A spherical body of density `rho` is floating half immersed in liquid of density d, if `alpha` is the surface tension of the liquid, then the diameter of the body is

Options

A. `sqrt((3alpha)/(g(2rho-d)))`
B. `sqrt((6sigma)/(g(2rho-d)))`
C. `sqrt((4sigma)/(g(2p-d)))`
D. `sqrt((12 sigma)/(g(2rho-d)))`

Answer

A

Explanation

weight of body
= Bouyant force + Force of suface tension
`(4)/(3)pi r^3 rho xx g = 2/3 pi r^3 dg + 2 pi r delta `
`2/3 pi r^3 g (2 rho -d)=2 pi alpha`
`So, r^2= (3 alpha)/(g(rho2p-d)`
So, `r = sqrt(3 alpha)/(g(2rho -d))`