A piece of aluminum has density 2.70 g/cm³ and mass 834g.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 the aluminum = volume of aluminum submerged in oil * density of oil * gravitational acceleration
Volume of aluminum submerged in oil = Mass of aluminum / density of aluminum = 834g / 2.70 g/cm^3 = 309.3 cm^3
Buoyant force on the aluminum = 309.3 cm^3 * 0.650 g/cm^3 * 9.81 m/s^2 = 1976.07 g

The reading on the spring balance will be equal to the weight of the aluminum minus the buoyant force acting on it.
Weight of aluminum = Mass of aluminum * gravitational acceleration = 834g * 9.81 m/s^2 = 8173.54 g
Reading on spring balance = Weight of aluminum - Buoyant force on the aluminum = 8173.54 g - 1976.07 g = 6197.47 g

Therefore, the spring balance will register a reading of 633 g (to the nearest gram).

A Piece of Aluminum Has Density 2