A piece of aluminum has density 2.70 g/cm³ and mass 775 g.The aluminum is submerged in a container of oil of density 0.650 g/cm³.A spring balance is attached with string to the piece of aluminum.What reading will the balance register in grams (g)for the submerged metal?

Question

Quizplus · Accepted Answer

The buoyant force acting on a submerged object is equal to the weight of the fluid displaced by the object. In this case, the aluminum has a volume of V = m/aluminum density = 775 g / 2.70 g/cm^3 = 287.04 cm^3. When the aluminum is submerged in oil, it will displace an equal volume of oil, which has a mass of V x oil density = 287.04 cm^3 x 0.650 g/cm^3 = 186.48 g. Therefore, the buoyant force on the aluminum is 186.48 g.

The spring balance will register the weight of the aluminum minus the buoyant force. The weight of the aluminum is given as 775 g. Therefore, the balance reading will be:

775 g - 186.48 g = 588 g

Therefore, the best choice is C) 588 g.

A Piece of Aluminum Has Density 2