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Question

(D) When two drops of radius $r$ merge to form a bigger drop of radius $R$, the volume remains conserved.
Volume of two small drops = Volume of one b

Vedclass · Accepted Answer

$2^{8/3} \pi r^2 T$

Vedclass · Answer

(D) When two drops of radius $r$ merge to form a bigger drop of radius $R$, the volume remains conserved.
Volume of two small drops = Volume of one big drop
$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 2r^3$
$R = 2^{1/3} r$
The surface energy $E$ of a drop is given by $E = 	ext{Surface Area} 	imes 	ext{Surface Tension} = 4 \pi R^2 T$.
Substituting the value of $R$:
$E = 4 \pi (2^{1/3} r)^2 T$
$E = 4 \pi (2^{2/3} r^2) T$
$E = 2^2 	imes 2^{2/3} \pi r^2 T$
$E = 2^{2 + 2/3} \pi r^2 T$
$E = 2^{8/3} \pi r^2 T$.

Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

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