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(C) Let the radius of the big drop be $R$. Since the volume remains constant:$27 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 27r^3 \Righ

Vedclass · Accepted Answer

$(B, C, D)$

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(C) Let the radius of the big drop be $R$. Since the volume remains constant:$27 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 27r^3 \Rightarrow R = 3r$. (Statement $D$ is correct.)Excess pressure inside a drop is given by $P = \frac{2T}{r}$.$P_{	ext{small}} = \frac{2T}{r}$ and $P_{	ext{big}} = \frac{2T}{R} = \frac{2T}{3r} = \frac{1}{3} P_{	ext{small}}$. (Statement $A$ is incorrect.)Initial surface energy $U_i = 27 	imes (4 \pi r^2 T) = 108 \pi r^2 T$.Final surface energy $U_f = 4 \pi R^2 T = 4 \pi (3r)^2 T = 36 \pi r^2 T$.Since $U_f Energy released $\Delta U = U_i - U_f = 108 \pi r^2 T - 36 \pi r^2 T = 72 \pi r^2 T$. (Statement $C$ is correct.)Therefore,statements $B, C,$ and $D$ are correct.

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