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

Answer

D

Explanation

In equilibrium force due to surface tension `+` force of buoyancy `=` Weight of the spherical liquid drop
`2pirT+(2)/(3)+(2)/(3)pir^(3)d_(2)g=(4)/(3)pir^(3)d_(1)g`
`T=sigma,d_(1)=rho,d_(2)=d`
`2pirsigma+(2)/(3)pir^(3)dg=(4)/(3)pir^(3)rhog`
`impliesr^(2)=(3sigma)/(g(2rho-d))implies[(D)/(2)]^(2)=(3sigma)/(g(2rho-d))`