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`        A big drop of radius 1mm is divided in to 1000 small drop if the surface tension of liquid is 72 dyne/cm calculate the amount of work`
7 months ago

```							A liquid drop breaks into 1000 droplets of equal size .Means , volume of big drop = 1000 × volume of each droplet Let R is the radius of big drop and r is the radius of small droplet.4/3πR³ = 1000 × 4/3 πr³ R³ = (10r)³ ⇒ R = 10r 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= 1000T.a {T is surface tension and a is area}E₂= 1000T × 4πr² Now, change in energy = E₂ - E₁ = 1000T × 4πr² - T × 4πR²= T × 4π [ 1000r² - R² ] = T × 4π [ 1000(R/10)² - R² ] { ∵R = 10r }= T × 4π [ 10R² - R²]= T × 4π × 9R² = 36πR²T Now, put the values of T and R = 36 × 3.14 × (1 × 10⁻³)² × 0.07 J/m²=0.791 × 10⁻⁵ J/m²
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7 months ago
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