A) \[\frac{2T}{rJ}\]
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 :
Surface tension (T) of a liquid is equal to the work required to increase the surface area of the liquid film by unity at constant temperature. Hence, \[W=T\Delta A=\]energy liberated ... (i) Also, shape of water drop is spherical, hence its surface area is\[4\pi {{r}^{2}},\]where Volume of n drops = volume of big drop \[n\times \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[R={{n}^{1/3}}.r\] ...(ii) Energy liberated\[=mc\Delta \theta \] ... (iii) where c is specific heat and A9 the temperature variation. \[E=T[n4\pi {{r}^{2}}-4\pi {{R}^{2}}]=mc\Delta \theta J\] For water c = 1, \[\rho =1\] \[\Rightarrow \] \[\Delta \theta =\frac{T}{J}\left[ \frac{n4\pi {{r}^{2}}}{n4\pi {{r}^{3}}/3}-\frac{4\pi {{R}^{2}}}{4\pi {{R}^{3}}/3} \right]\] \[\Delta \theta =\frac{3T}{J}\left[ \frac{1}{r}-\frac{1}{R} \right]\]
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