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Question: Bernoulli’s theorem is a consequence of the law of conservation of: A. Angular momentum B. Mass ...

Bernoulli’s theorem is a consequence of the law of conservation of:
A. Angular momentum
B. Mass
C. Energy
D. Momentum

Explanation

Solution

Bernoulli’s theorem states that when the speed of a fluid increases simultaneously the static pressure decreases or there is a decrease in the potential energy of the fluid and vice-versa. It is applicable to the fluid in an ideal state. Bernoulli’s equation can be used to find the quantity which is conserved. Then, we can say, Bernoulli’s theorem is a consequence of the law of conservation of that quantity.

Complete step by step answer:
Bernoulli’s equation gives the relation between the pressure, kinetic energy, and the gravitational potential energy of the fluid in a container. The fluid can be a liquid or a gas. The equation for Bernoulli’s principle is given by,
P+12ρv2+ρgh=constantP + \dfrac {1}{2} \rho {v}^{2} + \rho gh= constant …(1)
Where, P is the pressure exerted by the fluid
ρ\rho is the density of the fluid
vv is the velocity of the fluid
gg is the gravitational acceleration
hh is the height of the container containing fluid
In equation. (1), second term i.e. 12ρv2\dfrac {1}{2} \rho {v}^{2} gives the kinetic energy per unit volume and the third term i.e. ρgh\rho gh gives the gravitational potential energy per unit volume. As the addition of both these terms is constant, the energies are conserved.
Hence, Bernoulli’s theorem is a consequence of the law of conservation of energy.

So, the correct option is C i.e. energy.

Note:
Bernoulli’s theorem is ideally applied to incompressible fluids only. The conservation of energy is an assumption. When the fluid flows from one point to another, there is some loss in the energy due to the internal friction. Thus, for a compressible fluid, elastic energy has to be added to Bernoulli's equation. Bernoulli’s equation at constant depth is given by,
P1+12ρv12=P2+12ρv22{P}_{1} + \dfrac {1}{2} \rho {v}_{1}^{2}={P}_{2} + \dfrac {1}{2} \rho {v}_{2}^{2}
This equation is applicable for only a small volume of fluid.