A) \[1:1\]
B) \[5:3\]
C) \[2:3\]
D) \[3:2\]
Correct Answer: D
Solution :
Excess pressure inside a liquid drop \[\Delta p=\frac{2T}{R}\] where T is surface tension and R is radius of liquid drop. \[\therefore \] \[\frac{\Delta {{p}_{1}}}{\Delta {{p}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{0.75}{0.50}\] \[\Rightarrow \] \[\frac{\Delta {{p}_{1}}}{\Delta {{p}_{2}}}=\frac{3}{2}\]
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