Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 14 - Fluids and Elasticity - Exercises and Problems: 22

Answer

The length of steel that is above the surface is 8.4 cm

Work Step by Step

Let $V_c$ be the volume of the cylinder. We can find the mass of the cylinder as: $M = V_c~\rho_{steel}$ $M = (\pi)(0.05~m)^2(0.20~m)(7900~kg/m^3)$ $M = 12.4~kg$ The buoyant force on the cylinder is equal to the cylinder's weight $Mg$. The buoyant force is equal to the weight of the mercury that is displaced. Let $\rho_m$ be the density of mercury. We can find the volume $V_m$ of mercury that is displaced. Therefore; $F_B = Mg$ $\rho_m~V_m~g = Mg$ $V_m = \frac{M}{\rho_m}$ $V_m = \frac{12.4~kg}{13.6\times 10^3~kg/m^3}$ $V_m = 9.12\times 10^{-4}~m^3$ We can find the height $h$ such that the bottom part of the cylinder which is submerged has the volume $V_m$ as: $\pi~r^2~h = V_m$ $h = \frac{V_m}{\pi~r^2}$ $h = \frac{9.12\times 10^{-4}~m^3}{(\pi)~(0.05~m)^2}$ $h = 0.116~m = 11.6~cm$ The length of steel that is below the surface is 11.6 cm. Therefore, the length of steel that is above the surface is 20 cm - 11.6 cm which is 8.4 cm.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.