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III. Long Answer Questions · Q6

Q.State and prove Archimedes principle.

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Step 1. Statement. Archimedes' principle states that a body partially or wholly immersed in a fluid experiences an upward thrust equal to the weight of the fluid it displaces, acting through the centre of gravity of the displaced fluid.

Step 2. Proof. Consider a rectangular block of cross-sectional area A fully immersed in a liquid of density ρ\rho, with its top face at depth h1h_1 and bottom face at the greater depth h2h_2 (so the block's height is h2−h1h_2-h_1).

Step 3. The pressure at the top face is P1=Pa+ρgh1P_1=P_a+\rho g h_1, producing a downward force F1=P1AF_1=P_1A; the pressure at the bottom face is P2=Pa+ρgh2P_2=P_a+\rho gh_2, producing an upward force F2=P2AF_2=P_2A. (The horizontal pressure forces on the side faces cancel by symmetry.)

Step 4. The net upward force is F2−F1=(P2−P1)A=ρg(h2−h1)A=ρg[A(h2−h1)]=ρgVF_2-F_1=(P_2-P_1)A=\rho g(h_2-h_1)A=\rho g[A(h_2-h_1)]=\rho gV, where V=A(h2−h1)V=A(h_2-h_1) is exactly the volume of the block (and hence the volume of liquid it displaces). …

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