The work done in splitting a water drop of ra | vedclass

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(B) Given the radius of the large drop is $R$. Let $r$ be the radius of each smaller droplet. Since the total volume remains constant:$\frac{4}{3} \pi

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$12$

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(B) Given the radius of the large drop is $R$. Let $r$ be the radius of each smaller droplet. Since the total volume remains constant:$\frac{4}{3} \pi R^3 = 64 	imes \frac{4}{3} \pi r^3$$	herefore r^3 = \frac{R^3}{64} \implies r = \frac{R}{4}$Initial surface energy $E_1 = 4 \pi R^2 T$.Final surface energy $E_2 = 64 	imes (4 \pi r^2 T) = 64 	imes 4 \pi 	imes (\frac{R}{4})^2 	imes T = 64 	imes 4 \pi 	imes \frac{R^2}{16} 	imes T = 16 \pi R^2 T$.Work done $W = E_2 - E_1 = 16 \pi R^2 T - 4 \pi R^2 T = 12 \pi R^2 T$.

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ Surface tension of water). (in $\pi TR^2$)

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