$\frac{\mathrm{dp}}{\mathrm{dx}}=\mathrm{\rho }\left({\mathrm{\omega }}^2\mathrm{x}\right)\Rightarrow \mathrm{dp}={\mathrm{\rho \omega }}^2\mathrm{xdx}$
 Integrating both sides, we get                     
     
    
  $\underset{{\mathrm{p}}_1}{\overset{{\mathrm{p}}_3}{\int }}\mathrm{dp}={\mathrm{\rho \omega }}^2\underset{-\mathrm{h}}{\overset{2\mathrm{h}}{\int }}\mathrm{xdx}$

${\mathrm{p}}_3-{\mathrm{p}}_1={\mathrm{\rho \omega }}^2\left[\frac{{\mathrm{x}}^2}{2}\right]=\frac{{\mathrm{\rho \omega }}^2}{2}\left[4{\mathrm{h}}^2-{\mathrm{h}}^2\right]$
$\mathrm{\rho g}\left({\mathrm{h}}_3-{\mathrm{h}}_1\right)=\frac{3{\mathrm{\rho \omega }}^2{\mathrm{h}}^2}{2}=\frac{3\mathrm{\rho }}{2}{\mathrm{h}}^2\left(\frac{\mathrm{g}}{\mathrm{h}}\right)$
${\mathrm{h}}_3-{\mathrm{h}}_1=\frac{3\mathrm{h}}{2}$        ... (i)
But   ${\mathrm{h}}_3+{\mathrm{h}}_1=2\mathrm{h}$      ... (ii)
 After solving, ${\mathrm{h}}_3=\frac{7\mathrm{h}}{4} \text{ and }{\mathrm{h}}_1=\frac{\mathrm{h}}{4}\therefore \frac{{\mathrm{h}}_3}{{\mathrm{h}}_1}=7$