A drop of liquid of density $$\rho$$ is floating half immersed in a liquid of density $$\sigma$$ and surface tension $$7.5 \times 10^{-4} \text{ N cm}^{-1}$$. The radius of drop in cm will be: (Take: $$g = 10 \text{ m s}^{-2}$$)

A liquid drop of density $$\rho$$ floats half-immersed in a liquid of density $$\sigma$$ with surface tension $$T = 7.5 \times 10^{-4} \text{ N cm}^{-1} = 7.5 \times 10^{-2} \text{ N m}^{-1}$$.

Set up the equilibrium condition.

For the floating drop, three forces act: weight (downward), buoyancy (upward), and surface tension (upward, acting along the contact circle of radius $$r$$).

$$\text{Weight} = \text{Buoyancy} + \text{Surface tension force}$$ $$\frac{4}{3}\pi r^3 \rho g = \frac{1}{2} \times \frac{4}{3}\pi r^3 \sigma g + T \times 2\pi r$$

The buoyancy equals the weight of liquid displaced by the immersed half-sphere (volume $$\frac{2}{3}\pi r^3$$), and the surface tension acts along the circular contact line of circumference $$2\pi r$$.

Simplify the equation.

$$\frac{4}{3}\pi r^3 \rho g - \frac{2}{3}\pi r^3 \sigma g = 2\pi r T$$ $$\frac{2}{3}\pi r^3 (2\rho - \sigma) g = 2\pi r T$$ $$r^2 = \frac{3T}{g(2\rho - \sigma)}$$

Substitute values and find r in cm.

Working in SI units:

$$r^2 = \frac{3 \times 7.5 \times 10^{-2}}{10 \times (2\rho - \sigma)} = \frac{0.225}{10(2\rho - \sigma)} = \frac{0.0225}{(2\rho - \sigma)}$$ $$r = \frac{0.15}{\sqrt{2\rho - \sigma}} \text{ m} = \frac{15}{\sqrt{2\rho - \sigma}} \text{ cm}$$

The correct answer is Option A: $$\dfrac{15}{\sqrt{2\rho - \sigma}}$$.