An iron block of density ρ_Fe and of volume l³ is immersed in a fluid of density ρ_fluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is A) W = (ρ_Fe − ρ_fluid)ghl². B) W = (ρ_fluid − ρ_Fe)gl³. C) W = (ρ_Fe − ρ_fluid)gl³. D) W = (ρ_Fe + ρ_fluid)ghl². E) W = (ρ_Fe + ρ_fluid)gl³.

Question

Quizplus · Accepted Answer

When an object is immersed in a fluid, it experiences a buoyant force equal to the weight of the fluid displaced by the object. This buoyant force reduces the weight measured by the scale. The buoyant force on the iron block is given by the formula F = ρghA, where ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth of the block below the surface, and A is the area of the block. The weight of the block, which is measured by the scale, is given by the formula W = mg, where m is the mass of the block and g is the acceleration due to gravity. Since the density of the block is given by ρ = m/V, where V is the volume of the block, we have:
m = ρV = ρ(l/2^3) = ρ(l/8)

Substituting this expression for m into the formula for W and adding the buoyant force, we get:
W = mg + F = ρ(l/8)g + ρghA

Substituting the given densities ρFe and ρfluid and expressing the area A in terms of the volume and height of the block, we obtain:
W = ρFe(gl^3/8) + ρfluidgh(l^2/4)

Simplifying this expression, we get:
W = (ρFe - ρfluid)gh(l^3/12)

which is the same as option C.

An Iron Block of Density ρFe and of Volume L3

An Iron Block of Density ρ_Fe and of Volume L³