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A liquid drop of density ' Q ' is floating half immersed in a liquid of density ' d '. Diameter of the liquid drop is ( Q > d , g = acceleration due to gravity, T = surface tension)

Options

  1. A[ 3 T g(2 Q-d) ]^ 1 2
  2. B[ 6 T g(Q-d) ]^ 1 2
  3. C[ 12 ~T ~g (2 Q - d ) ]^ 1 2
  4. D[ 9 T g(Q-d) ]^ 1 2

Correct answer

B. [ 6 T g(Q-d) ]^ 1 2

Step-by-step solution

The force balance invovles, surface-tension force and boyancy force balance the weight of the droplet, so the equation is: (2 r) T+ ( 4 3 r^3 ) dg = ( 4 3 r^3 ) Qg Where, r is the radius of the drop, T the interfacial tension between the liquids, g is acceleration due to gravity. Therefore, T = ( 2 3 r ^2 )( Q - d ) g Diameter =2 r= [ 6 T g(Q-d) ]^ 1 2 .

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