Question

The total energy of the free surface of a liquid drop is 2 x 10⁻⁴ π times the surface tension of the liquid. What is the diameter of the drop? (Assume all terms in Si units)

Answer 1

Given :

E = 2πT  

E = T∆A  

∆A = 4πr ²

E = T∆A  

∴ E = 4πr²T

∴ 2πT = 4πr ²T [. . . E = 2πT]  

∴ 2r² = 1  

∴ r ²= 1/ 2  

∴ r = 1 /√2

d = 2r

2 × 1/√2 = √2  

∴ d = 1.414 m

∴the diameter of the drop is 1.414m

Answer 2

Given :

E = 2πT  

E = T∆A  

∆A = 4πr ²

E = T∆A  

∴ E = 4πr²T

∴ 2πT = 4πr ²T [. . . E = 2πT]  

∴ 2r² = 1  

∴ r ²= 1/ 2  

∴ r = 1 /√2

d = 2r

2 × 1/√2 = √2  

∴ d = 1.414 m

∴the diameter of the drop is 1.414m