The excess pressure inside a spherical water drop A is four times that of another water drop B. Then, the ratio of the mass of water drop A to that of drop B is

Step-by-step solution

The excess pressure inside a spherical water drop of radius R and surface tension T is given by P = 2T R Given that P_A = 4 P_B 2T R_A = 4 ( 2T R_B ) R_B = 4R_A R_A R_B = 1 4 The mass of a spherical drop is M = 4 3 R^3 The ratio of their masses is M_A M_B = ( R_A R_B )^3 M_A M_B = ( 1 4 )^3 = 1 64 The ratio of the mass of water drop A to that of drop B is 1 : 64 Answer: 1 : 64