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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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प्रश्न

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? 

योग

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Upthrust in Fluids, Archimedes’ Principle and Floatation - Exercise 5 (C) [पृष्ठ १२४]

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सेलिना Concise Physics [English] Class 9 ICSE
अध्याय 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (C) | Q 3 | पृष्ठ १२४
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