Two drops of same radius are falling through | vedclass

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Question

When the two drops coalesce the volume is conserved, thus if $r$ is the radius of the individual drops and R of the coalesced drop, we get

$2 	ime

Vedclass · Accepted Answer

$(4)^{1/3}$ $v$

Vedclass · Answer

When the two drops coalesce the volume is conserved, thus if $r$ is the radius of the individual drops and R of the coalesced drop, we get

$2 	imes \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$

$R=(2)^{1 / 3} r$

We know that the terminal velocity of each individual drop is given as $v=\frac{\left(\frac{2}{9}ight) r^{2} g(ho-\sigma)}{\eta}$

Thus the terminal velocity of the coalesced drop is given as $v^{\prime}=\frac{\left(\frac{2}{9}ight)\left(2^{1 / 3} right)^{2} g(ho-\sigma)}{\eta}$

$v^{\prime}=2^{2 / 3} v=4^{1 / 3} v$

Two drops of same radius are falling through air with steady velocity of $v $ $cm/s$. If the two drops coalesce, what would be the terminal velocity?

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