A vessel has a liquid of density $ \rho = \alpha x +\beta $ , Where x is the depth from the free surface $ \alpha = 4 \times 10^3 gm^{-4} $ and $ \beta = 10^3 Kgm^{-3} $. A solid cylindrical shape, length (3/4)m and density $ 0.5 \times 10^3 Kgm^{-3} $ floats vertically. Find the length of the cylinder, beneath the free surface of liquid.

Applying Archimedes' principle for vertical equilibrium, the weight of the cylinder equals the weight of the displaced liquid. The cylinder mass is its length of 3/4 m multiplied by its cross-sectional area and its density of 500 kg/m^3. The displaced liquid mass is the cross-sectional area multiplied by the submerged length and the integral of the liquid density from 0 to that submerged length. Setting up the equation 500 * 0.75 = the integral from 0 to h of (4000x + 1000) dx results in 375 = 2000h^2 + 1000h. Solving this quadratic equation 2h^2 + h minus 0.75 = 0 yields a positive submerged length of 0.25 m.

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