question_answer

Two liquids of densities \[{{\rho }_{1}}\]and \[{{\rho }_{2}}\]\[\left( {{\rho }_{2}}\text{=}2{{\rho }_{1}} \right)\]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) [JEE MAIN Held on 08-01-2020 Evening]

A) 1/4       

B) 1/2

C) 2/3                   

D) 1/3

Correct Answer: A

Solution :

\[{{F}_{1}}=\frac{\left( {{P}_{1}}+{{P}_{2}} \right)}{2}A\] \[{{F}_{2}}=\frac{\left( {{P}_{2}}+{{P}_{3}} \right)}{2}A\] \[\therefore \,\,\,\,\,\,\,\,\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{5\rho g}{20\rho g}=\frac{5}{20}=\frac{1}{4}\]