A water drop of diameter 2,cm is broken into | vedclass

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(A) Given: Diameter $D = 2\,cm$,so radius $r = 1\,cm = 0.01\,m$. Surface tension $T = 0.075\,N/m$. Number of droplets $n = 64$.By volume conservation:

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$2.8 	imes 10^{-4}$

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(A) Given: Diameter $D = 2\,cm$,so radius $r = 1\,cm = 0.01\,m$. Surface tension $T = 0.075\,N/m$. Number of droplets $n = 64$.By volume conservation: $\frac{4}{3} \pi r^3 = n 	imes \frac{4}{3} \pi r_0^3$,where $r_0$ is the radius of each small droplet.$r^3 = 64 r_0^3 \implies r_0 = \frac{r}{4} = \frac{0.01}{4} = 0.0025\,m$.Initial surface area $A_i = 4 \pi r^2$.Final surface area $A_f = n 	imes 4 \pi r_0^2 = 64 	imes 4 \pi (r/4)^2 = 64 	imes 4 \pi (r^2/16) = 16 \pi r^2$.Gain in surface energy $\Delta SE = T 	imes (A_f - A_i) = T 	imes (16 \pi r^2 - 4 \pi r^2) = T 	imes 12 \pi r^2$.Substituting the values: $\Delta SE = 0.075 	imes 12 	imes 3.14159 	imes (0.01)^2$.$\Delta SE = 0.075 	imes 12 	imes 3.14159 	imes 0.0001 = 0.9 	imes 3.14159 	imes 10^{-4} \approx 2.827 	imes 10^{-4}\,J$.

$A$ water drop of diameter $2\,cm$ is broken into $64$ equal droplets. The surface tension of water is $0.075\,N/m$. In this process,the gain in surface energy will be ...........$J$.

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