Question

Two liquids of densities ${\rho }_1\text{and}{\rho }_2({\rho }_2=2{\rho }_1)$ are filled up behind a square wall of side $10\text{m}$ as shown in figure. Each liquid has a height of $5\text{m}$. 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)

• A. $1/3$
• B. $2/3$
• C. $1/2$
• D. $1/4$

Answer 1

The correct option is D $1/4$

$\text{Let}{P}_1,P_2\text{and}{P}_3\text{be the pressure at points}M,N\text{and}O\text{respectively.}$

$\text{Pressure is given by P}=\rho gh$

$\text{Now,}{P}_1=0(\because h=0)$

$P_2=\rho g\left(5\right)$

$P_3=\rho g\left(5\right)+2\rho g\left(5\right)=15\rho g$

$\text{Force on upper part,}{F}_1=\frac{(P_1+P_2)}{2}A$

$\text{Force on lower part,}{F}_2=\frac{(P_2+P_3)}{2}A$

$\therefore \frac{F_1}{F_2}=\frac{5\rho g}{20\rho g}=\frac{5}{20}=\frac{1}{4}$