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Statement

$4.2.6$ A piece of iron weighs 9.8 N in water. Determine its volume. The density of iron is 7.8 $\cdot$ 10$^3$ m$^3$.

Solution

Let's consider the following figure

The normal force $N$ is what the equipment measures as apparent weight. From Newton Second Law $$N = mg-F_A$$ $$N = \rho~g~V-\rho_w~g~V$$ $$N = (\rho-\rho_w)~g~V$$ and taking $\rho_w = 1000$ kg/m$^3$ $$\boxed{V = \frac{N}{(\rho-\rho_w)~g}=147 \rm{cm}^3}$$

BSc. Luis Daniel Fernández Quintana
Physics Department (FCNE)
Universidad de Oriente, Cuba