The energy needed for breaking a liquid drop | vedclass

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Question

(A) The volume of the large drop is equal to the sum of the volumes of the '$n$' small droplets.$V_{large} = n 	imes V_{small}$$\frac{4}{3} \pi R^3 =

Vedclass · Accepted Answer

$4\pi T R^2 [\frac{R}{r} - 1]$

Vedclass · Answer

(A) The volume of the large drop is equal to the sum of the volumes of the '$n$' small droplets.$V_{large} = n 	imes V_{small}$$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$$R^3 = n r^3 \implies n = \frac{R^3}{r^3}$The energy required is equal to the increase in surface area multiplied by the surface tension '$T$'.$E = T 	imes (	ext{Final Surface Area} - 	ext{Initial Surface Area})$$E = T 	imes (n 	imes 4\pi r^2 - 4\pi R^2)$Substitute $n = \frac{R^3}{r^3}$:$E = 4\pi T (\frac{R^3}{r^3} 	imes r^2 - R^2)$$E = 4\pi T (\frac{R^3}{r} - R^2)$$E = 4\pi T R^2 (\frac{R}{r} - 1)$

The energy needed for breaking a liquid drop of radius '$R$' into '$n$' droplets each of radius '$r$' is [$T$ = surface tension of the liquid]

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