## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 9 - Problems - Page 362: 34

#### Answer

(a) 91.7% of the ice's volume is submerged. (b) The specific gravity of ice is $0.917$

#### Work Step by Step

(a) If the ice is floating in the water, then the buoyant force is equal in magnitude to the weight of the ice. According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V_i$ be the volume of the ice. Let $V_w$ be the volume of water that is displaced by the ice. Let $m_b$ be the mass of the ice. Let $m_w$ be the mass of the water that is displaced by the ice. We can find the ratio $\frac{V_w}{V_i}$: $m_w~g = m_i~g$ $m_w = m_i$ $\rho_w~V_w = \rho_i~V_i$ $\frac{V_w}{V_i} = \frac{\rho_i}{\rho_w}$ $\frac{V_w}{V_i} = \frac{917~kg/m^3}{999.87~kg/m^3}$ $\frac{V_w}{V_i} = 0.917$ 91.7% of the ice's volume is submerged. (b) We can find the specific gravity of ice: $\frac{917~kg/m^3}{1000~kg/m^3} = 0.917$ The specific gravity of ice is $0.917$.

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