Two liquids of densities ${\mathrm{\rho }}_1 \mathrm{an} {\mathrm{\rho }}_2 \left({\mathrm{\rho }}_{2 }=2{\mathrm{\rho }}_1\right)$ are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing )__________$\times {10}^{-1}$

 

Let,

${\mathrm{\rho }}_1=\mathrm{\rho }, {\mathrm{\rho }}_2=2\mathrm{\rho }$

Let $P_A$ be the top pressure. Let $P_B$ represent the pressure at the liquid-liquid contact and $P_C$ be the pressure at the wall's bottom.

Let $F_1$ and $F_2$ stand for the forces that are applied to the upper and lower portions of the wall, respectively.

${\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_0=0$

${\mathrm{P}}_{\mathrm{B}}={\mathrm{\rho }}_1\mathrm{gh}=\mathrm{\rho gh}=\mathrm{\rho g}\times 5=5\mathrm{\rho g}$

${\mathrm{P}}_{\mathrm{C}}={\mathrm{\rho }}_1\mathrm{gh}5+{\mathrm{\rho }}_2\mathrm{g}5=5\mathrm{\rho g}+2\times 5\mathrm{\rho gC}=15\mathrm{\rho g}$

${\mathrm{F}}_1=\left(\frac{{\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}}{2}\right)=\left(\frac{0+5\mathrm{\rho g}}{2}\right)\mathrm{A}=\frac{5\mathrm{\rho gA}}{2}$

${\mathrm{F}}_2=\left(\frac{{\mathrm{P}}_{\mathrm{B}}+{\mathrm{P}}_{\mathrm{C}}}{2}\right)\times \mathrm{A}=\left(\frac{5\mathrm{\rho g}+15\mathrm{\rho g}}{2}\right)\mathrm{A}=\frac{20\mathrm{\rho gA}}{2}$

$\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{{\displaystyle \frac{5\mathrm{\rho gA}}{2}}}{{\displaystyle \frac{20\mathrm{\rho gA}}{2}}}=\frac{1}{4}=25\times {10}^{-1}.$

Hence the correct answer is 25.