Two water drops of the same radius are falling through air with a velocity 5cm/s. If the two drops coalesce to form one drop, the terminal velocity of the drop will be

Options

A. 10 cm/s
B. `5sqrt2cm//s`
C. 2.5 cm/s
D. `5xx4^(1//3)cm//s`

Explanation

To solve the problem of determining the terminal velocity of a single drop formed by the coalescence of two smaller drops, we can follow these steps:

### Step 1: Understand the relationship between volume and radius
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For two drops of radius \( r \), the total volume \( V_{total} \) is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]
When these two drops coalesce, they form a new drop with radius \( R \). The volume of the new drop is:
\[ V = \frac{4}{3} \pi R^3 \]
Setting the two volumes equal gives:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 2: Simplify the equation
We can cancel out \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]

### Step 3: Solve for the new radius \( R \)
Taking the cube root of both sides, we find:
\[ R = (2)^{1/3} r \]

### Step 4: Relate terminal velocity to radius
The terminal velocity \( V_t \) of a sphere falling through a fluid is proportional to the square of its radius:
\[ V_t \propto r^2 \]
Let \( V_1 \) be the terminal velocity of the smaller drops (given as 5 cm/s) and \( V_2 \) be the terminal velocity of the larger drop:
\[ \frac{V_1}{V_2} = \frac{r^2}{R^2} \]

### Step 5: Substitute for \( R \)
Substituting \( R = (2)^{1/3} r \) into the equation gives:
\[ \frac{V_1}{V_2} = \frac{r^2}{((2)^{1/3} r)^2} = \frac{r^2}{\frac{4}{3} r^2} = \frac{3}{4} \]

### Step 6: Solve for \( V_2 \)
Rearranging the equation gives:
\[ V_2 = V_1 \cdot \frac{4}{3} \]
Substituting \( V_1 = 5 \) cm/s:
\[ V_2 = 5 \cdot \frac{4}{3} = \frac{20}{3} \approx 6.67 \text{ cm/s} \]

### Step 7: Final calculation
To find the exact value:
\[ V_2 = 5 \cdot \frac{4}{3} = 6.67 \text{ cm/s} \]

### Conclusion
The terminal velocity of the new drop formed by the coalescence of the two smaller drops is approximately **6.67 cm/s**.

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