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A block of wood of mass 24 kg floats on water. The volume of wood is 0.032 m3. Find: the volume of block below the surface of water, the density of wood. (Density of water = 1000 kg m−3) - Physics

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Question

A block of wood of mass 24 kg floats on water. The volume of wood is 0.032 m3. Find: 

  1. the volume of block below the surface of water,
  2. the density of wood.

(Density of water = 1000 kg m−3)

Numerical

Solution

Mass of block of wood = 24 kg

Volume of wood = 0.032 m3

(a) Upthrust = Volume of block below the surface of water (v) × density of liquid × g

Now for floatation , Upthrust  = weight of the body = 24 kgf

or , 24 kgf = v × 1000 × g

or , v = `24/1000 = 0.024  "m"^3`

(b) According to the law of floatation, 

`"Volume of the submerged block"/"Total volume of block" = "Density of wood"/"Density of water"`

or , `0.024/0.032 = "Density of wood"/1000`

or , Density of wood = `1000 xx 0.024/0.032 = 750  "kgm"^-3` 

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Relation Between Volume of Submerged Part of a Floating Body, the Densities of Liquid and the Body
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Chapter 4: Fluids - Exercise 2 [Page 174]

APPEARS IN

Frank Physics [English] Class 9 ICSE
Chapter 4 Fluids
Exercise 2 | Q 33 | Page 174
Selina Concise Physics [English] Class 9 ICSE
Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (C) | Q 2 | Page 124
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