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A Wooden Cube of Side 10 Cm Has Mass 700 G. What Part of It Remains Above the Water Surface While Floating Vertically on Water Surface?

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Question

A wooden cube of side 10 cm has mass 700 g. What part of it remains above the water surface while floating vertically on water surface? 

Sum
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Solution

Given, Side of wooden cube = 10 cm

Hence, Volume of wooden cube = 10 cm × 10 cm × 10 cm = 1000 cm3

Mass = 700 g

Density = `"mass"/"volume"`

∴ Density of wooden cube = `700/1000` = 0.7 g cm-3

By the principle of floatation,

`"Volume of immersed part"/"Total volume" = "Density of wood"/"Density of water"`

Density of water = 1 g cm-3

Density of wooden cube = 0.7 g cm-3 

∴ `"Volume of immersed part"/"Total volume" = 0.7/1`

Hence, fraction submerged = 0.7

Height of wooden cube = 10 cm

Part of wooden cube which is submerged = 10 x 0.7 = 7 cm

Therefore, part above water = 10 - 7 = 3 cm

Hence, 3 cm of height of wooden cube remains above water while floating.

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Relation Between Volume of Submerged Part of a Floating Body, the Densities of Liquid and the Body
  Is there an error in this question or solution?
Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation - Exercise 5 (C) [Page 124]

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Selina Concise Physics [English] Class 9 ICSE
Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (C) | Q 3 | Page 124

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