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Small water droplets of radius $0.01 \mathrm{~mm}$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \mathrm{~cm} / \mathrm{s}$. Due to condensation, if $8 \mathrm{such}$ droplets are coalesced and formed a larger drop, the new terminal velocity will be ........... $\mathrm{cm} / \mathrm{s}$.
$20$
$40$
$50$
$70$
Solution
$\mathrm{m}=\text { mass of small drop }$
$\mathrm{M}=\text { mass of bigger drop }$
$\mathrm{V}_{\mathrm{t}}=\frac{2}{9} \frac{\mathrm{R}^2(\rho-\sigma) \mathrm{g}}{\eta}$
$8 \propto \mathrm{m}=\mathrm{M}$
$8 \mathrm{r}^3=\mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{R}$
$\text { as } \mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \because \text { Radius double so } \mathrm{V}_{\mathrm{t}} \text { becomes } 4 \text { time }$
$\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}$
