Two rain drops of same radius r falling with terminal velocity v merge and from a bigger drop of radius R. The terminal velocity of the bigger drop is

Options

A. `v(R )/(r )`
B. `v(R^(2))/(r^(2))`
C. v
D. 2v

Explanation

To solve the problem of finding the terminal velocity of a bigger drop formed by merging two smaller drops, we can follow these steps:

### Step 1: Understand the relationship between terminal velocity and radius
The terminal velocity \( V \) of a spherical drop falling through a fluid is given by the formula:

\[ V = \frac{2}{9} \frac{r^2 ( \rho - \sigma ) g}{\eta} \]

where:
- \( r \) is the radius of the drop,
- \( \rho \) is the density of the liquid,
- \( \sigma \) is the density of the substance (the drop),
- \( g \) is the acceleration due to gravity,
- \( \eta \) is the coefficient of viscosity of the fluid.

### Step 2: Write the terminal velocity for the smaller drops
For the smaller drops of radius \( r \), the terminal velocity \( v \) is:

\[ v = \frac{2}{9} \frac{r^2 ( \rho - \sigma ) g}{\eta} \]

### Step 3: Write the terminal velocity for the bigger drop
When two smaller drops merge to form a bigger drop of radius \( R \), we can express the terminal velocity \( V_R \) of the bigger drop as:

\[ V_R = \frac{2}{9} \frac{R^2 ( \rho - \sigma ) g}{\eta} \]

### Step 4: Relate the radii of the drops
When two drops of radius \( r \) merge, the volume of the new drop must equal the sum of the volumes of the two smaller drops:

\[ \frac{4}{3} \pi R^3 = 2 \left( \frac{4}{3} \pi r^3 \right) \]

This simplifies to:

\[ R^3 = 2r^3 \quad \Rightarrow \quad R = r \cdot 2^{1/3} \]

### Step 5: Substitute \( R \) in terms of \( r \) into the terminal velocity equation
Now, substituting \( R = r \cdot 2^{1/3} \) into the equation for \( V_R \):

\[ V_R = \frac{2}{9} \frac{(r \cdot 2^{1/3})^2 ( \rho - \sigma ) g}{\eta} \]

This simplifies to:

\[ V_R = \frac{2}{9} \frac{r^2 \cdot 2^{2/3} ( \rho - \sigma ) g}{\eta} \]

### Step 6: Relate \( V_R \) to \( v \)
We can express \( V_R \) in terms of \( v \):

\[ V_R = v \cdot 2^{2/3} \]

### Conclusion
Thus, the terminal velocity \( V_R \) of the bigger drop is:

\[ V_R = v \cdot 2^{2/3} \]