Question

Eight droplets of water, each of radius 0.2 mm, coalesce into a single drop. Find the change in total surface energy. (Surface tension of water = 0.072N/rn) (Ans : 1.446 x 10⁻⁷ J)

Answer 1

Given :

n = 8

T = 0.072 N/m

r = 0.2 mm = 0.2 × 10⁻³ m = 2 × 10⁻⁴ m

R = radius of drop  

: E = T∆A

Volume of 8 drops of radius ‘r’ is equal to volume of single drop of radius ‘R’

[4/ 3 ]πr³n = 4/ 3 πR³

 nr³ = R³

∴ R = r.∛ n = r. ∛8 = 2r  

R = 2 × 2 × 10⁻⁴m

Total work done in drops = change in surface energy.

∴ E = T × 4π (nr² – R² )  

= T × 4π [8 × (2 × 10⁻⁴) ² – (4 × 10⁻⁴) ² ]  

= 0.072 × 4 × 3.14 [8 × 4 × 10⁻⁸ – 16 × 10⁻⁸]  

= 0.072 × 4 × 3.14 × 16 × 10⁻⁸

∴ E = 1.446 × 10⁻⁷ J

Answer 2

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