Question

one cubic meter of Aluminum has a mass of 2.70 x10^3 kg, and 1.00 m^3 of iron has a mass of 7.87 x 10^3 kg. Find the radius of a solid Aluminum sphere that will balance a solid iron sphere of radius 2.00cm on an equal-arm balance

Answer 1

$$m_a=m_i$$

$$\rho_a V_a=\rho_i V_i$$

$$\rho_a \frac{4}{3}\pi r_a^3=\rho_i \frac{4}{3}\pi r_i^3$$

$$r_a=r_i\left( \frac{\rho_i}{\rho_a}\right)^{\frac{1}{3}}$$

$$r_a=2\left( \frac{7870}{2700}\right)^{\frac{1}{3}}=2.86\ cm$$