Solid weighs 0.040 N in air and 0.024 N when fully immersed in a liquid of density 800 kg m-3. What is the volume of the solid? (g = 1Oms2)

Answer Details

When an object is immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces. Let V be the volume of the solid. The weight of the solid in air is its actual weight, W_a = 0.040 N. The weight of the solid in the liquid is reduced by the upthrust, which is equal to the weight of the displaced liquid. The weight of the displaced liquid is equal to its volume times its density times the acceleration due to gravity, or Vρg. Thus, the weight of the solid in the liquid is W_l = 0.040 - Vρg. Since the solid is fully immersed in the liquid, it displaces a volume of liquid equal to its own volume. Therefore, W_l = Vρg - Vρ_lg, where ρ_l is the density of the liquid. Setting the two expressions for W equal, we have: 0.040 - Vρg = Vρg - Vρ_lg Simplifying and solving for V, we get: V = (0.040)/(2ρg - ρ_lg) Substituting the given values, we get: V = (0.040)/(2 x 800 x 9.81 - 9.81) V ≈ 2.0 x 10^-6 m³ Therefore, the volume of the solid is approximately 2.0 x 10^-6 m³. The correct option is (i) 2.0 x 10^-6m³.