Question

Small water droplets of radius 0.01 mm are formed in the upper atmosphere and falling with a terminal velocity of 10 cm/s. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be ………..cm/s.

Answer 1

The terminal velocity ($V_t$) of a droplet is given by:
V_t =$ \frac{2R^2 (\rho - \sigma) g}{9\eta}$,
where $R$ is the radius of the droplet.
When 8 small droplets coalesce:
Mass of larger drop $M \propto 8m$, \quad M = \(\frac{4}{3}\pi R^3 (\rho)).
The radius of the larger drop is:
$R_{\text{large}}^3 = 8R_{\text{small}}^3 \implies R_{\text{large}} = 2R_{\text{small}}.$
Since terminal velocity ($V_t$) is proportional to $R^2$:
$V_t \propto R^2 \implies V_t^{\text{large}} = 4V_t^{\text{small}}.$
For the smaller droplet:
$V_t^{\text{small}} = 10 \, \text{cm/s}.$
Thus:
$V_t^{\text{large}} = 4 \cdot 10 = 40 \, \text{cm/s}.$