A) \[\frac{3T}{RJ}\]
B) \[\frac{3T}{rJ}\]
C) \[\frac{3T}{J}\left( \frac{1}{r}+\frac{1}{R} \right)\]
D) \[\frac{3T}{J}\left( \frac{1}{r}-\frac{1}{R} \right)\]
Correct Answer: D
Solution :
Let there are n droplets which coalesces to form a bigger drop. Then. \[\frac{4}{3}\pi {{r}^{3}}.\,n=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[n=\frac{{{R}^{3}}}{{{r}^{3}}}\] \[\therefore \] Increase in area \[\Delta A=4\pi {{r}^{2}}n-4\pi {{R}^{2}}\] \[\Delta \,A=4\pi {{r}^{2}}{{R}^{3}}/{{r}^{3}}-4\pi {{R}^{2}}\] \[\Delta \,A=4\pi {{R}^{3}}\left( \frac{1}{r}-\frac{1}{R} \right)\] Work done in this process \[W=T\,\Delta \,A\] ?..(i) But \[W=JH=ms\Delta \theta J\] ?..(ii) From Eqs. (i) and (ii), we get \[m\Delta \theta J=T\Delta A\] \[\therefore \] \[\Delta \theta =\frac{T\Delta A}{mJ}\] (\[\because \] \[s=1\]) Putting \[m=\frac{4}{3}\pi {{R}^{3}},\], and the value of \[\Delta A\] it, \[\Delta \theta =\frac{T.4\pi {{R}^{3}}\left( \frac{1}{r}-\frac{1}{R} \right)}{\frac{4}{3}\pi {{R}^{3}}J}\] or \[\Delta \theta =\frac{3T}{J}\left( \frac{1}{r}-\frac{1}{R} \right)\]
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