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. \[\therefore \]\[\frac{\Delta {{p}_{1}}}{\Delta {{p}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}\] \[=\frac{0.75}{0.50}\] \[\Rightarrow \]\[\frac{\Delta {{p}_{1}}}{\Delta {{p}_{2}}}=\frac{3}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
