The terminal velocity `(v_t)` of the spherical rain drop depends on the radius (r) of the spherical rain drop as:

Options

A. `r^(1//2)`
B. `r`
C. `r^2`
D. `r^3`

Explanation

To determine how the terminal velocity \( v_t \) of a spherical raindrop depends on its radius \( r \), we can use the formula for terminal velocity derived from Stokes' law. The terminal velocity of a sphere falling through a viscous fluid is given by:

\[ v_t = \frac{2}{9} \cdot \frac{g r^2 (\rho_b - \rho_l)}{\eta} \]

Where:
- \( v_t \) = terminal velocity
- \( g \) = acceleration due to gravity
- \( r \) = radius of the raindrop
- \( \rho_b \) = density of the raindrop (body)
- \( \rho_l \) = density of the fluid (liquid)
- \( \eta \) = coefficient of viscosity of the fluid

### Step-by-step Solution:

1. **Identify the Variables**:
- The terminal velocity \( v_t \) depends on the radius \( r \) of the raindrop, the gravitational acceleration \( g \), the densities \( \rho_b \) and \( \rho_l \), and the viscosity \( \eta \).

2. **Write the Formula**:
- The formula for terminal velocity is given as:
\[ v_t = \frac{2}{9} \cdot \frac{g r^2 (\rho_b - \rho_l)}{\eta} \]

3. **Analyze the Dependence on Radius**:
- From the formula, we can see that \( v_t \) is proportional to \( r^2 \). This means that if the radius \( r \) of the raindrop increases, the terminal velocity \( v_t \) increases with the square of the radius.

4. **Conclusion**:
- Therefore, we conclude that the terminal velocity \( v_t \) of a spherical raindrop depends on the square of its radius \( r \).

### Final Expression:
\[ v_t \propto r^2 \]