A homogeneous solid cylinder of length L and cross-sectional area A /5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L /4 in the denser liquid as sh

Question

A homogeneous solid cylinder of length L and cross-sectional area $A /5$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $L /4$ in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure $p_{0}$ . Then, density of the solid cylinder is

TestHub · Accepted Answer

Let the density of the solid cylinder be $ρ$ .

For the cylinder to be in equilibrium, its weight must be equal to the total buoyant force from both liquids.

Weight of the cylinder, $W = (\frac{A}{5}) L ρ g$

Since length $\frac{L}{4}$ is in the denser liquid (density $2 d$ ), the remaining length $L - \frac{L}{4} = \frac{3 L}{4}$ is in the lighter liquid (density $d$ ).

Total buoyant force, $F_{B} = Weight of displaced liquids$

$F_{B} = (\frac{A}{5}) (\frac{L}{4}) (2 d) g + (\frac{A}{5}) (\frac{3 L}{4}) (d) g$

Equating $W$ and $F_{B}$ :

$(\frac{A}{5}) L ρ g = (\frac{A}{5}) g [\frac{2 d L}{4} + \frac{3 d L}{4}]$

$ρ = \frac{2 d}{4} + \frac{3 d}{4} = \frac{5 d}{4}$

Answer: $\frac{5 d}{4}$

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