A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5 × 10⁻⁴ Ncm ⁻¹. The radius of drop in cm will be: (g = 10 ms ⁻²) (a) (15)/(√((2ρ - σ))) (b) (15)/(√((ρ - σ))) (c) (3)/(2√((ρ - σ))) (d) (3)/(20√((2ρ - σ)))

Answer: (a). Weight = buoyancy + surface force gives r² = (3T)/((2ρ - σ)g). T = 7.5×10⁻⁴ N cm⁻¹ = 7.5×10⁻² N m⁻¹, g = 10. r² = (0.225)/(10(2ρ - σ)) = (0.0225)/(2ρ-σ) m² ⇒ r = (0.15)/(√(2ρ - σ)) m. r = (15)/(√(2ρ - σ)) cm (densities in SI). So the answer is (a).

A drop of liquid of density ρ is floating half immersed in a liquid… | JEE Main 2022 (25 Jul, Shift 2)

Q13JEE Main 2022 · 25 JUL · SHIFT 2Difficult

Force due to Surface Tension

A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5 \times 10^{-4}$ Ncm ${}^{-1}$ . The radius of drop in $\text{cm}$ will be:

(g = 10 ms ${}^{-2}$ )

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Answer: (a)

Weight = buoyancy + surface force gives $r^2 = \dfrac{3T}{(2\rho - \sigma)g}$ .
$T = 7.5\times10^{-4}\text{ N cm}^{-1} = 7.5\times10^{-2}\text{ N m}^{-1}$ , $g = 10$ .
$r^2 = \dfrac{0.225}{10(2\rho - \sigma)} = \dfrac{0.0225}{2\rho-\sigma}\text{ m}^2 \Rightarrow r = \dfrac{0.15}{\sqrt{2\rho - \sigma}}\text{ m}$ .
$r = \dfrac{15}{\sqrt{2\rho - \sigma}}\text{ cm}$ (densities in SI). So the answer is (a).

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