Q.
Two liquids of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{2}=2\rho_{1}\right)$ 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):
Solution:
$F_{1}=\frac{\rho gh}{2}\times A$
$F_{2}\left(\rho gh+\frac{2\rho gh}{2}\right)A$
$\frac{F_{1}}{F_{2}}=\frac{1}{4}$
