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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
19 days ago

Arun
16143 Points

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
19 days ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions