To solve this problem, we need to apply the principle of buoyancy and equilibrium.$\\$Given:$\\$• Two non-mixing liquids with densities $\rho$ and $n\rho$ (where $n > 1). \\$• Heights of each liquid are $h. \\$• A solid cylinder of length $L$ and density $d$ floats with $pL$ in the denser liquid.$\\$Let's analyze the situation:$\\$• The cylinder is floating, so the buoyant force equals the weight of the cylinder.$\\$• The buoyant force is the sum of the forces due to the two liquids.$\\$The volume of the cylinder submerged in the lighter liquid is $(1-p)L \cdot A, \\$where $A$ is the cross-sectional area of the cylinder.$\\$The volume of the cylinder submerged in the denser liquid is $pL \cdot A. \\$The buoyant force due to the lighter liquid is:$\\ F_{b1} = (1-p)L \cdot A \cdot \rho \cdot g \\$The buoyant force due to the denser liquid is:$\\ F_{b2} = pL \cdot A \cdot n\rho \cdot g \\$The total buoyant force $F_b$ is:$\\ F_b = F_{b1} + F_{b2} \\ F_b = (1-p)L \cdot A \cdot \rho \cdot g + pL \cdot A \cdot n\rho \cdot g \\$The weight of the cylinder $W$ is:$\\ W = L \cdot A \cdot d \cdot g \\$For equilibrium, $F_b = W. \\$Equating the forces:$\\ (1-p)L \cdot A \cdot \rho \cdot g + pL \cdot A \cdot n\rho \cdot g = L \cdot A \cdot d \cdot g \\$Cancel $L \cdot A \cdot g$ from both sides:$\\ (1-p)\rho + p \cdot n\rho = d \\$Simplify the expression:$\\ \rho - p\rho + pn\rho = d \\ d = \rho + p(n\rho - \rho) \\ d = \rho + p(n-1)\rho \\ d = [1 + (n-1)p]\rho \\$Therefore, the density $d$ is equal to $[1 + (n-1)p]\rho. \\$This corresponds to Option 2.