IMAT 2017 Q58 [Upthrust]

A solid wooden cube has sides of length a. The density of the wood is ρ.

The cube is completely immersed in a beaker of oil, which has a density σ. The top surface of the cube is horizontal. The gravitational field strength is g.

What is the upward force (upthrust) on the cube due to the oil?

[Assume that no oil is absorbed by the wood.]

A. (σ – ρ)a^3g
B. ρa^3g
C. σa^3
D. σa^3g
E. ρa^3

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We can assume the use of Archimedes Principle for this question, which states that the up thrust or more accurately- Buoyant Force on the cube due to the fluid is equal to the product of the fluid’s density, the object’s volume and the gravitational field strength.

We can represent this by expressing:

F_B= PxgxV

Where F_B is the Buoyant Force, P is the fluid density (σ), V is the object volume (a^3) and g is the gravitational field strength.

Simply replacing the mentioned values in the question, we can collect that F_B= σa^3g, making our answer D.