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  • Two small drops of mercury each of radius r from a single large drop. The ratio of surface energy before and after this change is:Option: 1 &nbs

Two small drops of mercury each of radius r from a single large drop. The ratio of surface energy before and after this change is:

Option: 1

 2:2\frac{2}{3}


Option: 2

2\tfrac{2}{3}:1 


Option: 3

2:1


Option: 4

1:2


Answers (1)

best_answer

Suppose, R is the radius of bigger drop. Then by equating volumes, we have,

 2\left ( \frac{4}{3}\Pi r^{3} \right ) = \frac{4}{3}\Pi R^{3} 

R=2\tfrac{1}{3}r

Now, Surface energy ∝ Surface area

 Therefore, \frac{U1}{U2}=\frac{A1}{A2}=\frac{2r^{2}}{R^{2}}=\frac{2}{2^{2/3}}

Posted by

Divya Prakash Singh

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