Analytic balances are calibrated to give correct mass values for such items as steel objects of density \(\rho_{s}=\) \(8000.00 \mathrm{~kg} / \mathrm{m}^{3}\). The calibration compensates for the buoyant force arising because the measurements are made in air, of density \(\rho_{\mathrm{a}}=1.205 \mathrm{~kg} / \mathrm{m}^{3}\). What compensation must be made to measure the masses of objects of a different material, of density \(\rho\) ? Does the buoyant force of air matter?

The buoyant force formula is given by F = Vρg, where V is the volume of the submerged object, ρ is the density of the fluid (air), and g is the acceleration due to gravity.

We know that the mass of an object is given by m = Vρ, where m is the mass, V is the volume, and ρ is the density of the object. We can rewrite the buoyant force formula as follows: F = m(ρ_g/ρ) where m is the mass, ρ_g is the density of the fluid, and ρ is the density of the object. Now, let's denote the mass measured by the balance without buoyancy compensation as m0 and mass after compensation as mc, we get: mc = m0 - m(ρ_a/ρ) Since the balance is calibrated for steel objects having density ρ_s, let's find the expression for mass without buoyancy compensation in terms of ρ_s. m0 = m(1 - ρ_a/ρ_s) Now, we need to find the compensation needed to measure the mass of objects with a different material density ρ.

We can find the compensation needed by solving for mc in terms of ρ and ρ_s: mc = m(1 - ρ_a/ρ_s) - m(ρ_a/ρ) Factoring out m and simplifying, we get: mc = m(1 - ρ_a(ρ - ρ_s)/(ρ_sρ))

The buoyant force depends on the density of the fluid and the density of the object. Since the buoyant force formula involves subtracting the product of the densities, the importance of the buoyant force for materials having density close to steel (ρ ≈ ρ_s) will be very small, as the subtraction leads to a negligible difference. However, for objects with significantly lower or higher densities than steel, the buoyant force of air will matter.