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Q.State and prove Bernoulli's theorem. OR Derive expression for height of the liquid rise

(h) in a capillary tube.
Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2026Subjective· 5mImportance★★★★★
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Applying the work-energy theorem to a fluid element flowing through a tube of varying cross-section and height gives P+12ρv2+ρgh=constantP+\tfrac12\rho v^2+\rho gh=\text{constant}.

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

P+12ρv2+ρgh=constantP + \frac12\rho v^2 + \rho gh = \text{constant}

Proof: Consider fluid flowing through a pipe of varying cross-section, entering at point 1 (area A1A_1, speed v1v_1, height h1h_1, pressure P1P_1) and leaving at point 2 (area A2A_2, speed v2v_2, height h2h_2, pressure P2P_2), in a small time interval dtdt.

By the equation of continuity (mass conservation for incompressible flow), the volume of fluid entering equals the volume leaving:

A1v1 dt=A2v2 dt=dVA_1v_1\,dt = A_2v_2\,dt = dV

and the mass moved is dm=ρ dVdm = \rho\,dV at both ends.

Work done by pressure forces:

  • At the inlet, the fluid behind pushes the element in, doing positive work W1=P1A1(v1 dt)=P1 dVW_1 = P_1A_1(v_1\,dt) = P_1\,dV.
  • At the outlet, the element pushes fluid ahead of it, so work W2=P2A2(v2 dt)=P2 dVW_2 = P_2A_2(v_2\,dt) = P_2\,dV is done AGAINST it (i.e. the fluid loses this much energy pushing forward).

Net work done on the fluid element by pressure forces:

Wnet=P1 dV−P2 dV=(P1−P2) dVW_{\text{net}} = P_1\,dV - P_2\,dV = (P_1-P_2)\,dV

By the work-energy theorem, this net work equals the sum of the changes in kinetic energy and potential energy of the fluid element (mass dm=ρ dVdm=\rho\,dV) as it moves from 1 to 2:

(P1−P2) dV=12 dm (v22−v12)+dm g(h2−h1)(P_1-P_2)\,dV = \frac12\,dm\,(v_2^2-v_1^2) + dm\,g(h_2-h_1)

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