# What fraction of an iceberg of density 910 kgm-3 will be above the surface of sea water of density 1170 kgm-3 ? - Physics

Sum

What fraction of an iceberg of density 910 kgm-3 will be above the surface of sea water of density 1170 kgm-3 ?

#### Solution

Let volume of iceberg = Vi = x
Volume of iceberg inside sea water = Vw
Density of iceberg = ρi = 910 kmg-3
Density of sea waer = ρw = 1170 kgm-3
By law of floatation:
Weight of iceberg = Weight of sea water displaced by the iceberg

"V"_"i" xx rho_"i" xx "g" = "V"_"w" xx rho_"w" xx "g"

"x" xx 910 = "V"_"w" xx 1170

"V"_"w" = "x" xx 910/1170

Volume of iceberg inside seawater = Volume of sea water dispalced by iceberg

= "V"_"w" = 7/9"x" = 7/9"x"

Volume of an iceberg above the sea water = "x" - 7/9x

= (1 - 7/9)"x" = (9-7)/9x

= 2/9 xx "Volume of iceberg"

Fraction of iceberg above sea water = 2/9"x" xx 1/"x" = 2/9 part

=> 2/9th part of iceberg is above the sea water.

Concept: Archimedes' Principle
Is there an error in this question or solution?

#### APPEARS IN

Goyal Brothers Prakashan Class 9 A New Approach to ICSE Physics Part 1
Chapter 5 Upthrust and Archimedes’ Principle
Practice Problems 4 | Q 1
Share