MHT CET202619 April 2026Evening ShiftPhysicsMechanical Properties of FluidsActual
The energy needed for breaking a liquid drop of radius 'R' into 'n' droplets each of radius 'r' is [ T = surface tension of the liquid]
Options
- A4 T R^2 [ R r - 1 ]
- B4 T R [ R r - 1 ]
- C4 T [ R^2 r^2 - 1 ]
- D4 T [1 + R^3 r ]
Correct answer
A. 4 T R^2 [ R r - 1 ]
Step-by-step solution
Volume of the initial drop is equal to the total volume of n droplets. 4 3 R^3 = n 4 3 r^3 n = R^3 r^3 Initial surface area of the drop, A₁ = 4 R^2 Final total surface area of n droplets, A₂ = n 4 r^2 Substituting the value of n : A₂ = ( R^3 r^3 ) 4 r^2 = 4 R^3 r Increase in surface area, A = A₂ - A₁ = 4 R^3 r - 4 R^2 A = 4 R^2 [ R r - 1 ] Energy required to break the drop, E = T A E = 4 T R^2 [ R r - 1 ] Answer: 4 T R^2 [ R r - 1 ]