Two drops of equal radius r coalesce to form a single drop under isothermal conditions . The radius of such a drop would be

Options

A. r
B. `2^^ `(1/3) r
C. `r/2`
D. `3sqrt3 `

Explanation

To find the radius of the single drop formed when two drops of equal radius \( r \) coalesce, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Volume of a Sphere**:
The volume \( V \) of a sphere with radius \( r \) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

2. **Calculate the Total Volume of Two Drops**:
Since we have two drops of equal radius \( r \), the total volume \( V_{\text{total}} \) of the two drops is:
\[ V_{\text{total}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

3. **Set the Volume of the Larger Drop**:
Let the radius of the larger drop formed after coalescence be \( R \). The volume of this larger drop is:
\[ V_{\text{larger}} = \frac{4}{3} \pi R^3 \]

4. **Equate the Volumes**:
Since the volume before coalescence must equal the volume after coalescence, we can set the two volumes equal to each other:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

5. **Simplify the Equation**:
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 2 r^3 = R^3 \]

6. **Solve for \( R \)**:
To find \( R \), we take the cube root of both sides:
\[ R = (2 r^3)^{1/3} = 2^{1/3} r \]

7. **Final Result**:
Therefore, the radius of the larger drop formed by the coalescence of the two smaller drops is:
\[ R = 2^{1/3} r \]