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

Correct Answer: 40

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:
\[ V_t^{\text{large}} = 4 \cdot 10 = 40 \, \text{cm/s}. \]