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.

Updated On: Mar 22, 2025
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Correct Answer: 40

Solution and Explanation

The terminal velocity (VtV_t) of a droplet is given by:
\[ V_t =2R2(ρσ)g9η \frac{2R^2 (\rho - \sigma) g}{9\eta}, \]
where RR is the radius of the droplet.
When 8 small droplets coalesce:
Mass of larger drop M8mM \propto 8m, \quad \(M = 43\frac{4}{3}\pi R^3 (\rho)\).
The radius of the larger drop is:
Rlarge3=8Rsmall3    Rlarge=2Rsmall. R_{\text{large}}^3 = 8R_{\text{small}}^3 \implies R_{\text{large}} = 2R_{\text{small}}.
Since terminal velocity (VtV_t) is proportional to R2R^2:
VtR2    Vtlarge=4Vtsmall. V_t \propto R^2 \implies V_t^{\text{large}} = 4V_t^{\text{small}}.
For the smaller droplet:
Vtsmall=10cm/s. V_t^{\text{small}} = 10 \, \text{cm/s}.
Thus:
Vtlarge=410=40cm/s. V_t^{\text{large}} = 4 \cdot 10 = 40 \, \text{cm/s}.

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