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Step-by-Step Solution
Step 1: Identify the forces acting on the droplet
1. Weight of the droplet (downward): If the droplet has density $d$, radius $r$, then its volume is $ \frac{4}{3}\pi r^3$. Therefore, its weight is
$ W = \frac{4}{3}\pi r^3\,d\,g.$
2. Buoyant force (upward): The droplet is half-immersed in the liquid of density $ \rho $. Hence, the submerged volume is half of the droplet’s total volume, i.e.
$ \frac{1}{2}\times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3.$
Thus, the buoyant force is
$ F_{b} = \left(\frac{2}{3}\pi r^3\right)\,\rho\,g.$
3. Surface tension force (upward): The droplet makes a circular contact of radius $r$ with the liquid. The surface tension $T$ acts along that contact circumference, providing an upward pull:
$ F_{s} = (2\pi r)\,T.$
(The factor $2\pi r$ is the length of the contact line where tension acts.)
Step 2: Write the equilibrium condition
As the droplet floats in equilibrium, the total upward force (buoyant force + surface tension force) balances the total downward force (weight of the droplet). Mathematically:
$ F_{b} + F_{s} = W. $
Substitute the expressions for each force:
$ \left(\frac{2}{3}\pi r^3\,\rho\,g\right) + \left(2\pi r\,T\right) = \frac{4}{3}\pi r^3\,d\,g. $
Step 3: Rearrange and simplify
Bring the terms involving $r^3$ together and keep the surface tension term on one side:
$ 2\pi r\,T = \frac{4}{3}\pi r^3\,d\,g - \frac{2}{3}\pi r^3\,\rho\,g. $
Factor out the common factors on the right-hand side:
$ 2\pi r\,T = \frac{2}{3}\pi r^3 (2d - \rho)\,g. $
Cancel out $2\pi r$ (assuming $r \neq 0$):
$ T = \frac{1}{3} \, r^2 (2d - \rho)\,g. $
Step 4: Solve for the radius $r$
Rearrange to isolate $r^2$:
$ r^2 = \frac{3T}{(2d - \rho)\,g }. $
Taking the square root:
$ r = \sqrt{\frac{3T}{(2d - \rho)\,g}}. $
This matches the given correct option.
