Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 14 - Fluids - Problems - Page 409: 41a


Fraction of iceberg visible $=0.10 $

Work Step by Step

Let $V'$ be the volume of iceberg below water and $V$ be the total volume of iceberg. Since the iceberg is floating, its weight is equal to the weight of salt water that it displaces (Archimedes Principle). This can be written as: $m_{(ice)} g = V'\rho_{w} g $ . . . . . . . . . . . . . . . .(1) But mass of iceberg can be written in terms of its volume and density : $mass=volume\times density$ $m_{(ice)}=V \times\rho_{(ice)}$ Thus replacing $m_{(ice)}$ by $V \rho_{(ice)} $ in equation (1) gives : $V \rho_{(ice)} g = V'\rho_{w} g $ Substituting the known values and solving gives: $V\times917$ $kg/m^{3}\times g=V'\times1024$ $kg/m^{3}\times g$ $\frac{V'}{V}=\frac{917}{1024}=0.895$ This is the fraction of ice berg inside salt water. To find the fraction outside water (visible part) subtract this fraction from $whole$ : Fraction of iceberg visible $=1-0.895= 0.105\approx0.10$
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