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

(A) $\frac{15}{\sqrt{2 \rho-\sigma}}$

(B) $\frac{15}{\sqrt{\rho-\sigma}}$

(C) $\frac{3}{2 \sqrt{\rho-\sigma}}$

(D) $\frac{3}{20 \sqrt{2 \rho-\sigma}}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Fluid Mechanics

A drop of liquid of density

ρ

is floating half immersed in a liquid of density

σ

and surface tension

7.5 \times 1 0^{- 4} Ncm^{- 1}

. The radius of drop in cm will be : (Take :

g = 10 m / s^{2}

)

Answer: (A)

For equilibrium, the buoyant force plus surface tension force equals the weight of the drop:

σ \frac{V}{2} g + 2 π R T = ρ V g

where

V = \frac{4}{3} π R^{3}

. Substituting:

2 π R T = \frac{( 2 ρ - σ )}{2} \cdot \frac{4}{3} π R^{3} g

Simplifying:

R^{2} = \frac{3 T}{( 2 ρ - σ ) g}

Substituting

T = 7.5 \times 1 0^{- 4} N/cm = 7.5 \times 1 0^{- 2} N/m

and

g = 10 m/s^{2}

R = \frac{3 \times 7.5 \times 1 0 ^{- 2}}{( 2 ρ - σ ) \times 10} = \frac{3}{20 2 ρ - σ} m = \frac{15}{2 ρ - σ} cm

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option