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9-1.Fluid Mechanics
hard
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?
A
$4$ $ v$
B
$(4)^{1/3}$ $v$
C
$2$ $ v$
D
$64$ $ v$
Solution
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
$2 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$
$R=(2)^{1 / 3} r$
We know that the terminal velocity of each individual drop is given as $v=\frac{\left(\frac{2}{9}\right) r^{2} g(\rho-\sigma)}{\eta}$
Thus the terminal velocity of the coalesced drop is given as $v^{\prime}=\frac{\left(\frac{2}{9}\right)\left(2^{1 / 3} r\right)^{2} g(\rho-\sigma)}{\eta}$
$v^{\prime}=2^{2 / 3} v=4^{1 / 3} v$
Standard 11
Physics
