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: 37

Answer

The inner diameter of the sphere is $~~57.3~cm$

Work Step by Step

The weight of the water displaced by the sphere is equal to the weight of the sphere. Then the mass of the water displaced by the sphere is equal to the mass of the sphere. We can find the mass of the displaced water: $m = \rho~V$ $m = \rho~\frac{4}{3}\pi~r_2^3$ $m = (1000~kg/m^3)~(\frac{4}{3})(\pi)~(0.300~m)^3$ $m = 113.1~kg$ Then the mass of the sphere is $113.1~kg$ We can find the volume of the iron used to make the sphere: $V = \frac{m}{\rho}$ $V = \frac{113.1~kg}{7870~kg/m^3}$ $V = 0.0144~m^3$ We can find $r_1$, the inner radius: $V = \frac{4}{3}\pi~r_2^3-\frac{4}{3}\pi~r_1^3$ $\frac{4}{3}\pi~r_1^3 = \frac{4}{3}\pi~r_2^3 - V$ $r_1^3 = r_2^3 - \frac{3V}{4\pi}$ $r_1 = \sqrt[3] {r_2^3 - \frac{3V}{4\pi}}$ $r_1 = \sqrt[3] {(0.300~m)^3 - \frac{(3)(0.0144~m^3)}{4\pi}}$ $r_1 = 0.2867~m$ $r_1 = 28.67~cm$ Then the inner diameter of the sphere is $~~57.3~cm$
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