A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `sigma` is the surface tension the diameter of the drop of the liquid is

Options

A. ` R = sqrt((3T ) /(g (2 rho - sigma ))) `
B. `R = sqrt ((3 T )/(2g (2 rho - sigma ))) `
C. ` R = sqrt ((3T)/(2g rho)) `
D. `R = sqrt ((2T)/(3g rho))`

Answer

A

Explanation

Let's consider equilibrium of the drop ` F_("surfacetension") + F_("Buotantforce") = w `
` 2pi R T + (2)/(3) pi R ^ 3 sigma g = (4)/(3) pi R ^ 3 rho g , ` Solving ` R = sqrt((3 T)/((2rho - sigma )g) `