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ICSE Class 9 Physics Question 11 of 16

Upthrust in Fluids, Archimedes' Principle and Floatation — Question 4

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4
Question

Question 1(iv)

For a floating body of volume V, the relation between the volume of its submerged part v, the densities of liquid (ρL) and the body (ρS) is :

  1. vV\dfrac{\text v}{\text V} = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}

  2. Vv\dfrac{\text V}{\text v} = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}

  3. v×ρs=V×ρL\text v \times \text ρ_\text s = \text V \times \text ρ_\text L

  4. none of these

Answer

vV\dfrac{\text v}{\text V} = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}

Reason —

Given,

Volume of body = V

Volume of body submerged in liquid = v

Density of body = ρs

Density of liquid = ρL

Let weight of the body be W.

W = volume of the body x density of the body x g = V ρs g

Weight of liquid displaced by the body will be equal to upthrust. Let it be FB.

FB = volume of the liquid displaced x density of the liquid x g = v ρL g

From principle of floatation,

W = FB

⇒ V ρs g = v ρL g

⇒ vV\dfrac{\text v}{\text V} = ρsρL\dfrac{\text ρ_\text s}{\text ρ_\text L}

Hence proved.