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(B) Let the radius of the large drop be $R = D/2$. Let the radius of each small drop be $r$.Since the volume remains constant,the volume of the large

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$2\pi D^2 \sigma$

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(B) Let the radius of the large drop be $R = D/2$. Let the radius of each small drop be $r$.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of $27$ small drops:$\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3$$R^3 = 27 r^3 \implies r = R/3 = D/6$.The initial surface energy is $E_i = 4\pi R^2 \sigma$.The final surface energy is $E_f = 27 	imes 4\pi r^2 \sigma = 27 	imes 4\pi (R/3)^2 \sigma = 27 	imes 4\pi (R^2/9) \sigma = 3 	imes 4\pi R^2 \sigma = 12\pi R^2 \sigma$.The change in surface energy is $\Delta E = E_f - E_i = 12\pi R^2 \sigma - 4\pi R^2 \sigma = 8\pi R^2 \sigma$.Substituting $R = D/2$:$\Delta E = 8\pi (D/2)^2 \sigma = 8\pi (D^2/4) \sigma = 2\pi D^2 \sigma$.

$A$ liquid drop of diameter $D$ breaks into $27$ small drops of equal size. If the surface tension of the liquid is $\sigma$,then the change in surface energy is

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