The ratio of the terminal velocities of two drops of radii R and `R//2` is

Options

A. `2`
B. `1`
C. `1//2`
D. `4`

Explanation

To find the ratio of the terminal velocities of two drops of radii R and R/2, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
Terminal velocity (Vt) is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. For a droplet falling through a viscous medium, the forces acting on it are gravitational force and drag force.

2. **Forces Acting on the Droplet**:
- The gravitational force (Fg) acting on the droplet is given by:
\[ F_g = mg = \rho V g = \rho \left(\frac{4}{3} \pi r^3\right) g \]
where \( \rho \) is the density of the liquid, \( V \) is the volume of the droplet, and \( g \) is the acceleration due to gravity.

- The drag force (Fd) acting on the droplet due to viscosity is given by:
\[ F_d = 6 \pi \eta r V_t \]
where \( \eta \) is the viscosity of the fluid, \( r \) is the radius of the droplet, and \( V_t \) is the terminal velocity.

3. **Setting Up the Equation**:
At terminal velocity, the gravitational force equals the drag force:
\[ F_g = F_d \]
Therefore, we have:
\[ \rho \left(\frac{4}{3} \pi r^3\right) g = 6 \pi \eta r V_t \]

4. **Solving for Terminal Velocity (Vt)**:
Rearranging the equation gives:
\[ V_t = \frac{\rho \left(\frac{4}{3} \pi r^3\right) g}{6 \pi \eta r} \]
Simplifying this, we find:
\[ V_t = \frac{2 g r^2 \rho}{9 \eta} \]
This shows that the terminal velocity is directly proportional to the square of the radius of the droplet:
\[ V_t \propto r^2 \]

5. **Calculating the Ratio of Terminal Velocities**:
Let’s denote the terminal velocities of the two droplets as \( V_{t1} \) for radius \( R \) and \( V_{t2} \) for radius \( \frac{R}{2} \):
\[ V_{t1} \propto R^2 \quad \text{and} \quad V_{t2} \propto \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
Now, we can find the ratio:
\[ \frac{V_{t1}}{V_{t2}} = \frac{R^2}{\frac{R^2}{4}} = 4 \]

### Final Answer:
The ratio of the terminal velocities of the two drops of radii \( R \) and \( \frac{R}{2} \) is:
\[ \frac{V_{t1}}{V_{t2}} = 4 \]