Question

A liquid drop of diameter d breaks up into 27 tiny drops find the resulting change in energy take surface tension of the liquid as sigma

Answer 1

A liquid drop breaks into 27 droplets of equal size .
Means , volume of big drop = 27 × volume of each droplet
Let R is the radius of big drop and r is the radius of small droplet.
4/3πR³ = 27 × 4/3 πr³
R³ = (3r)³ ⇒ R = 3r

intial energy of liquid drop = E₁ = T.A { here T is surface tension and A is area}
E₁ = T × 4πR²

final energy of 1000 droplets = E₂= 27T.a {T is surface tension and a is area}
E₂= 27T × 4πr²

Now, change in energy = E₂ - E₁
= 27T × 4πr² - T × 4πR²
= T × 4π [ 27r² - R² ]
= T × 4π [ 27(R/3)² - R² ] { ∵R = 10r }
= T × 4π [ 3R² - R²]
= T × 4π × 2R²
= 8πR²T
∵ surface tension is σ And diameter of drop is d e.g., R = d/2
so, change in energy = 2πd²σ

Answer 2

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