Q.How does an aeroplane take off?

Bernoulli's theorem: P + ½ρv² + ρgh = constant along a streamline, for steady, incompressible, non-viscous flow.

Bernoulli's theorem follows from applying the work-energy theorem to a fluid element flowing along a streamline. It states that, for an ideal fluid (incompressible, non-viscous) in steady (streamline) flow, the total mechanical energy per unit volume — the sum of pressure energy (P), kinetic energy per unit volume (½ρv²), and gravitational potential energy per unit volume (ρgh) — remains constant at every point along the streamline: P + ½ρv² + ρgh = constant. A key consequence is that where the fluid speed v is higher (e.g. a narrower part of a pipe), the pressure P must be lower, and vice versa — this explains phenomena like the lift on an aircraft wing and the Magnus effect on a spinning ball.

Bernoulli's theorem is a statement of conservation of energy for an ideal, incompressible, non-viscous fluid in streamline flow.

For steady, streamline flow of an ideal (non-viscous, incompressible) fluid, Bernoulli's theorem states that the sum of pressure energy, kinetic energy and potential energy per unit volume remains constant along a streamline:

$P + \dfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}$

Here $P$ is the pressure energy per unit volume, $\dfrac{1}{2}\rho v^{2}$ the kinetic energy per unit volume, and $\rho g h$ the potential energy per unit volume. This is a direct consequence of the work–energy theorem applied to fluid flow, so it is the total energy (per unit volume) that stays constant, not momentum, angular momentum or mass individually (mass is conserved separately via the continuity equation).

The pressure energy per unit volume of a liquid is simply equal to the pressure P.

Bernoulli's equation expresses conservation of energy per unit volume along a streamline:

$P + \dfrac{1}{2}\rho v^2 + \rho gh = \text{constant}$

Here, $\dfrac{1}{2}\rho v^2$ is the kinetic energy per unit volume and $\rho gh$ is the potential energy per unit volume. By the same logic (and dimensionally, since $[P] = [\text{energy}]/[\text{volume}] = \text{N/m}^2 = \text{J/m}^3$), the pressure energy per unit volume is just $P$ itself.

(Note: $P/\rho$ would instead be the pressure energy per unit mass.)

The statement is False: Bernoulli's theorem follows from conservation of energy, not conservation of momentum.

Bernoulli's theorem states that for an ideal (incompressible, non-viscous) fluid in streamline flow, the sum P + (1/2)rhov^2 + rhogh remains constant along a streamline. It is derived by applying the work-energy theorem to a fluid element — the net work done on the fluid element by pressure forces equals its change in kinetic energy plus change in gravitational potential energy. This is fundamentally an application of the law of conservation of mechanical energy to fluid flow, not conservation of momentum.

Bernoulli's theorem is a statement of energy conservation for flowing fluids.

Bernoulli's equation, $P+\tfrac12\rho v^2+\rho gh=\text{constant}$, states that the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline for an ideal (non-viscous, incompressible) fluid in steady flow. This is derived directly from the work-energy theorem applied to a fluid element — i.e. from the principle of conservation of energy.

Bernoulli's theorem (P + ½ρv^2 + ρgh = constant along a streamline) is derived from the work-energy theorem for an ideal (incompressible, non-viscous) fluid in streamline flow — it is fundamentally an energy-conservation statement.

Bernoulli's equation is obtained by equating the net work done on a fluid element (by pressure differences) to its change in kinetic and potential energy as it flows along a streamline — exactly the work-energy theorem, which itself is a form of the conservation of mechanical energy. It assumes the fluid is incompressible and non-viscous (no energy lost to friction), so total mechanical energy per unit volume is conserved along the flow.

It is not a statement of conservation of mass (that's the continuity equation, A1v1 = A2v2) nor directly conservation of momentum — it's about energy.

Bernoulli's theorem is a statement of the conservation of energy for a flowing ideal fluid.

Bernoulli's equation states that for steady, incompressible, non-viscous flow along a streamline,

$P + \dfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$

Each term here represents an energy density (energy per unit volume): $P$ is associated with the fluid's pressure (flow) energy, $\dfrac{1}{2}\rho v^2$ is kinetic energy per unit volume, and $\rho g h$ is gravitational potential energy per unit volume. The theorem is derived by applying the work–energy theorem to a fluid element moving through a pipe of varying cross-section and height, and it asserts that the total mechanical energy per unit volume of the fluid remains constant along a streamline (in the absence of viscous losses). This is exactly the principle of conservation of energy applied to fluid flow.

A Venturi-meter measures flow rate using Bernoulli's theorem (option c).

Bernoulli's theorem states that along a streamline of an ideal (incompressible, non-viscous) fluid in steady flow, the sum P + (1/2)pv^2 + p g h is constant, where P is pressure, p is density, v is flow speed, and h is height.

A Venturi-meter is a tube with a constriction (narrower section) fitted into a pipe. By continuity, the fluid speeds up as it passes through the narrower section. By Bernoulli's theorem, since the speed increases at the constriction, the pressure there must decrease. Measuring this pressure difference (via manometer tubes) lets the flow speed/rate be calculated — this is a direct, textbook application of Bernoulli's theorem.

The other options are unrelated: a voltmeter measures potential difference (electrical, not fluid flow); a screw gauge and spherometer are mechanical length-measuring instruments based on the principle of a micrometer screw, with no connection to fluid dynamics.