If n identical water droplets falling under gravity with terminal velocity v coalesce to form a single drop which has the terminal velocity $4$ v, find the number n.

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Solution

No of identical water dropletts falling falling under gravity $= n$
Thermal velocity $= V$
And then coalesce to form a single drop
The thermal velocity $== 4 V$
Find n
The terminal velocity of an object is directly proportional to the square of the radius of the drop. The volume of the small drop is $\frac{4}{3} π r^{3}$ . The volume of the bigger drop formed by collection of $N$ drops is $N \frac{4}{3} π R^{3}$
(R be the radius of the bigger droplet and r be the radius of the smaller droplet)
Thus the radius of the bigger drop is $R = N^{1 / 3} \times r (\frac{V_{2}}{V_{1}}) = (\frac{R}{r})^{2}) (\frac{V_{2}}{V_{1}}) = (\frac{r - N^{1 / 3}}{r})^{2} \times \frac{V_{2}}{V_{1}} = N^{2 / 3} 4 V = N^{2 / 3} V N = 8$

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If n identical water droplets falling under gravity with terminal velocity v coalesce to form a single drop which has the terminal velocity 4v, find the number n.

If n identical water droplets falling under gravity with terminal velocity v coalesce to form a single drop which has the terminal velocity $4$ v, find the number n.