Question

Drops Of Liquid Of Density D Are Floating Half Immersed In A Liquid Of Density. If The Surface Tension Of Liquid Is T, Then Radius Of The Drop Will Be

Answer 1

answer : For the drops to be in equilibrium,

Upward force on drop= Downward force on drop

⟹T(2πR)=d34πR3g−ρ32πR3g

                          =32πR3(2d−ρ)g

⟹R2=g(2d−ρ)3T

⟹R=g(2d−ρ)3T

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Answer 2

Answer:

Radius = $\sqrt{\frac{3T}{g(2d-P)} }$.

Explanation:

For the drops to be in equilibrium,

Upward force on drop= Downward force on drop

Let the density of the liquid be P.

Let the radius be R.

$T(2\pi R)=d\frac{4}{3}\pi R^3g-P\frac{2}{3} \pi R^3g$

            $=\frac{2}{3} \pi R^3(2d-P)g$

$R^2=\frac{3T}{g(2d-P)}$

$R=\sqrt{\frac{3T}{g(2d-P)} }$  

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