MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass
        
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

Answers : (1)

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
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

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


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details