A liquid drop of diameter 4mm breaks into 1000 droplets of equal size.calculate the resultant change in the energy.the surface tension of the liquid is 0.07N/m
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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 × (2 × 10⁻³)² × 0.07 J/m²
=3.165 × 10⁻⁵ J/m²
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 × (2 × 10⁻³)² × 0.07 J/m²
=3.165 × 10⁻⁵ J/m²
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