The excess pressure inside a spherical drop of water is four times that of another drop. Then, their respective mass ratio is

Answer

64

Explanation

Excess pressure inside a spherical drop of water ` P = (2T)/( R ) `
given ` P _ 1 = 4 P_2, (2 T)/(R_1) = 4 xx (2T ) /(R_2) or R_2 = 4 R _ 1 ` Now, ` ( M_1 )/(M_2 ) = ((4)/(3) pi R_1 ^3 rho ) /((4)/(3) pi R_2^3 rho ) , (M_1)/(M_2) = (R_1 ^3 ) /(R_2^3 ) , ( M_1 ) /(M_2) = ( 1)/(64) `