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(C) The change in surface energy is given by $E = T(\Delta A)$,where $\Delta A$ is the change in surface area.Initial surface area $A_i = n(4\pi r^2)$

Vedclass · Accepted Answer

$E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is released.

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(C) The change in surface energy is given by $E = T(\Delta A)$,where $\Delta A$ is the change in surface area.Initial surface area $A_i = n(4\pi r^2)$ and final surface area $A_f = 4\pi R^2$.Since volume is conserved,$n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,which implies $n = \frac{R^3}{r^3}$.The change in surface area is $\Delta A = A_i - A_f = 4\pi(nr^2 - R^2)$.Substituting $n = \frac{R^3}{r^3}$,we get $\Delta A = 4\pi(\frac{R^3}{r} - R^2) = 4\pi R^3(\frac{1}{r} - \frac{1}{R})$.Since $V = \frac{4}{3}\pi R^3$,we have $4\pi R^3 = 3V$.Thus,$\Delta A = 3V(\frac{1}{r} - \frac{1}{R})$.Since $r  0$ represents a decrease in surface area,meaning energy is released.Therefore,$E = 3VT(\frac{1}{r} - \frac{1}{R})$ is released.

$A$ certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,which one of the following statements is true for the energy $(E)$ in the process?

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