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,