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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: (i) volume of the body (ii) upthrust due to liquid (iii) relative density of the solid and (iv) relative density of the liquid.

Answer in Brief

Solution

(i) Volume of the body = W- 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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