A number of small water droplets of surface t | vedclass

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Question

(C) Let $n$ be the number of small droplets of radius $r$ that combine to form a large drop of radius $R$.By conservation of volume: $n \cdot \frac{4}

Vedclass · Accepted Answer

$\sqrt{\frac{6 T}{\rho}\left(\frac{R-r}{rR}\right)}$

Vedclass · Answer

(C) Let $n$ be the number of small droplets of radius $r$ that combine to form a large drop of radius $R$.By conservation of volume: $n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$,which implies $n = \frac{R^3}{r^3}$.The change in surface area is $\Delta A = n(4 \pi r^2) - 4 \pi R^2 = 4 \pi (n r^2 - R^2)$.Substituting $n = \frac{R^3}{r^3}$,we get $\Delta A = 4 \pi (\frac{R^3}{r} - R^2) = 4 \pi R^2 (\frac{R}{r} - 1)$.The energy released is $E = T \cdot \Delta A = 4 \pi T R^2 (\frac{R-r}{r})$.This energy is converted into kinetic energy: $E = \frac{1}{2} M v^2$,where $M$ is the mass of the large drop.$M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3$.Equating the two: $4 \pi T R^2 (\frac{R-r}{r}) = \frac{1}{2} (\rho \cdot \frac{4}{3} \pi R^3) v^2$.Simplifying for $v^2$: $v^2 = \frac{2 \cdot 3 \cdot 4 \pi T R^2 (R-r)}{4 \pi \rho R^3 r} = \frac{6 T (R-r)}{\rho R r}$.Therefore,$v = \sqrt{\frac{6 T}{\rho} \left( \frac{R-r}{rR} \right)}$.

$A$ number of small water droplets of surface tension '$T$',each of radius '$r$',are combined to form a single drop of radius '$R$'. If the released energy is converted into kinetic energy,then the velocity acquired by the bigger drop is . . . . . . ($\rho$ - density of water)

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