(C) Let $r$ be the radius of each small drop and $R$ be the radius of the big single drop.Since the total volume remains constant,$n \times \frac{4}{3

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energy released will be $[E(n-n^{2/3})]$

(C) Let $r$ be the radius of each small drop and $R$ be the radius of the big single drop.Since the total volume remains constant,$n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$.Therefore,$R = n^{1/3} r$ ... $(i)$.Initial surface energy,$E_1 = n \times (4 \pi r^2 T) = nE$ (where $T$ is surface tension).Final surface energy,$E_2 = 4 \pi R^2 T = 4 \pi (n^{1/3} r)^2 T = n^{2/3} (4 \pi r^2 T) = n^{2/3} E$.Energy released = $E_1 - E_2 = nE - n^{2/3} E = E(n - n^{2/3})$.

Class 11 Physics Fluid Mechanics and Surface Tension Excess Pressure and coalesce of Bubble and drop

Medium

Let $n$ be the number of liquid drops,each with surface energy $E$. These drops join to form a single drop. In this process:

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A
some energy will be absorbed
B
energy absorbed is $[E(n-n^{2/3})]$
C
energy released will be $[E(n-n^{2/3})]$
D
energy released will be $[E(2^{2/3}-1)]$

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Fluid Mechanics and Surface Tension

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Excess Pressure and coalesce of Bubble and drop

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Let $n$ be the number of liquid drops,each with surface energy $E$. These drops join to form a single drop. In this process:

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