A spherical drop of water of radius 1mm and surface tension 72 x 10-3 N/m is split in to 1000 identical spherical droplets. Calculate the energy released in the process
Answers
Answered by
0
Answer:
The energy released in the process is 8.95×10^−5 Joules
Explanation:
Correct statement:
What is the work done in splitting a drop of water 1 mm radius into 106 droplets (surface tension of water =72×10−3J/m^2)?
Solution:
- Radius of big drop = R
- Radius of smaller drops = r
- 4 / 3πR^3 = n 4 / 3πr^3
R^3 = n . r^3
n=(R^3 / r^3)
n = ( R/r)^3
- Initial area of drop =4πR^2
- Final area of drops =n 4πr^2
Changes in area =ΔA= nπr^2 − 4πR^2
= 4π( n r^2−R^2)
Work done = 4πT (n.r^2−R^2)
W = 4πR^3.T(1 / r − 1 /R)
or = 4πR^2.T[n^1/3 − 1]
= 4π × (10^−3)^2×72×10^−3[10^6/3 − 1]
= 8.95×10^−5 Joules
Hence the energy released in the process is 8.95×10^−5 Joules
Also learn more
The radius of a sphere is 5 cm. Calculate the radius of gyration about it's diameter and about any tangent ?
https://brainly.in/question/1933979
Similar questions
Math,
5 months ago
Social Sciences,
5 months ago
Math,
5 months ago
Social Sciences,
9 months ago
Math,
9 months ago
India Languages,
11 months ago
Math,
11 months ago