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The gauge pressure inside a soap bubble of surface tension T is equal to the pressure exerted by a liquid column of density and height h . What is the work done in blowing this soap bubble from zero radius? (Let g be the acceleration due to gravity)

Options

  1. A64 T^3 ^2 g^2 h^2
  2. B128 T^3 ^2 g^2 h^2
  3. C32 T^3 ^2 g^2 h^2
  4. D16 T^3 ^2 g^2 h^2

Correct answer

B. 128 T^3 ^2 g^2 h^2

Step-by-step solution

Excess pressure inside a soap bubble is given by P = 4T R , where R is the radius of the bubble. Gauge pressure exerted by a liquid column of height h and density is P = g h . Equating the two pressures as given: 4T R = g h R = 4T g h The work done in blowing a soap bubble from zero radius is equal to its surface energy. Since a soap bubble has two free surfaces (inner and outer), the work done is: W = T 2 4 R^2 = 8 R^2 T Substituting the value of R into the work equation: W = 8 T ( 4T g h )^2 W = 8 T ( 16T^2 ^2 g^

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