Two liquids of densities ρ1 and ρ2 (ρ2= 2ρ1) are filled up behind a square wall of side 10 m as sh скачать в хорошем качестве

Two liquids of densities ρ1 and ρ2 (ρ2= 2ρ1) are filled up behind a square wall of side 10 m as sh

Two liquids of densities ρ1 and ρ2 (ρ2= 2ρ1) 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) (a) 1/4 (b) 2/3 (c) 1/3 (d) 1/2 *YouTube Description* 🌟 *Hydrostatic Pressure: Force Ratio of Two Liquids* 🌟 In this fascinating fluid mechanics problem, we dive into how two non-mixing liquids with different densities exert forces on the upper and lower sections of a square wall. This conceptual question is perfect for JEE and NEET aspirants aiming to master hydrostatics. --- **Problem Statement**: Two liquids of densities \( \rho_1 \) and \( \rho_2 \) (\( \rho_2 = 2\rho_1 \)) are filled behind a square wall with a side length of \( 10 \, \text{m} \). Each liquid has a height of \( 5 \, \text{m} \). What is the ratio of the forces exerted by these liquids on the upper part (MN) to the lower part (NO) of the wall? ✅ **Correct Answer**: \(\boxed{\frac{1}{4}}\). --- *Theory and Explanation* When liquids exert pressure on a vertical surface, the force is calculated using the formula: \[ F = \rho g h_{\text{avg}} A, \] where: \( \rho \) = density of the liquid, \( g \) = acceleration due to gravity, \( h_{\text{avg}} \) = average height of the liquid column affecting the force, \( A \) = area of the section considered. #### *Breaking Down the Problem* 1️⃣ **For the upper section (MN)**: The height of the liquid above MN is smaller, leading to a smaller average pressure. The force is proportional to \( \rho_1 h_{\text{avg}} A \). 2️⃣ **For the lower section (NO)**: The height of the liquid above NO is larger, resulting in a greater average pressure. The force is proportional to \( \rho_2 h_{\text{avg}} A \). By substituting the given density ratio \( \rho_2 = 2\rho_1 \), and analyzing the pressure difference across the wall sections, the ratio of the forces is: \[ F_{\text{MN}} : F_{\text{NO}} = \frac{1}{4}. \] --- *Key Concepts to Understand* 1️⃣ **Hydrostatic Pressure Distribution**: Pressure increases linearly with depth in a liquid. 2️⃣ **Force on a Vertical Surface**: Integrating the pressure over the height gives the force. 3️⃣ **Effect of Density**: Denser liquids exert more force for the same height. --- *Why This Problem Matters* This question is a classic example of hydrostatics, testing your ability to relate pressure, force, and density in practical scenarios. Such problems are frequent in JEE and NEET exams, emphasizing both conceptual clarity and numerical application. --- *Challenge for You* What would happen to the force ratio if \( \rho_2 = 3\rho_1 \)? Drop your answers in the comments! --- *Hashtags & Links* #FluidMechanics #Hydrostatics #PhysicsChallenge #JEEMains2025 #NEET2025 #LearnPhysics 🌐 Follow us on Instagram for more physics insights: [Instagram Link](https://www.instagram.com/physics_beh...) 🎯 *Like, share, and subscribe for more conceptual problem-solving videos!* do just one support to us that you don't forget to like and share our video..( if you like our content )... also don't forget to subscribe our channel... it will give a huge support to us from your end... It will be helpful for iit, neet and as well as for cbse, icse,and other state boards examination. for any further query contact us at : 7843900879 #physics #onlinemyschool #physicsbehindeverything facebook :   / pbevns   twitter : https://twitter.com/behind_physics?s=09 Instagram : https://www.instagram.com/physics_beh... @pbe @physicsbehind everything