JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Fluid Mechanics

The force a liquid exerts on a vertical wall equals the average pressure over the strip times its area, so for a strip of width $w$ the thrust is $\int Pwdy$. The upper strip MN carries only the lighter liquid ${\rho }_1$ over a depth of $5$ m, giving $F_1=\frac{1}{2}{\rho }_{1}g(5{)}^{2}w=\frac{1}{2}{\rho }_{1}g(25)(10)=125{\rho }_{1}g$. The lower strip NO lies between depths $5$ m and $10$ m; here the pressure already begins at ${\rho }_{1}g(5)$ from the liquid above and then grows through the denser liquid ${\rho }_2=2{\rho }_1$, so $F_2=[{\rho }_{1}g(5)](5)(10)+\frac{1}{2}{\rho }_{2}g(5{)}^2(10)=250{\rho }_{1}g+250{\rho }_{1}g=500{\rho }_{1}g$. The ratio of the thrusts is $\frac{F_1}{F_2}=\frac{125}{500}=\frac{1}{4}$.