A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

Question

A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet): (A) r=√(2T/3(d+ρ)g) (B) r=√(T/(d+ρ)g) (C) r=√(T/(d-ρ)g) (D) r=√(3T/(2d-ρ)g)

Tardigrade Admin · Accepted Answer

FBD of droplet B + F = mg B = ((2/3)π R3)ρ g F = T(2π R) m =d ((4/3)π R3) ((2/3)π R3)ρ g + T(2π R) = ((4/3)π R3)g T(2π R) = ((2/3)π R3)g [2d-ρ] R = √(3T/(2d-ρ)g)

Q. A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $ρ$ and surface tension $T$ . The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

$r = \frac{2 T}{3 ( d + ρ ) g}$

$r = \frac{T}{( d + ρ ) g}$

$r = \frac{T}{( d - ρ ) g}$

$r = \frac{3 T}{( 2 d - ρ ) g}$

FBD of droplet
$B + F = m g$
$B = (\frac{2}{3} π R^{3}) ρ g$
$F = T (2 π R)$
$m = d (\frac{4}{3} π R^{3})$
$(\frac{2}{3} π R^{3}) ρ g + T (2 π R) = (\frac{4}{3} π R^{3}) g$
$T (2 π R) = (\frac{2}{3} π R^{3}) g [2 d - ρ]$
$R = \frac{3 T}{( 2 d - ρ ) g}$

Q. A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

r=3(d+ρ)g2T​​

r=(d+ρ)gT​​

r=(d−ρ)gT​​

r=(2d−ρ)g3T​​