Two exactly simlar rain drops falling with terminal velocity of `(2)^(1//3) ms^(-1) `coalese to from a bigger drop Find the new terminal velocity of the bigger drop.

Answer

To solve the problem of finding the new terminal velocity of a larger drop formed by the coalescence of two smaller raindrops, we can follow these steps:

### Step 1: Understand the Relationship Between Volume and Radius
The volume \( V \) of a sphere (raindrop) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the drop.

### Step 2: Calculate the Volume of Two Smaller Drops
Since we have two identical raindrops, the total volume \( V_{total} \) of the two smaller drops is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 3: Set Up the Volume of the Bigger Drop
Let \( R \) be the radius of the larger drop formed by the coalescence of the two smaller drops. The volume of the larger drop is:
\[ V_{big} = \frac{4}{3} \pi R^3 \]

### Step 4: Equate the Volumes
Since the total volume of the two smaller drops equals the volume of the larger drop, we can set up the equation:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
By simplifying, we can cancel out \( \frac{4}{3} \pi \):
\[ 2r^3 = R^3 \]

### Step 5: Solve for the Radius of the Bigger Drop
Taking the cube root of both sides gives us:
\[ R = (2)^{1/3} r \]

### Step 6: Understand the Terminal Velocity Formula
The terminal velocity \( v \) of a drop is given by:
\[ v = \frac{2}{9} \frac{r^2 \rho g}{\eta} \]
where \( \rho \) is the density of the drop, \( g \) is the acceleration due to gravity, and \( \eta \) is the viscosity of the fluid.

### Step 7: Relate Terminal Velocity to Radius
Since the terminal velocity is directly proportional to the square of the radius, we can express the relationship as:
\[ \frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 \]

### Step 8: Substitute the Radius Relationship
From Step 5, we know \( R = (2)^{1/3} r \). Therefore:
\[ \frac{v_2}{v_1} = \left(\frac{(2)^{1/3} r}{r}\right)^2 = (2)^{2/3} \]

### Step 9: Calculate the New Terminal Velocity
Given that the terminal velocity of the smaller drops \( v_1 = (2)^{1/3} \) m/s, we can find \( v_2 \):
\[ v_2 = v_1 \cdot (2)^{2/3} = (2)^{1/3} \cdot (2)^{2/3} = (2)^{3/3} = 2 \text{ m/s} \]

### Final Answer
The new terminal velocity of the bigger drop is:
\[ \boxed{2 \text{ m/s}} \]

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