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A Body Weighs W1gf in Air and When Immersed in a Liquid It Weighs W2gf, While It Weights W3gf on Immersing It in Water. Find: (I) Volume of the Body (Ii) Upthrust Due to Liquid (Iii) Relative Den - Physics

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Question

A body weighs W1gf in air and when immersed in a liquid it weighs W2gf, while it weights W3gf on immersing it in water. Find:

  1. volume of the body
  2. upthrust due to liquid
  3. relative density of the solid
  4. relative density of the liquid
Long Answer
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Solution

(i) Volume of the body = W1 − Wcm3

(ii) Upthrust due to liquid = loss in weight when immersed in liquid = W1  Wgf

(iii)

Weight of a body in air = W1gf

Weight of that body in liquid = W2gf

Weight of that body in water = W3gf 

RD of solid = `"Weight of solid in air"/"Weight in air - Weight in water"`

= `"W"_1/("W"_1 - "W"_3)`

(iv)

Weight of a body in air = W1gf

Weight of that body in liquid = W2gf

Weight of that body in water = W3gf 

RD of Liquid = `(W_1 - W_2)/(W_1 - W_3)`

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Determination of Relative Density of a Solid Substance by Archimedes’ Principle
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Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation - Exercise 5 (B) [Page 116]

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Selina Concise Physics [English] Class 9 ICSE
Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (B) | Q 15 | Page 116
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