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The R.D. of ice is 0.92 and that of sea water is 1.025. Find the total volume of an iceberg which floats with its volume 800 cm3 above water. - Physics

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Question

The R.D. of ice is 0.92 and that of sea water is 1.025. Find the total volume of an iceberg which floats with its volume 800 cm3 above water.
Answer in Brief

Solution

Relative density of Ice = 0.92
Relative density of sea water = 1.025
Let the total volume of iceberg = X cm3.
The volume of the iceberg above water = 800 cm3.
The volume of the iceberg is submerged in the water = (X - 800) cm3.
Fraction of iceberg submerged = (X- 800)/X
Now we know that the fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of ice / Density of sea water) = fraction submerged
0.92/1.025 = (X-800)/X
0.8975 X = X - 800
X - 0.8975 X = 800
0.1025 X = 800
X = 800/0.1025 = 7804.8 cm3.
Total volume of iceberg = 7804.8 cm3.

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Determination of Relative Density of a Solid Substance by Archimedes’ Principle
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Chapter 4: Fluids - Exercise 2 [Page 174]

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Frank Physics [English] Class 9 ICSE
Chapter 4 Fluids
Exercise 2 | Q 26 | Page 174
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