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(A) Let the radius of the bigger drop be $R$. Since the volume is conserved,the volume of the bigger drop equals the sum of the volumes of the two sma

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$4 \pi r^2 T[2-2^{2/3}]$

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(A) Let the radius of the bigger drop be $R$. Since the volume is conserved,the volume of the bigger drop equals the sum of the volumes of the two smaller drops: $2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.This simplifies to $R^3 = 2r^3$,or $R = 2^{1/3} r$.The initial surface energy of the two drops is $U_i = 2 	imes (4 \pi r^2 T) = 8 \pi r^2 T$.The final surface energy of the bigger drop is $U_f = 4 \pi R^2 T = 4 \pi (2^{1/3} r)^2 T = 4 \pi r^2 T (2^{2/3})$.The energy released is $\Delta U = U_i - U_f = 8 \pi r^2 T - 4 \pi r^2 T (2^{2/3}) = 4 \pi r^2 T (2 - 2^{2/3})$.

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

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