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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? - Physics

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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

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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