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(A) The terminal velocity $V$ of a drop of radius $r$ is given by $V = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.Since $V \propto r^2$,we have $\f

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$\frac{VR^2}{r^2}$

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(A) The terminal velocity $V$ of a drop of radius $r$ is given by $V = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.Since $V \propto r^2$,we have $\frac{V_{big}}{V_{small}} = \frac{R^2}{r^2}$.Given that $n$ drops of radius $r$ coalesce to form one big drop of radius $R$,the volume is conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $R^3 = nr^3$ or $R = n^{1/3}r$.Substituting $R^2 = n^{2/3}r^2$ into the velocity ratio,we get $V_{big} = V \cdot \frac{R^2}{r^2} = V \cdot n^{2/3}$.However,expressing the result in terms of $R$ and $r$ as per the options,we use the proportionality $V \propto r^2$. Thus,$V_{big} = V \cdot \frac{R^2}{r^2}$.

$n$ small water drops of same size (radius $r$) fall through air with constant terminal velocity $V$. They coalesce to form a big drop of radius $R$. The terminal velocity of the big drop is

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