Let ${\rho }_{\mathrm{Al}}$ represent the density of aluminum and ${\rho }_{\mathrm{Fe}}$ that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius $r_{\mathrm{Fe}}$ on an equal-arm balance.

Key Concepts: Density, Mass, Weight, Balance, Equilibrium, Sphere Explanation: When a sphere is balanced, it must be the case that the masses of each sphere are equal. Since the material of the spheres are different, the masses can be compared by comparing their weights. When the forces are in equilibrium, the weight of one sphere acting downwards is balanced by the weight of the other sphere acting upwards. The weight of an object can be found through the relation $W=mg$, where $W$ is the weight, $m$ is the mass, and $g$ is the acceleration due to gravity. We also know that the volume of a sphere is $V=\frac{4}{3}\pi r^3$, and the density is defined as $\rho =\frac{m}{V}$. Solving for mass and substituting it into the weight equation, we can set up an equation for the radius of the aluminum sphere in terms of the known variables. Step by Step Solution: Step 1. Using the density formula, we can express the mass of the iron sphere $m_{\mathrm{Fe}}$ in terms of the radius $r_{\mathrm{Fe}}$ and the density ${\rho }_{\mathrm{Fe}}$ as follows: $m_{\mathrm{Fe}}={\rho }_{\mathrm{Fe}}\cdot \frac{4}{3}\pi r_{\mathrm{Fe}}^3$. Step 2. The weight of the iron sphere can be found using $W_{\mathrm{Fe}}=m_{\mathrm{Fe}}g$. Here, we assume the acceleration due to gravity $g$ to be constant and equal to 9.81 $m/{\mathrm{s}}^2$. Thus, $W_{\mathrm{Fe}}={\rho }_{\mathrm{Fe}}\cdot \frac{4}{3}\pi r_{\mathrm{Fe}}^3\cdot 9.81$. Step 3. Since the aluminum sphere has the same weight when in equilibrium with the iron sphere, we can set $W_{\mathrm{Al}}=W_{\mathrm{Fe}}$. Plugging in the expression for $W_{\mathrm{Fe}}$ from step 2, we get ${\rho }_{\mathrm{Fe}}\cdot \frac{4}{3}\pi r_{\mathrm{Fe}}^3\cdot 9.81={\rho }_{\mathrm{Al}}\cdot \frac{4}{3}\pi r_{\mathrm{Al}}^3\cdot 9.81$. Step 4. Canceling out the common factor of $9.81$ and solving for $r_{\mathrm{Al}}$, we get $r_{\mathrm{Al}}={\left(\frac{{\rho }_{\mathrm{Fe}}{r}_{\mathrm{Fe}}^3}{{\rho }_{\mathrm{Al}}}\right)}^{1/3}$. Final Answer: $r_{\mathrm{Al}}={\left(\frac{{\rho }_{\mathrm{Fe}}{r}_{\mathrm{Fe}}^3}{{\rho }_{\mathrm{Al}}}\right)}^{1/3}$