In S.I. system,the total energy of the free s | vedclass

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(B) The surface energy $E$ of a liquid drop is given by the formula $E = T 	imes A$,where $T$ is the surface tension and $A$ is the surface area of t

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$2$

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(B) The surface energy $E$ of a liquid drop is given by the formula $E = T 	imes A$,where $T$ is the surface tension and $A$ is the surface area of the drop.The surface area $A$ of a spherical drop of radius $r$ is $4 \pi r^2$.Thus,$E = T 	imes 4 \pi r^2$.According to the problem,the total energy is $2 \pi$ times the surface tension,so $E = 2 \pi T$.Equating the two expressions for $E$: $4 \pi r^2 T = 2 \pi T$.Dividing both sides by $2 \pi T$ (assuming $T 
eq 0$),we get $2 r^2 = 1$,which means $r^2 = 1/2$.Therefore,$r = 1/\sqrt{2}$.The diameter $d$ of the drop is $2r = 2 	imes (1/\sqrt{2}) = \sqrt{2} \ m$.

In $S.I.$ system,the total energy of the free surface of a liquid drop is $2 \pi$ times the surface tension of the liquid. The diameter of the drop is . . . . . . . (in $m$)

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