A liquid drop of diameter D breaks into 27 ti | vedclass

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Question

(A) Let the radius of the large drop be $R = D/2$. The volume of the large drop is $V = \frac{4}{3}\pi R^3$.When it breaks into $n = 27$ smaller drops

Vedclass · Accepted Answer

$2\pi TD^2$

Vedclass · Answer

(A) Let the radius of the large drop be $R = D/2$. The volume of the large drop is $V = \frac{4}{3}\pi R^3$.When it breaks into $n = 27$ smaller drops of radius $r$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$.Thus,$R^3 = 27r^3$,which gives $r = R/3$.The initial surface area is $A_i = 4\pi R^2$.The final surface area is $A_f = n 	imes 4\pi r^2 = 27 	imes 4\pi (R/3)^2 = 27 	imes 4\pi (R^2/9) = 12\pi R^2$.The change in surface area is $\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$.The change in energy (surface energy) is $\Delta E = T 	imes \Delta A = T 	imes 8\pi R^2$.Substituting $R = D/2$,we get $\Delta E = T 	imes 8\pi (D/2)^2 = T 	imes 8\pi (D^2/4) = 2\pi TD^2$.

$A$ liquid drop of diameter $D$ breaks into $27$ tiny drops. The change in energy is

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