Highly selective Backlog Qs for JEE MainPhysicsMechanical Properties of Fluids
A certain number of spherical liquid drops of radius r coalesce to form a single drop of radius R and volume V . If T is the surface tension of the liquid, which one of the following statements is true for the energy (E) in the process?
Options
- AE=3 V T [ 1 r - 1 R ] is absorbed.
- BE=4 V T [ 1 r - 1 R ] is released.
- CE=3 V T [ 1 r - 1 R ] is released.
- DE=4 V T [ 1 r - 1 R ] is absorbed.
Correct answer
C. E=3 V T [ 1 r - 1 R ] is released.
Step-by-step solution
Change in surface energy is given by E=T( A) ---(1) The initial area is given by, A = (4 r^2 ) n The final area is given by, a =4 R^2 Therefore, the change in area is given by, aligned & A=a-A & A=4 (n r^2-R^2 )---(2) aligned Now, using volume conservation: ( 4 3 r^2 ) n= 4 3 R^3 n= R^3 r^3 ---(3) A=4 [ R^3 r^3 r^2-R^2 ]=4 [ R^3 r - R^3 R ]= ( 4 3 R^3 ) 3 [ 1 r - 1 R ]=3 V [ 1 r - 1 R ] Introducing above value in equation (1) E=3 V T [ 1 r - 1 R ]