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

Question

Drops of liquid of density d are floating half immersed in a liquid of density ρ. If the surface tension of the liquid is T, then the radius of the drop is (A) √(3 T/g(3 d-ρ)) (B) √(6 T/g(2 d-ρ)) (C) √(3 T/g(2 d-ρ)) (D) √(3 T/g(4 d-3 ρ))

Tardigrade Admin · Accepted Answer

According to the question, or (4/3) π r3 d g =(2/3) π r3 ρ g+T × 2 π r r =√(3 T/g(2 d-ρ))

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

$\frac{3 T}{g ( 3 d - ρ )}$

$\frac{6 T}{g ( 2 d - ρ )}$

$\frac{3 T}{g ( 2 d - ρ )}$

$\frac{3 T}{g ( 4 d - 3 ρ )}$

According to the question,
or $\frac{4}{3} π r^{3} d g = \frac{2}{3} π r^{3} ρ g + T \times 2 π r$
$r = \frac{3 T}{g ( 2 d - ρ )}$

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

g(3d−ρ)3T​​

g(2d−ρ)6T​​

g(2d−ρ)3T​​

g(4d−3ρ)3T​​