Answer to Question #86165 in Quantum Mechanics for Denis

Question

When an object of mass 200g is submerged in methanol, its apparent weight is 1.052N. When submerged in benzene, its apparent weight is 0.951N. if the density of methanol is 0.8g/cm3, what is the density of benzene?

Answer 1

According to the Archimedes law, an object immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object. Thus, if the mass of the object is m and its volume is V, its apparent weight is mg – ρVg = g(m – ρV), where ρ is the fluid density and g = 9.807 m/s² is the acceleration of gravity. Denoting by ρ_b the density of benzene, and by ρ_m the density of methanol, we write the conditions of the problem as

g(m – ρ_mV) = F_m = 1.052 N , g(m – ρ_bV) = F_b = 0.951 N .

From these equations, we have, respectively,

ρ_mV = m – F_m/g , ρ_bV = m – F_b/g .

Dividing the second equation by the first one, we obtain

$\dfrac{\rho_{\rm b}}{\rho_{\rm m}} = \dfrac{m - F_{\rm b}/g}{m - F_{\rm m}/g}$ or $\rho_{\rm b} = \dfrac{m - F_{\rm b}/g}{m - F_{\rm m}/g} \rho_{\rm m}$ .

Substituting the numerical values, and taking into account that m = 0.2 kg in the International System of Units (SI), we have

$\rho_{\rm b} = \dfrac{ 0.2 - 0.951/9.807}{ 0.2 - 1.052/9.807} \rho_{\rm m} = 1.111 \rho_{\rm m} = 1.111 \times 0.8~ \text{g/cm}^{3 } = 0.889~\text{g/cm}^{3 }$ .

Answer: 0.889 g/cm³.