A homogeneous solid cylinder of length $L$ ($L < H/2$), 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 $D$ of solid is given by

Step 1: Identify the given parameters and dimensions.

The cylinder of length $L$ and cross-sectional area $A/5$ floats vertically at the interface of two liquids.

• Upper liquid of density $d$ contains length $\dfrac{3L}{4}$ of the cylinder.
• Lower liquid of density $2d$ contains length $\dfrac{L}{4}$ of the cylinder.

Step 2: Express the weight of the cylinder.

The weight $W$ of the cylinder acts downwards through its center of gravity.

$$W = D \cdot \frac{A}{5} \cdot L \cdot g$$

Step 3: Calculate the buoyant forces.

The buoyant force $F_1$ exerted by the upper liquid of density $d$ is:

$$F_1 = d \cdot \frac{A}{5} \cdot \frac{3L}{4} \cdot g$$

The buoyant force $F_2$ exerted by the lower liquid of density $2d$ is:

$$F_2 = 2d \cdot \frac{A}{5} \cdot \frac{L}{4} \cdot g$$

Step 4: Find the total buoyant force.

The total upward buoyant force $F_b$ is the sum of the buoyant forces from both liquids.

$$F_b = F_1 + F_2$$ $$F_b = \left(\frac{3}{4} + \frac{2}{4}\right) d \cdot \frac{A}{5} \cdot L \cdot g$$ $$F_b = \frac{5}{4} d \cdot \frac{A}{5} \cdot L \cdot g$$

Step 5: Apply the condition for equilibrium.

For the cylinder to float in equilibrium, the downward weight must equal the upward buoyant force.

$$W = F_b$$ $$D \cdot \frac{A}{5} \cdot L \cdot g = \frac{5}{4} d \cdot \frac{A}{5} \cdot L \cdot g$$

Step 6: Solve for the density $D$ of the solid.

Cancel the common terms $\dfrac{A}{5}$, $L$, and $g$ from both sides of the equation.

$$D = \frac{5}{4}d$$

Answer: The density $D$ of the solid is $\dfrac{5}{4}d$.