Drops of liquid of density d are floating half immersed in a liquid of density p. If the surface tension of the liquid is T, then the radius of the drop is

Options

A. `sqrt((3T)/(g(3d-p)))`
B. `sqrt((6T)/(g(2d-p)))`
C. `sqrt((3T)/(g(2d-p)))`
D. `sqrt((3T)/(g(4d-p)))`

Answer

A

Explanation

Given
`4/3 pi r^(3) dg=2/3 pi r^(3) rho g + Txx2pir`
`T=(4/3 pir^(3)dg-2/3 pi r^(3) rhog)/(2pir)`
`=1/3pir^(3)g(4d-2p)xx1/(2pir)`
`therefore r= sqrt((6T)/((4d-2p)g))=sqrt((3T)/((2d-rho)g))`