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# Please answer this

Asked by Prashant DIGHE 20th October 2019, 7:21 AM
Answered by 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,

Answered by Expert 20th October 2019, 8:47 PM
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