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

Option 1 -

$\frac{15}{\sqrt[]{(2 ρ - σ)}}$

Option 2 -

$\frac{15}{\sqrt[]{(ρ - σ)}}$

Option 3 -

$\frac{3}{2 \sqrt[]{(ρ - σ)}}$

Option 4 -

$\frac{3}{20 \sqrt[]{(2 ρ - σ)}}$

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

Since liquid drop is in equilibrium, so

mg = F_B + 2πRT

$\Rightarrow \frac{4}{3} π R^{3} ρ g = \frac{2}{3} π R^{3} σ g + 2 π R T$

$\Rightarrow R = \frac{3 T}{(2 ρ - σ) g} = \frac{15}{\sqrt[]{(2 ρ - σ)}} \times 1 0^{- 2} m = \frac{15}{\sqrt[]{(2 ρ - σ)}} c m$

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