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

Step-by-step solution

Excess pressure inside 1^ st spherical drop (A) is given by, P _A= 2 T r₁ For 2^ nd drop (B), P_B= 2 T r₂ - 2 ~T r ₁ =4 ( 2 ~T r ₂ ) (given, P _ A =4 P _ B ) r₁ r₂ = 1 4 Now, m₁ m₂ = V₁ ₁ V₂ ₂ As both are drops of water, ₁= ₂ m ₁ ~m ₂ = V ₁ ~V ₂ = 4 3 r ₁^3 4 3 ₂^3 = ( 1 4 )^3= 1 64