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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the bigger drop.Surface area of one small drop is $A = 4 \pi r^2$.Since the tota

Vedclass · Accepted Answer

$180$

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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the bigger drop.Surface area of one small drop is $A = 4 \pi r^2$.Since the total volume remains constant,$216 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$.$R^3 = 216 r^3 \Rightarrow R = 6r$.Initial surface area of $216$ drops = $216 	imes A$.Final surface area of the bigger drop = $4 \pi R^2 = 4 \pi (6r)^2 = 36 	imes (4 \pi r^2) = 36A$.Energy released = (Initial surface area - Final surface area) $	imes T$.Energy released = $(216A - 36A) 	imes T = 180 AT$.

$216$ small identical liquid drops,each of surface area $A$,coalesce to form a bigger drop. If the surface tension of the liquid is $T$,what is the energy released in the process (in $AT$)?

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