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Please answer this

Asked by Prashant DIGHE 20th October 2019, 7:21 AM
Answered by Expert
surface tension force acting upwards on the floating drop along the circumference that equals ( π D σ) ,
where D is diameter of the drop and σ is surface tension .
bouyance force acting upwards on the drop = (1/12)πD3 d g
weight of the drop acting downwards = (1/6)πD3 ρ g
Hence ,  ( π D σ) + (1/12)πD3 d g = (1/6)πD3 ρ g  ..................(1)
By simplifying above expression,  we get, begin mathsize 14px style D space equals space square root of fraction numerator 12 space sigma over denominator open parentheses 2 space rho space minus space d close parentheses space g end fraction end root end style
Answered by Expert 20th October 2019, 8:47 PM
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