Question

Two small drops mercury, each of radius r, coalesces the form a single large drop. The ratio of the total surface energies before and after the change is. 

(A) 1 : 2^(1/2) 

(B) 2^(1/3) : 1 

(C) 2 : 1 

(D) 1 : 2

Answer 1

Correct option: (B) 2^(1/3) : 1

Explanation:

If n small drops (each of radius r) combines to form drop of radius R.

As mass is same

(4/3)πn ∙ r³ = (4/3)πr³ 

∴ R³ = nr³ 

Here

n = 2

R³ = 2r³ 

R = 2^(1/3) ∙ r

∴ (r² / R²) = 2^–(2/3)          (1)

[{Initial surface energy} / {final surface energy}]

= {(2 × 4πr²T) / (4πR²T)} ------ initially there are two drops of radius r.

= (r² / R²) × 2

= 2 × 2^–(2/3) from (1)

= 2^(1/3) 

Ratio = 2^(1/3) : 1