A large water drop falls through air and attains its terminal velocity. It then breaks into 27 identical smaller droplets, which eventually attain their own steady terminal velocities. Assuming the buoyant force from air is negligible, the ratio of the rate of heat generation by viscous drag for the original large drop to that for one individual smaller droplet is

Step-by-step solution

Let the radius of the large drop be R and that of a smaller droplet be r . By conservation of volume: 4 3 R^3 = 27 4 3 r^3 R = 3r The rate of heat generation (power dissipated) by viscous drag is P = F_v v_T . At terminal velocity, the viscous drag force F_v equals the weight of the drop (since buoyancy is negligible). F_v = mg = 4 3 R^3 g F_v R^3 The terminal velocity is given by v_T = 2 R^2 g 9 v_T R^2 Therefore, the power dissipated is P R^3 R^2 P R^5 . The ratio of the rate of heat generation by the large drop