1. **Stating the Problem:** We have a solid aluminum sphere and a solid iron sphere balanced on an equal-arm balance. The iron sphere has a radius of $r_{Fe} = 2.00$ cm. Given the density of aluminum $\rho_{Al} = 2.70 \times 10^{3}$ kg/m³ and the density of iron $\rho_{Fe} = 7.86 \times 10^{3}$ kg/m³, find the radius $r_{Al}$ of the aluminum sphere such that their masses are equal. 2. **Formula for Mass:** Mass $m$ of a sphere is given by its density $\rho$ multiplied by its volume $V$. The volume of a sphere is $$V = \frac{4}{3} \pi r^{3}.$$ Therefore, mass is $$m = \rho \times \frac{4}{3} \pi r^{3}.$$ 3. **Setting the masses equal for balance:** $$m_{Al} = m_{Fe}$$ $$\rho_{Al} \times \frac{4}{3} \pi r_{Al}^{3} = \rho_{Fe} \times \frac{4}{3} \pi r_{Fe}^{3}$$ 4. **Simplify by canceling common factors:** $$\rho_{Al} r_{Al}^{3} = \rho_{Fe} r_{Fe}^{3}$$ 5. **Solve for $r_{Al}$:** $$r_{Al}^{3} = \frac{\rho_{Fe}}{\rho_{Al}} r_{Fe}^{3}$$ $$r_{Al} = \left( \frac{\rho_{Fe}}{\rho_{Al}} \right)^{\frac{1}{3}} r_{Fe}$$ 6. **Plug in values:** Convert $r_{Fe}$ to meters: $$r_{Fe} = 2.00\,\mathrm{cm} = 0.0200\,\mathrm{m}$$ Calculate the ratio: $$\frac{\rho_{Fe}}{\rho_{Al}} = \frac{7.86 \times 10^{3}}{2.70 \times 10^{3}} = 2.911...$$ Calculate the cube root: $$\left(2.911\right)^{\frac{1}{3}} \approx 1.43$$ Calculate $r_{Al}$: $$r_{Al} = 1.43 \times 0.0200 = 0.0286\,\mathrm{m} = 2.86\,\mathrm{cm}$$ **Final answer:** The radius of the aluminum sphere is approximately **2.86 cm** to balance the iron sphere of 2.00 cm radius.