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The Density of Ice is 0.92 G Cm-3 and that of Sea Water is 1.025 G Cm-3. Find the Total Volume of an Iceberg Which Floats with Its Volume 800 Cm3 Above Water.

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Question

The density of ice is 0.92 g cm-3 and that of sea water is 1.025 g cm-3. Find the total volume of an iceberg which floats with its volume 800 cm3 above water. 

Sum
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Solution

Let V be the volume of the iceberg. 

Volume of iceberg above water = 800 cm3 

Volume of iceberg submerged in water = v 

Density of ice (ρice) = 0.92 g cm-3

Density of sea water `(ρ_"sea water")` = 1.025 gcm-3

According to the law of floatation,

`"Volume of immersed part of body"/"Total volume of body" = "Density of body"/"Density of liquid"`

`v/"V" = ρ_"ice"/ρ_"seawater"`

⇒ `v/"V" = 0.92/1.025`

⇒ `v/"V" = 0.92/1.025`

⇒ `v/"V"` = 0.8976

⇒ `v` = 0.89756 V

Now,

∴ Floating Volume of iceberg = V - `v` 

⇒ 800 = V - 0.89756 V 

⇒ 800 = V(1 - 0.89756)

⇒ `800/0.10244` = V

⇒ V = 7809.45 cm3 ≈ 7809.5 cm3

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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 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 8 | Page 124

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