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Q.State and prove Bernoulli's theorem. OR State and prove Newton's law of cooling.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2024Subjective· 5mImportance★★★★★
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Applying the work-energy theorem to a fluid element moving through a pipe of changing cross-section and height gives Bernoulli's equation, P + ½ρv² + ρgh = constant.

Statement: For the steady, streamline flow of an ideal (incompressible, non-viscous) fluid, the sum of pressure energy, kinetic energy, and potential energy, per unit volume, remains constant along any streamline:

P+12ρv2+ρgh=constantP + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}

Proof: Consider an ideal fluid flowing steadily through a pipe of varying cross-section. Take a fluid element between cross-sections of area A1A_1 (at height h1h_1, speed v1v_1, pressure P1P_1) and A2A_2 (at height h2h_2, speed v2v_2, pressure P2P_2). In a time Δt\Delta t, a volume ΔV=A1v1Δt=A2v2Δt\Delta V = A_1 v_1 \Delta t = A_2 v_2 \Delta t of fluid enters at 1 and an equal volume leaves at 2 (by conservation of mass, since the fluid is incompressible).

Work done by pressure forces: The fluid behind pushes the element forward at 1, doing work P1A1(v1Δt)=P1ΔVP_1 A_1 (v_1\Delta t) = P_1 \Delta V; the fluid ahead pushes back at 2, doing work −P2A2(v2Δt)=−P2ΔV-P_2 A_2 (v_2\Delta t) = -P_2\Delta V. Net work by pressure:

W=(P1−P2)ΔVW = (P_1 - P_2)\Delta V

Change in kinetic and potential energy: The net effect (since the flow is steady) is as if a mass Δm=ρΔV\Delta m = \rho \Delta V was moved from position 1 (height h1h_1, speed v1v_1) to position 2 (height h2h_2, speed v2v_2):

ΔKE=12Δm(v22−v12),ΔPE=Δm g(h2−h1)\Delta KE = \frac{1}{2}\Delta m (v_2^2 - v_1^2), \qquad \Delta PE = \Delta m\, g(h_2 - h_1)

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