Two equal drops of water are falling through air with a steady velocity of `10cms^(-1)` If drops recombine tofrom a single drop , what will be new terminal velocity ?

Answer

To solve the problem, we need to determine the new terminal velocity of a single drop formed by the recombination of two equal drops of water that were initially falling with a steady velocity of 10 cm/s.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two equal drops of water falling through the air with a terminal velocity of 10 cm/s. When these two drops combine, they form a single larger drop. We need to find the terminal velocity of this new drop.

2. **Volume Conservation**:
The volume of the two smaller drops must equal the volume of the larger drop. The volume \( V \) of a sphere (drop) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For two drops of radius \( r \):
\[ V_{\text{two drops}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]
For the larger drop of radius \( R \):
\[ V_{\text{single drop}} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

3. **Simplifying the Volume Equation**:
Canceling \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]
Taking the cube root of both sides:
\[ R = 2^{1/3} r \]

4. **Finding the Relationship of Terminal Velocity**:
The terminal velocity \( V \) of a drop is proportional to the square of its radius:
\[ V \propto R^2 \]
Therefore, if \( V_1 \) is the terminal velocity of the smaller drop and \( V_2 \) is the terminal velocity of the larger drop:
\[ \frac{V_2}{V_1} = \frac{R^2}{r^2} \]

5. **Substituting for \( R \)**:
Substitute \( R = 2^{1/3} r \):
\[ \frac{V_2}{V_1} = \frac{(2^{1/3} r)^2}{r^2} = \frac{2^{2/3} r^2}{r^2} = 2^{2/3} \]

6. **Calculating the New Terminal Velocity**:
Now, substituting \( V_1 = 10 \) cm/s:
\[ V_2 = V_1 \times 2^{2/3} = 10 \times 2^{2/3} \]
To calculate \( 2^{2/3} \):
\[ 2^{2/3} \approx 1.5874 \]
Thus:
\[ V_2 \approx 10 \times 1.5874 \approx 15.874 \text{ cm/s} \]

7. **Final Answer**:
The new terminal velocity of the single drop is approximately:
\[ V_2 \approx 15.88 \text{ cm/s} \]