Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of `6cms^(-1)`. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air

Options

A. `1.5 cms^(-1)`
B. `6 cms^(-1)`
C. `24 cms^(-1)`
D. `32 cms^(-1)`

Explanation

To solve the problem, we need to find the terminal speed of a larger drop formed by the coalescence of eight smaller drops, each having a terminal speed of 6 cm/s.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( v \) of a spherical droplet is given by the formula:
\[ v = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta} \]
where \( r \) is the radius of the droplet, \( g \) is the acceleration due to gravity, \( \rho \) is the density of the droplet, \( \sigma \) is the density of the fluid (air), and \( \eta \) is the viscosity of the fluid.

2. **Relationship Between Terminal Velocity and Radius**:
From the formula, we can see that terminal velocity \( v \) is directly proportional to the square of the radius \( r^2 \):
\[ v \propto r^2 \]

3. **Volume Conservation**:
When eight smaller droplets coalesce to form one larger droplet, the volumes must be equal. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For eight smaller droplets, the total volume is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]
For the larger droplet, the volume is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Simplifying, we find:
\[ 32 r^3 = 4 R^3 \implies R^3 = 8 r^3 \implies R = 2r \]

4. **Finding the Terminal Speed of the Larger Droplet**:
Let \( v_s \) be the terminal speed of the smaller droplets (which is given as 6 cm/s) and \( v_L \) be the terminal speed of the larger droplet. From the relationship established earlier:
\[ \frac{v_L}{v_s} = \left(\frac{R}{r}\right)^2 \]
Substituting \( R = 2r \):
\[ \frac{v_L}{v_s} = \left(\frac{2r}{r}\right)^2 = 4 \]
Therefore:
\[ v_L = 4 v_s \]
Substituting \( v_s = 6 \) cm/s:
\[ v_L = 4 \times 6 \text{ cm/s} = 24 \text{ cm/s} \]

5. **Final Answer**:
The terminal speed of the larger droplet is \( 24 \) cm/s.