Two drops of the same radius are falling through air with a steady velcoity of `5 cm s^(-1).` If the two drops coalesce, the terminal velocity would be

Options

A. `10 cm s^(-1)`
B. `2.5 cm s^(-1)`
C. `5xx(4)^(1//3) cm s^(-1)`
D. `5xxsqrt(3) cm s^(-1)`

Answer

To solve the problem of finding the terminal velocity of two coalesced drops, we can follow these steps:

### Step 1: Understand the relationship between terminal velocity and radius
The terminal velocity \( V_t \) of a spherical drop falling through a fluid is given by the formula:
\[ V_t = \frac{2}{9} \frac{R^2 g (\rho_s - \rho_f)}{\eta} \]
where:
- \( R \) is the radius of the drop,
- \( g \) is the acceleration due to gravity,
- \( \rho_s \) is the density of the drop,
- \( \rho_f \) is the density of the fluid,
- \( \eta \) is the viscosity of the fluid.

### Step 2: Initial conditions
Given that the two drops have the same radius \( r \) and are falling with a steady velocity of \( 5 \, \text{cm/s} \), we can denote the initial terminal velocity \( V_{t1} \) as:
\[ V_{t1} = 5 \, \text{cm/s} \]

### Step 3: Volume conservation during coalescence
When the two drops coalesce, the volume is conserved. The volume of a single drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the total volume of two drops is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]
This volume will equal the volume of the new drop with radius \( R \):
\[ \frac{4}{3} \pi R^3 = \frac{8}{3} \pi r^3 \]

### Step 4: Solve for the new radius \( R \)
Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = 2r^3 \]
Taking the cube root:
\[ R = r \cdot 2^{1/3} \]

### Step 5: Relate new terminal velocity to the old one
Since the terminal velocity is proportional to the square of the radius, we can write the ratio of the new terminal velocity \( V_{t2} \) to the initial terminal velocity \( V_{t1} \):
\[ \frac{V_{t2}}{V_{t1}} = \left(\frac{R}{r}\right)^2 \]
Substituting \( R = r \cdot 2^{1/3} \):
\[ \frac{V_{t2}}{V_{t1}} = \left(2^{1/3}\right)^2 = 2^{2/3} \]

### Step 6: Calculate the new terminal velocity
Now, substituting \( V_{t1} = 5 \, \text{cm/s} \):
\[ V_{t2} = V_{t1} \cdot 2^{2/3} = 5 \cdot 2^{2/3} \, \text{cm/s} \]

### Conclusion
The new terminal velocity after the two drops coalesce is:
\[ V_{t2} = 5 \cdot 2^{2/3} \, \text{cm/s} \]