fluid to its velocity. Here’s the step-by-step solution:

### Step 1: Convert the radius of the raindrop to meters
The radius of the raindrop is given as 0.02 cm. We need to convert this to meters.

\[ r = 0.02 \, \text{cm} = 0.02 \times 10^{-2} \, \text{m} = 0.0002 \, \text{m} \]

### Step 2: Write down the formula for the viscous force
According to Stokes' law, the viscous force \( F \) acting on a sphere moving through a viscous fluid is given by:

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

where:
- \( \mu \) is the viscosity of the fluid,
- \( r \) is the radius of the sphere,
- \( v \) is the velocity of the sphere.

### Step 3: Write down the formula for the weight of the raindrop
The weight \( W \) of the raindrop can be expressed as:

\[ W = \text{mass} \times g = \text{density} \times \text{volume} \times g \]

The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3 \]

Thus, the weight becomes:

\[ W = \rho \left( \frac{4}{3} \pi r^3 \right) g \]

### Step 4: Equate the viscous force and the weight
At terminal velocity, the viscous force equals the weight of the raindrop:

\[ 6 \pi \mu r v = \rho \left( \frac{4}{3} \pi r^3 \right) g \]

### Step 5: Simplify the equation
We can cancel \( \pi \) and \( r \) from both sides of the equation:

\[ 6 \mu v = \frac{4}{3} r^2 \rho g \]

### Step 6: Solve for terminal velocity \( v \)
Rearranging the equation gives:

\[ v = \frac{4 r^2 \rho g}{18 \mu} = \frac{2 r^2 \rho g}{9 \mu} \]

### Step 7: Substitute the known values
Substituting the values:
- \( r = 0.0002 \, \text{m} \)
- \( \rho = 1000 \, \text{kg/m}^3 \)
- \( g = 9.9 \, \text{m/s}^2 \)
- \( \mu = 1.8 \times 10^{-5} \, \text{N s/m}^2 \)

\[ v = \frac{2 \times (0.0002)^2 \times 1000 \times 9.9}{9 \times 1.8 \times 10^{-5}} \]

Calculating this step-by-step:

1. Calculate \( (0.0002)^2 = 4 \times 10^{-8} \, \text{m}^2 \).
2. Calculate \( 2 \times 4 \times 10^{-8} \times 1000 \times 9.9 = 7.92 \times 10^{-5} \).
3. Calculate \( 9 \times 1.8 \times 10^{-5} = 1.62 \times 10^{-4} \).
4. Finally, divide \( 7.92 \times 10^{-5} \) by \( 1.62 \times 10^{-4} \):

\[ v \approx 0.488 \, \text{m/s} \]

### Final Answer
The estimated speed of the vertically falling raindrops is approximately **0.488 m/s**.