then force exerted by liquid 1 is of magnitude ρ1v1g. Statement II: Total buoyant force = ρ1v1g + ρ2v2g.

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Statement I: When a body floats such that its parts are immersed into two immiscible liquids, then force exerted by liquid 1 is of magnitude ρ1v1g. Statement II: Total buoyant force = ρ1v1g + ρ2v2g.
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Statement I: When a body floats such that its parts are immersed into two immiscible liquids, then force exerted by liquid 1 is of magnitude ρ1v1g. Statement II: Total buoyant force = ρ1v1g + ρ2v2g. Statement I: When a body floats such that its parts are immersed into two immiscible liquids then force exerted by [...]
Posted in: Cengage JEE Mains Physics by B.M Sharma , Chapter 12 - Fluid Mechanics, Surface Tension and Viscosity ,
Tags: Statement I: When a body floats such that its parts are immersed into two immiscible liquids , then force exerted by liquid 1 is of magnitude ρ1v1g. Statement II: Total buoyant force = ρ1v1g + ρ2v2g. ,