The energy needed for breaking a liquid drop | vedclass

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(D) The volume of the large drop is equal to the total volume of the $216$ small droplets.$V_{large} = 216 	imes V_{small}$$\frac{4}{3} \pi R^3 = 216

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) The volume of the large drop is equal to the total volume of the $216$ small droplets.$V_{large} = 216 	imes V_{small}$$\frac{4}{3} \pi R^3 = 216 	imes \frac{4}{3} \pi r^3$$R^3 = 216 r^3$$R = 6r \implies r = \frac{R}{6}$The energy needed is equal to the increase in surface area multiplied by the surface tension $T$.$E = T 	imes (A_{final} - A_{initial})$$A_{initial} = 4 \pi R^2$$A_{final} = 216 	imes (4 \pi r^2) = 216 	imes 4 \pi \left(\frac{R}{6}ight)^2 = 216 	imes 4 \pi 	imes \frac{R^2}{36} = 6 	imes 4 \pi R^2 = 24 \pi R^2$$E = T 	imes (24 \pi R^2 - 4 \pi R^2) = T 	imes 20 \pi R^2$Comparing this with $x 	imes T 	imes R^2$, we get $x = 20 \pi$.

The energy needed for breaking a liquid drop of radius $R$ into $216$ droplets, each of radius $r$, is $x$ times $TR^2$. The value of $x$ is [$T =$ surface tension of the liquid]. (in $\pi$)

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