Class 11th Overview Physics Physics Laws of Motion

**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 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>5</mn> </mrow> <mrow> <mroot> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mi>ρ</mi> <mo>−</mo> <mi>σ</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>5</mn> </mrow> <mrow> <mroot> <mrow> <mrow> <mo>(</mo> <mrow> <mi>ρ</mi> <mo>−</mo> <mi>σ</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> <mroot> <mrow> <mrow> <mo>(</mo> <mrow> <mi>ρ</mi> <mo>−</mo> <mi>σ</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> <mn>0</mn> <mroot> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mi>ρ</mi> <mo>−</mo> <mi>σ</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Payal Gupta | Contributor-Level 10

8 months ago

Correct Option - 1

Detailed Solution:

Since liquid drop is in equilibrium, so

mg = F_B + 2πRT

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

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

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