Question

A liquid drop of density ρ is floating half immersed in a liquid of density d. The diameter of the liquid drop is ______.

(ρ > d, g = acceleration due to gravity, T = surface tension)

Options

• `[(3T)/(g(2rho - d))]^{1/2}`

• `[(12T)/(g(2rho - d))]^{1/2}`

• `[(6T)/(g(rho - d))]^{1/2}`

• `[(9T)/(g(rho - d))]^{1/2}`

Answer 1

A liquid drop of density ρ is floating half immersed in a liquid of density d. The diameter of the liquid drop is `underlinebb([(12T)/(g(2rho - d))]^{1/2})`

Explanation:

The equation of surface tension of the liquid drop floating half-submerged is

`2pirT + 1/2 xx (4/3pir^3)d = 4/3pir^3rhog`

⇒ `2pirT = (2pir^3g)/3(2rho - d)`

⇒ `r^2 = (3T)/(g(2rho - d)) ⇒ r = sqrt((3T)/(g(2rho - d)))`

∴ Diameter, d = 2r = `[(12T)/(g(2rho - d))]^{1"/"2}`