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Question

(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the volume remains constant, the volume of $8$ small drops e

Vedclass · Accepted Answer

$40 \,cm \,s^{-1}$

Vedclass · Answer

(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the volume remains constant, the volume of $8$ small drops equals the volume of the big drop:$8 \cdot (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 8r^3 \implies R = 2r$.Terminal velocity $v_t$ is given by the formula $v_t = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$, which implies $v_t \propto r^2$.Let $v_1$ be the terminal velocity of the small drop and $v_2$ be the terminal velocity of the big drop.$\frac{v_2}{v_1} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$.Given $v_1 = 10 \,cm \,s^{-1}$, we have $v_2 = 4 \cdot 10 \,cm \,s^{-1} = 40 \,cm \,s^{-1}$.

Eight equal drops of water are falling through air with a steady velocity of $10 \,cm \,s^{-1}$. If the drops combine to form a single drop, then the terminal velocity of this big drop is:

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