A raindrop reaching the ground with terminal velocity has momentum p. Another drop of twice the radius, also reaching the ground with terminal velocity, will have momentum :

Options

A. ` 4 p `
B. `8 p `
C. `16 p `
D. `32 p `

Explanation

To solve the problem, we need to determine the momentum of a raindrop with twice the radius of another raindrop, both of which are falling with terminal velocity.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( V_t \) of a sphere falling through a viscous fluid is given by the equation:
\[ V_t = \frac{2}{9} \cdot \frac{r^2 (\rho - \sigma) g}{\eta} \]
where:
- \( r \) = radius of the drop
- \( \rho \) = density of the drop
- \( \sigma \) = density of the fluid
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity

2. **Momentum of the First Drop**:
The momentum \( p \) of the first drop (with radius \( r \)) is given by:
\[ p = m \cdot V_t \]
where \( m \) is the mass of the drop. The mass can be expressed as:
\[ m = \text{Volume} \cdot \text{Density} = \frac{4}{3} \pi r^3 \cdot \rho \]
Thus, the momentum becomes:
\[ p = \left(\frac{4}{3} \pi r^3 \cdot \rho\right) \cdot V_t \]

3. **Calculating Terminal Velocity for the Second Drop**:
For the second drop with radius \( 2r \):
\[ V'_t = \frac{2}{9} \cdot \frac{(2r)^2 (\rho - \sigma) g}{\eta} = \frac{2}{9} \cdot \frac{4r^2 (\rho - \sigma) g}{\eta} = 4 \cdot V_t \]
Therefore, the terminal velocity of the second drop is four times that of the first drop.

4. **Mass of the Second Drop**:
The mass of the second drop is:
\[ m' = \frac{4}{3} \pi (2r)^3 \cdot \rho = \frac{4}{3} \pi \cdot 8r^3 \cdot \rho = 8 \cdot m \]

5. **Calculating Momentum of the Second Drop**:
The momentum \( p' \) of the second drop is:
\[ p' = m' \cdot V'_t = (8m) \cdot (4V_t) = 32m \cdot V_t \]
Since \( p = m \cdot V_t \), we can substitute:
\[ p' = 32p \]

### Conclusion:
The momentum of the second drop (with twice the radius) is \( 32p \).