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(A) Let the radius of the large drop be $R = D/2$. When it breaks into $n = 27$ smaller drops of radius $r$,the volume remains constant.$V_{large} = n

Vedclass · Accepted Answer

$2\pi TD^2$

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(A) Let the radius of the large drop be $R = D/2$. When it breaks into $n = 27$ smaller drops of radius $r$,the volume remains constant.$V_{large} = n 	imes V_{small} \implies \frac{4}{3}\pi R^3 = 27 	imes \frac{4}{3}\pi r^3$.Taking the cube root,$R = 3r$,so $r = R/3$.The change in surface energy $\Delta E$ is given by the increase in surface area multiplied by surface tension $T$.$\Delta E = T 	imes (A_{final} - A_{initial}) = T 	imes (n 	imes 4\pi r^2 - 4\pi R^2)$.Substituting $r = R/3$ and $n = 27$:$\Delta E = T 	imes (27 	imes 4\pi (R/3)^2 - 4\pi R^2) = T 	imes (27 	imes 4\pi R^2/9 - 4\pi R^2) = T 	imes (12\pi R^2 - 4\pi R^2) = 8\pi R^2 T$.Since $R = D/2$,$R^2 = D^2/4$.$\Delta E = 8\pi (D^2/4) T = 2\pi TD^2$.

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

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