1. Use Stokes' law for terminal velocity

For a small spherical drop falling through air, the terminal velocity is

$v_t = \frac{2 r^2 (\rho - \sigma) g}{9\eta}$

where $r$ is the radius of the drop. Hence,

$v_t \propto r^2$

So terminal velocity is proportional to the square of the radius.

2. Find the radius of the new drop after coalescence

Let the radius of each small drop be $r$.

Since 8 equal drops coalesce, volume is conserved:

$8 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$

where $R$ is the radius of the new drop.

So,

$R^3 = 8r^3$

$R = 2r$

3. Compare terminal velocities

Since

$v \propto r^2$

we get

$\frac{V}{v} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$

Given the original steady speed,

$v = 10\,\text{cm/s}$

therefore,

$V = 4 \times 10 = 40\,\text{cm/s}$

4. Check options

• A: $40\,\text{cm/s}$ ✅
• B: $16\,\text{cm/s}$ ❌
• C: $10\,\text{cm/s}$ ❌
• D: $5\,\text{cm/s}$ ❌

Therefore, the correct answer is A.