Small water droplets of radius 0.01 ,mm are f | vedclass

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(B) The terminal velocity $V_t$ of a spherical droplet is given by Stokes' Law: $V_t = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.From this formula

Vedclass · Accepted Answer

$40$

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(B) The terminal velocity $V_t$ of a spherical droplet is given by Stokes' Law: $V_t = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$.From this formula, $V_t \propto r^2$, where $r$ is the radius of the droplet.Let $r$ be the radius of the small droplet and $R$ be the radius of the larger drop formed by coalescing $8$ small droplets.Since the volume is conserved, the volume of the larger drop equals the sum of the volumes of $8$ small droplets:$\frac{4}{3}\pi R^3 = 8 	imes \frac{4}{3}\pi r^3$.$R^3 = 8r^3 \Rightarrow R = 2r$.Now, the ratio of the new terminal velocity $V_t'$ to the initial terminal velocity $V_t$ is:$\frac{V_t'}{V_t} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$.Therefore, $V_t' = 4 	imes V_t = 4 	imes 10 \,cm/s = 40 \,cm/s$.

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 to form a larger drop, the new terminal velocity will be ........... $cm/s$.

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