27 identical liquid droplets, each of radius r and density , coalesce to form a single large drop. If the surface tension of the liquid is T and the surface energy released during this process is converted entirely into the kinetic energy of the resulting large drop, the speed of the large drop is:

Step-by-step solution

Let the radius of the large drop be R . By conservation of volume: 4 3 R^3 = 27 ( 4 3 r^3 ) R = 3r The initial total surface energy of the 27 droplets is: U_i = 27(4 r^2 T) = 108 r^2 T The final surface energy of the large drop is: U_f = 4 R^2 T = 4 (3r)^2 T = 36 r^2 T The surface energy released during coalescence is: U = U_i - U_f = 108 r^2 T - 36 r^2 T = 72 r^2 T The mass of the large drop is: M = 4 3 R^3 = 4 3 (3r)^3 = 36 r^3 The released energy is converted into kinetic energy of the large drop: 1 2 M v^2 = U