FLUIDS BYU 1

Analysis

We are dealing with a conservation of mass and volume problem involving a mixture of two substances (water and iron) with distinct densities. The key is to express the average density of the final contents in terms of the individual masses and total volume.

Step-by-Step Derivation

Let $A$ be the cross-sectional area of the vessel. The initial volume of water is $V_w = A h_0$. After adding the iron balls, the water level rises to $h$. The total volume of the contents is $V_{total} = A h$.

The volume occupied by the iron balls is the difference between the total volume and the initial water volume:

$$V_{iron} = V_{total} – V_{w} = A(h – h_0)$$

The total mass of the contents ($M_{total}$) is the sum of the mass of water ($M_w$) and the mass of iron ($M_i$):

$$M_{total} = M_w + M_i$$ $$\rho (Ah) = \rho_0 (Ah_0) + \rho_i [A(h – h_0)]$$

Here, $\rho$ is the average density, $\rho_0$ is the density of water, and $\rho_i$ is the density of iron. We can cancel $A$ from both sides:

$$\rho h = \rho_0 h_0 + \rho_i h – \rho_i h_0$$

Rearranging the terms to solve for $h$:

$$h(\rho_i – \rho) = h_0(\rho_i – \rho_0)$$ $$h = h_0 \left( \frac{\rho_i – \rho_0}{\rho_i – \rho} \right)$$