Questions & Answers

Question

Answers

Answer

Verified

129.3k+ views

First of all let us take a look at what actually Bernoulli’s theorem states. Bernoulli's principle is a concept in fluid dynamics. It says that as velocity of the fluid increases, pressure decreases. And also a higher pressure pushes or accelerates fluid toward lower pressure. So any variation in a fluid's speed must be matched by a change in pressure or force.

Here it is mentioned in the question that,

The holes are having a cross sectional area

$S=0.5c{{m}^{2}}$

The height difference between these holes are equivalent to,

$\Delta h=51cm$

Let the velocity of water, flowing through A be ${{v}_{a}}$ and ${{v}_{b}}$ that through point B, and also discharging rate through A

${{Q}_{a}}={{S}_{{{v}_{a}}}}$

And similarly through B

${{Q}_{b}}={{S}_{{{v}_{b}}}}$$4F=2N$

The force of reaction at the point A is,

${{F}_{a}}=\rho {{Q}_{a}}{{v}_{a}}=\rho S{{v}_{b}}^{2}$

As the both forces are antiparallel to each other, then the resultant force will be given as,

${{F}_{a}}=\rho S\left( {{v}_{a}}^{2}-{{v}_{b}}^{2} \right)$

Now let us apply Bernoulli’s theorem,

According to the theorem the liquid flowing out of A is,

${{P}_{0}}+\rho gh={{P}_{0}}+\dfrac{1}{2}\rho {{v}_{a}}^{2}$

And the liquid flowing out of B is,

${{P}_{0}}+\rho g\left( h+\Delta h \right)={{P}_{0}}+\dfrac{1}{2}\rho {{v}_{b}}^{2}$

Comparing both the equations will give rise to,

$\left( {{v}_{b}}^{2}-{{v}_{a}}^{2} \right)\dfrac{\rho }{2}=\Delta h\rho g$

Therefore the force will be,

$F=2\rho gS\Delta h=0.50N$

Hence the value of 4F will be,

$4F=2N$

Therefore the answer for the question has been calculated.

Fluid force is the force which results from liquid pressure experiencing over an area. Liquid pressure is the pressure at the depth of a liquid with weight. The mass density of a fluid is the mass of the fluid flowing per unit volume.