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Asked by Prashant DIGHE | 20 Oct, 2019, 07:21: AM
answered-by-expert Expert Answer
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 Thiyagarajan K | 20 Oct, 2019, 20:47: PM

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