A liquid drop falling through a viscous medium attains a terminal velocity v₀ . It is then broken into N identical smaller droplets. If the work done against surface tension during this process is equal to 3 times the initial surface energy of the unbroken drop, the terminal velocity of one of the smaller droplets will be

Step-by-step solution

Let the radius of the original drop be R and the radius of each smaller droplet be r . The initial surface energy is U_i = 4 R^2 T . The work done against surface tension is given as W = 3 U_i . The final surface energy is U_f = U_i + W = U_i + 3 U_i = 4 U_i . Since U_f = N(4 r^2 T) , we have: N(4 r^2 T) = 4(4 R^2 T) N r^2 = 4 R^2 From the conservation of volume, we know: 4 3 R^3 = N ( 4 3 r^3 ) N = ( R r )^3 Substituting N into the energy equation: ( R r )^3 r^2 = 4 R^2 R^3 r = 4 R^2 r = R 4 According to Stokes' l