Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 438: 51


The density of the liquid is $890~kg/m^3$

Work Step by Step

We can find the volume of liquid that is displaced by the tube. Let $h$ be the length of the part of the tube that is submerged. $V = \pi~r^2~h$ $V = (\pi)(0.020~m)^2~(0.25~m)$ $V = 3.14\times 10^{-4}~m^3$ The buoyant force on the tube is equal to the total weight of the tube and pellets. We can find the density of the liquid. $F_B = M_{tube}~g+M_{pellets}~g$ $\rho~V~g = M_{tube}~g+M_{pellets}~g$ $\rho = \frac{M_{tube}+M_{pellets}}{V}$ $\rho = \frac{(0.030~kg)+(0.25~kg)}{3.14\times 10^{-4}~m^3}$ $\rho = 890~kg/m^3$ The density of the liquid is thus $890~kg/m^3$.
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