Physics - Fluid Question with Solution | TestHub

Let $R$ be the radius of the liquid drop.

The weight of the drop acting downwards is:
$W=\frac{4}{3}\pi R^3\rho g$

Since the drop is half immersed, the volume of the displaced liquid is $\frac{2}{3}\pi R^3$. The buoyant force (upthrust) acting upwards is:
$F_B=\frac{2}{3}\pi R^{3}dg$

The surface tension force acts upwards along the circumference of the drop at the liquid surface. The length of the contact line is $2\pi R$. The force due to surface tension is:
$F_T=2\pi RT$

For the drop to be in equilibrium, the total downward force must equal the total upward force:
$W=F_B+F_T$

$\frac{4}{3}\pi R^3\rho g=\frac{2}{3}\pi R^{3}dg+2\pi RT$

$\frac{2}{3}\pi R^3(2\rho -d)g=2\pi RT$

$R^2=\frac{3T}{g(2\rho -d)}$

$R={\left[\frac{3T}{g(2\rho -d)}\right]}^{\frac{1}{2}}$

The diameter of the drop is $D=2R$:
$D=2{\left[\frac{3T}{g(2\rho -d)}\right]}^{\frac{1}{2}}={\left[\frac{12T}{g(2\rho -d)}\right]}^{\frac{1}{2}}$

Answer: ${\left[\frac{12\mathrm{T}}{\mathrm{g}(2\rho -\mathrm{d})}\right]}^{\frac{1}{2}}$