Excess presuure of soap bubble of radius 1 cm is balanced by a column of oil of specific gravity 0.8 , 2mm high , the surface tension will be ? i got the answer but decimal is coming in the wrong place so its ok if u show just the calculation part n thanks in advance :)

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2015-05-03T14:52:36+05:30

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pressure at the depth of 2 mm in oil
         = 0.8 * 1000 kg/m² * 9.81 m/sec² * 0.002 m
         = 15.696 N/m²

P = P2 - P1
   = excess pressure of a soap bubble = pressure inside - pressure outside
   =  4 S / R ,    where S = surface tension  and  R = radius of the bubble.

S = P * R / 4 = 15.696 *  0.01 m / 4 = 0.03924 N/m

if there is some different answer, perhaps the given values of radii  are not correct..

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