Two liquids of densities $\rho_1$ and $\rho_2$ $(\rho_2 = 2\rho_1)$ 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):
Answer: (D) $\dfrac{1}{4}$
Pressure varies linearly with depth in each layer, so force $=$ (average pressure) $\times$ area. Each part has area $5\times10 = 50\ \text{m}^2$.
Upper part MN: pressure goes from $0$ to $5\rho_1 g$, average $2.5\rho_1 g$:
$$F_1 = 2.5\rho_1 g\times50 = 125\rho_1 g$$
Lower part NO: pressure goes from $5\rho_1 g$ to $5\rho_1 g + 5\rho_2 g = 15\rho_1 g$, average $10\rho_1 g$:
$$F_2 = 10\rho_1 g\times50 = 500\rho_1 g$$
$$\frac{F_1}{F_2} = \frac{125}{500} = \frac{1}{4}$$
Solution by Sreeraj P, M.Sc Physics