MHT CET20249 May 2024Evening ShiftPhysicsMechanical Properties of FluidsActual
Let ' n ' is the number of liquid drops, each with surface energy ' E '. These drops join to form single drop. In this process
Options
- Asome energy will be absorbed
- Benergy absorbed is [E (n-n^ 2 / 3 ) ]
- Cenergy released will be [E (n-n^ 2 / 3 ) ]
- Denergy released will be [ E (2^ 2 / 3 -1 ) ]
Correct answer
C. energy released will be [E (n-n^ 2 / 3 ) ]
Step-by-step solution
Let r= radius of each small drop and R = radius of a big single drop. Then, n 4 3 r ^3= 4 3 R ^3 R = n ^ 1 / 3 r ...(i) Initial surface energy, E ₁= n 4 r ^2 T = nE Final surface energy, aligned E ₂ & =4 R ^2 T =4 r ^2 n ^ 2 / 3 T [ From ( i )] & = n ^ 2 / 3 E aligned Energy released =E₁-E₂= [E (n-n^ 2 / 3 ) ]