Eight equal drops of water are falling through air with a steady speed of 10 cm s$$^{-1}$$. If the drops coalesce, the new velocity is:-

Eight equal drops of water falling with steady speed (terminal velocity) of 10 cm/s coalesce into one drop. We need to find the new terminal velocity.

First, use conservation of volume to find the new radius.

When 8 equal spherical drops (each of radius $$r$$) merge into one larger drop (radius $$R$$), the total volume is conserved:

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

$$ R^3 = 8r^3 \implies R = 2r $$

Next, recall how terminal velocity depends on radius.

For a sphere falling through a viscous fluid, the terminal velocity is given by Stokes' law:

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

where $$\rho$$ is the density of the sphere, $$\sigma$$ is the density of the fluid, $$\eta$$ is the viscosity, and $$g$$ is gravitational acceleration. The key dependence is:

$$ v_t \propto r^2 $$

Now, calculate the new terminal velocity.

$$ \frac{v_{\text{new}}}{v_{\text{old}}} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4 $$

$$ v_{\text{new}} = 4 \times 10 = 40\;\text{cm/s} $$

The new terminal velocity is 40 cm/s.

The correct answer is Option 2: 40 cm s$$^{-1}$$.