A drop of liquid of density \rho is floating half immersed in a liquid of density \boldsymbol{\sigma} and surface tension 7.5 \times 10^{-4} {Ncm}^{-1}. The radius of drop in {cm} will be : (g=10 m s^{-2}) [25-Jul-2022-Shift-2]

Correct Answer is : 15√(2ρ−σ)

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

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

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

JEE Mains 25-July-2022 Shift 2 Solved Paper

Section: Physics

Question : 33 of 90

Marks: +1, -0

A drop of liquid of density ρ is floating half immersed in a liquid of density boldsymbol‌σ and surface tension 7.5×10−4Ncm−1. The radius of drop in cm will be :
(g=10ms−2)

[25-Jul-2022-Shift-2]

‌
15
√(2ρ−σ)
‌
15
√(ρ−σ)
‌
3
2√(ρ−σ)
‌
3
20√(2ρ−σ)

Validate

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