Question

a liquid drop of diameter d breakes into 27 drops . find the resulting change in energy .take surface tension as T

Answer 1

Let R is the radius of big drop
r is the radius of tiny drop
Then, volume of big drop = 27 × volume of tiny drop
4/3 πR³ = 27 × 4/3 πr³
R³ = 27r³ ⇒ R = 3r ----(1)

Now, we should find out increment of surface area of drops
= 27 × Surface area of tiny drop - surface area of big drop
= 27 × 4πr² - 4πR²
= 4π [27(R/3)² - R² ] , here r = R/3 from above equation (1)
= 4π[ 3R² - R²] = 8πR²

Now, Energy change = increases surface area × surface tension
Let surface tension of liquid is S
Then, Energy change =8πR² × S
but here given diameter of big drop is d , so d = 2R
hence, energy change = 2πd² × S