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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) 2021Subjective· 5mImportance★★★★★
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Bernoulli's theorem is the work-energy theorem applied to a flowing fluid: pressure energy + kinetic energy + potential energy per unit volume stays constant along a streamline.

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

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

where PP = pressure, ρ\rho = fluid density, vv = flow speed, hh = height above a reference level.

Proof (using the work-energy theorem):

Consider an ideal fluid flowing through a pipe of varying cross-section. Take a fluid element between cross-sections A1A_1 (at height h1h_1, pressure P1P_1, speed v1v_1) and A2A_2 (at height h2h_2, pressure P2P_2, speed v2v_2). In a small time Δt\Delta t, a volume ΔV\Delta V of fluid enters at section 1 and an equal volume ΔV\Delta V leaves at section 2 (by continuity/incompressibility, A1v1=A2v2A_1v_1 = A_2v_2).

Work done by pressure forces:

  • Work done ON the fluid by pressure at section 1 (pushing it in): W1=P1A1(v1Δt)=P1ΔVW_1 = P_1 A_1 (v_1\Delta t) = P_1 \Delta V
  • Work done BY the fluid against pressure at section 2 (pushing out): W2=P2A2(v2Δt)=P2ΔVW_2 = P_2 A_2(v_2 \Delta t) = P_2\Delta V
  • Net work done on the fluid element by pressure forces: W=(P1−P2)ΔVW = (P_1 - P_2)\Delta V

Change in kinetic energy:

ΔKE=12(Δm)v22−12(Δm)v12=12ρΔV(v22−v12)\Delta KE = \frac{1}{2}(\Delta m)v_2^2 - \frac{1}{2}(\Delta m)v_1^2 = \frac{1}{2}\rho\Delta V(v_2^2 - v_1^2)

Change in potential energy:

ΔPE=Δm g(h2−h1)=ρΔV g(h2−h1)\Delta PE = \Delta m \, g(h_2 - h_1) = \rho\Delta V\, g(h_2-h_1)

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