Twenty-seven rain drops of same diameter fall through air with terminal velocity ' $V$ '. If they coalesce forming a single drop, then the terminal velocity of the resultant drop is

V

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3 V

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9 V

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\frac{V}{9}

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Solution

The correct option is A $9 V$
Given, A drop falls with terminal velocity $V$ ,
$V = \frac{2 r^{2}}{9 η} (T_{s} - T) g$

After $27$ drops coalesce, volume become $27$ times

As, $V = \frac{1}{3} r^{3}$

$27 V = 27 r^{3} \frac{4}{3} π$

So, the radius of the resultant drop $= 3 r$ .

Terminal velocity of new drop $= \frac{2 (3 r)^{2} (t_{s} - T) g}{9 η}$

$= \frac{2 r^{2} (T_{s} - T) 9}{9 η} = 9 V$

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Twenty-seven rain drops of same diameter fall through air with terminal velocity 'V'. If they coalesce forming a single drop, then the terminal velocity of the resultant drop is

The correct option is A 9VGiven, A drop falls with terminal velocity V,V=2r29η(Ts−T)gAfter 27 drops coalesce, volume become 27 timesAs, V=13r327V=27r343πSo, the radius of the resultant drop=3r.Terminal velocity of new drop =2(3r)2(ts−T)g9η =2r2(Ts−T)99η=9V

Twenty-seven rain drops of same diameter fall through air with terminal velocity ' $V$ '. If they coalesce forming a single drop, then the terminal velocity of the resultant drop is