Question

Two mercury drops (each of radius 'r') merge to from bigger drop. The surface energy of the bigger drop,
the surface tension, is
(A) 4TP T
(B) 21tr2 T
(C) 28/3/2T (D) 25/340 2T
orshire remains​

Answer 1

Answer:

We know that,

Sum of volumes of 2 smaller drops = volume of the bigger drop

2×  

3

4

​  

πr  

3

=  

3

4

​  

πR  

3

 

R=2  

3

1

​  

 

r

Now,

Surface energy

=T4πR  

2

 

=Tπ2  

3

8

​  

 

r  

2

 

Hence, the surface energy is Tπ2  

3

8

​  

 

r  

2