A) Energy = \[4VT\left( \frac{1}{r}-\frac{1}{R} \right)\] is released
B) Energy = \[3VT\left( \frac{1}{r}+\frac{1}{R} \right)\] absorbed
C) Energy = \[3VT\left( \frac{1}{r}-\frac{1}{R} \right)\] is released
D) Energy is neither released nor absorbed
Correct Answer: C
Solution :
Here, if the surface area changes, it will change the surface energy as well. If the surface area decreases, it means that energy is released and vice-versa. Change in surface energy \[\Delta A\times T\,\,\,\,\,\,\,\,\,\,\,......(i)\] Let we have 'n' number of drops initially. So, \[\Delta A=4\pi {{R}^{2}}-n(4\pi {{r}^{2}})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(ii)\] Volume is constant So, \[n=\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}=V\,\,\,\,\,\,\,\,\,\,......(iii)\] From Eqs. (ii) and (iii) \[\Delta A=\frac{3}{R}\frac{4\pi }{3}\times {{R}^{3}}-\frac{3}{r}\left( n\frac{4\pi }{3{{r}^{3}}} \right)\] \[=\frac{3}{R}\times V-\frac{3}{r}V\] \[\Delta A=3V\left( \frac{1}{R}-\frac{1}{r} \right)=-ve\]value. As R > r, so \[\Delta A\] is negative. It means surface area is decreased, so energy must be released. Energy released = \[\Delta A\times T=-3VT\left( \frac{1}{r}-\frac{1}{R} \right)\] Above expression shows the magnitude of energy released.
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