Two identical drops falling with a terminal speed 'v' coalesce to form a bigger drop. The terminal velocity of the bigger drop will be

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Solution

The correct option is B
$\frac{4}{3} π R^{3} = 2. \frac{4}{3} π r^{3}$
$R = 2^{\frac{1}{3}} . r$
$v \propto r^{2}$
$\frac{v_{1}}{r_{1}^{2}} = \frac{v_{2}}{r_{2}^{2}} v^{'} = {(\frac{r_{2}}{r_{1}})}^{2} \Rightarrow v^{'} = v {[\frac{(r 2^{1 / 3})}{r}]}^{2} = v 2^{2 / 3}$

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Two identical drops falling with a terminal speed 'v' coalesce to form a bigger drop. The terminal velocity of the bigger drop will be

The correct option is B 43πR3=2.43πr3 R=213.r v∝r2 v1r21=v2r22v′=(r2r1)2⇒v′=v[(r21/3)r]2=v22/3

The correct option is B
$\frac{4}{3} π R^{3} = 2. \frac{4}{3} π r^{3}$
$R = 2^{\frac{1}{3}} . r$
$v \propto r^{2}$
$\frac{v_{1}}{r_{1}^{2}} = \frac{v_{2}}{r_{2}^{2}} v^{'} = {(\frac{r_{2}}{r_{1}})}^{2} \Rightarrow v^{'} = v {[\frac{(r 2^{1 / 3})}{r}]}^{2} = v 2^{2 / 3}$