A water drop of radius 1,cm is broken into 72 | vedclass

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(C) Initial radius $R = 1\,cm = 10^{-2}\,m$. Surface tension $T = 75\,dyne/cm = 75 	imes 10^{-3}\,N/m$.Initial surface area $A_i = 4\pi R^2 = 4\pi(10

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$7.5$

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(C) Initial radius $R = 1\,cm = 10^{-2}\,m$. Surface tension $T = 75\,dyne/cm = 75 	imes 10^{-3}\,N/m$.Initial surface area $A_i = 4\pi R^2 = 4\pi(10^{-2})^2 = 4\pi 	imes 10^{-4}\,m^2$.Initial surface energy $U_i = T 	imes A_i = 75 	imes 10^{-3} 	imes 4\pi 	imes 10^{-4} = 300\pi 	imes 10^{-7}\,J$.Let $r$ be the radius of each small droplet. By volume conservation: $\frac{4}{3}\pi R^3 = 729 	imes \frac{4}{3}\pi r^3$.$r^3 = \frac{R^3}{729} \implies r = \frac{R}{9} = \frac{1}{9}\,cm = \frac{1}{9} 	imes 10^{-2}\,m$.Final surface area $A_f = 729 	imes 4\pi r^2 = 729 	imes 4\pi 	imes (\frac{1}{9} 	imes 10^{-2})^2 = 729 	imes 4\pi 	imes \frac{1}{81} 	imes 10^{-4} = 9 	imes 4\pi 	imes 10^{-4} = 36\pi 	imes 10^{-4}\,m^2$.Final surface energy $U_f = T 	imes A_f = 75 	imes 10^{-3} 	imes 36\pi 	imes 10^{-4} = 2700\pi 	imes 10^{-7}\,J$.Gain in surface energy $\Delta U = U_f - U_i = (2700\pi - 300\pi) 	imes 10^{-7} = 2400\pi 	imes 10^{-7}\,J$.Using $\pi \approx 3.14$,$\Delta U = 2400 	imes 3.14 	imes 10^{-7} = 7536 	imes 10^{-7} = 7.536 	imes 10^{-4}\,J$.Rounding to the first decimal place,the gain is $7.5 	imes 10^{-4}\,J$.

$A$ water drop of radius $1\,cm$ is broken into $729$ equal droplets. If the surface tension of water is $75\,dyne/cm$,then the gain in surface energy up to the first decimal place will be $...\times 10^{-4}\,J$.

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