Question

A less-dense liquid of density r1 floats on top of a more-dense liquid of density r2. A uniform cylinder of length l and density r, with r1 ≤ r ≤ r2, floats at the interface with its long axis vertical. What fraction of the length is in the more-dense liquid?

Answer 1

To find the fraction of a floating cylinder's length submerged in a more-dense liquid, use the principle of buoyancy and Archimedes' principle to set up an equilibrium equation and solve for the submerged length fraction.

Step-by-step explanation:

To determine the fraction of the cylinder's length that is submerged in the more-dense liquid (density r2), we apply the principle of buoyancy. According to Archimedes' principle, the weight of the fluid displaced by the submerged part of the cylinder must equal the weight of the entire cylinder for it to float.

Let x be the length of the cylinder submerged in the denser liquid. The cylinder's total weight is r × A × l (where A is the cross-sectional area), and the weight of the liquid displaced in the denser section is r2 × A × x.

The cylinder is also partially submerged in the less-dense liquid (density r1), displacing a volume corresponding to the length (l - x). The weight of this displaced liquid is r1 × A × (l - x).

For equilibrium, the total weight of the displaced liquids must equal the weight of the cylinder:

r × A × l = r2 × A × x + r1 × A × (l - x)

Solving for x, we get:

x = {r × l - r1 × l} / {r2 - r1}

The fraction of the cylinder submerged in the denser liquid then becomes x / l.