College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 362: 40


The specific gravity of the alcohol is $0.785$

Work Step by Step

We can find the buoyant force exerted on the aluminum cylinder: $F_b = 1.03~N - 0.730~N = 0.30~N$ According to Archimedes' principle, the buoyant force is equal to the weight of the alcohol that is displaced. We can find the mass of the alcohol that is displaced: $mg = F_b$ $m = \frac{F_b}{g}$ $m = \frac{0.30~N}{9.80~m/s^2}$ $m = 0.0306~kg$ We can find the density of the alcohol: $\rho = \frac{m}{V} = \frac{0.0306~kg}{3.90\times 10^{-5}~m^3} = 785~kg/m^3$ We can find the specific gravity: $\frac{785~kg/m^3}{1000~kg/m^3} = 0.785$ The specific gravity of the alcohol is $0.785$.
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