The terminal velocity of a water drop of radius `0.01` mm falling through air is `1.12` cm/s. If the density of air is neglected, the coefficient of viscosity of air is
[density of water `=10^(3) kg//m^(3), g=9.8m//s^(2)]`

Options

A. `2xx10^(-5) N s//m^(2)`
B. `1.9xx10^(-5)N s//m^(2)`
C. `1.8xx10^(-5)N s//m^(2)`
D. `1.7xx10^(-5)N s//m^(2)`

Answer

To find the coefficient of viscosity of air given the terminal velocity of a water drop, we can use the formula for terminal velocity in a viscous medium. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Droplet
When a droplet falls through a fluid (in this case, air), three main forces act on it:
1. The gravitational force (weight) acting downward.
2. The buoyant force acting upward due to the displacement of air.
3. The drag force acting upward due to the viscosity of the air.

At terminal velocity, these forces balance each other, so the net force is zero.

### Step 2: Write the Equation for Terminal Velocity
The terminal velocity \( v_t \) of a spherical droplet can be expressed as:
\[ v_t = \frac{2}{9} \frac{(ρ - σ) g r^2}{η} \]
Where:
- \( ρ \) = density of the droplet (water)
- \( σ \) = density of the fluid (air, which we neglect)
- \( g \) = acceleration due to gravity
- \( r \) = radius of the droplet
- \( η \) = coefficient of viscosity of the fluid (air)

### Step 3: Substitute Known Values
Given:
- Radius \( r = 0.01 \) mm = \( 0.01 \times 10^{-3} \) m = \( 10^{-5} \) m
- Terminal velocity \( v_t = 1.12 \) cm/s = \( 1.12 \times 10^{-2} \) m/s
- Density of water \( ρ = 10^3 \) kg/m³
- Neglecting the density of air \( σ = 0 \)

Substituting these values into the equation:
\[ v_t = \frac{2}{9} \frac{(10^3 - 0) \cdot 9.8 \cdot (10^{-5})^2}{η} \]

### Step 4: Rearranging to Solve for Viscosity \( η \)
Rearranging the equation to solve for \( η \):
\[ η = \frac{2}{9} \frac{(10^3) \cdot 9.8 \cdot (10^{-5})^2}{v_t} \]

### Step 5: Substitute the Terminal Velocity
Now substituting \( v_t = 1.12 \times 10^{-2} \) m/s into the equation:
\[ η = \frac{2}{9} \frac{(10^3) \cdot 9.8 \cdot (10^{-5})^2}{1.12 \times 10^{-2}} \]

### Step 6: Calculate the Coefficient of Viscosity
Calculating the values:
1. Calculate \( (10^{-5})^2 = 10^{-10} \).
2. Substitute into the equation:
\[ η = \frac{2}{9} \cdot \frac{(10^3) \cdot 9.8 \cdot 10^{-10}}{1.12 \times 10^{-2}} \]
3. Simplifying:
\[ η = \frac{2 \cdot 9.8 \cdot 10^{-7}}{9 \cdot 1.12} \]
4. Calculate \( 9.8 \cdot 10^{-7} \) and divide by \( 10.08 \) (which is \( 9 \cdot 1.12 \)):
\[ η \approx 1.98 \times 10^{-5} \text{ N·s/m²} \]

### Final Answer
The coefficient of viscosity of air is approximately:
\[ η \approx 2 \times 10^{-5} \text{ N·s/m²} \]