Answer
425.1k+ views
Hint: In the fluid dynamics, Bernoulli's principle states that an increase in the velocity of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. This theorem is to be used to solve this question.
Complete step by step answer:
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}}}}$$4F=2N$
The force of reaction at the point A is,
${{F}_{a}}=\rho {{Q}_{a}}{{v}_{a}}=\rho S{{v}_{b}}^{2}$
As the both forces are antiparallel to each other, then the resultant force will be given as,
${{F}_{a}}=\rho S\left( {{v}_{a}}^{2}-{{v}_{b}}^{2} \right)$
Now let us apply Bernoulli’s theorem,
According to the theorem the liquid flowing out of A is,
${{P}_{0}}+\rho gh={{P}_{0}}+\dfrac{1}{2}\rho {{v}_{a}}^{2}$
And the liquid flowing out of B is,
${{P}_{0}}+\rho g\left( h+\Delta h \right)={{P}_{0}}+\dfrac{1}{2}\rho {{v}_{b}}^{2}$
Comparing both the equations will give rise to,
$\left( {{v}_{b}}^{2}-{{v}_{a}}^{2} \right)\dfrac{\rho }{2}=\Delta h\rho g$
Therefore the force will be,
$F=2\rho gS\Delta h=0.50N$
Hence the value of 4F will be,
$4F=2N$
Therefore the answer for the question has been calculated.
Note:
Fluid force is the force which results from liquid pressure experiencing over an area. Liquid pressure is the pressure at the depth of a liquid with weight. The mass density of a fluid is the mass of the fluid flowing per unit volume.
Complete step by step answer:
First of all let us take a look at what actually Bernoulli’s theorem states. Bernoulli's principle is a concept in fluid dynamics. It says that as velocity of the fluid increases, pressure decreases. And also a higher pressure pushes or accelerates fluid toward lower pressure. So any variation in a fluid's speed must be matched by a change in pressure or force.
Here it is mentioned in the question that,
The holes are having a cross sectional area
$S=0.5c{{m}^{2}}$
The height difference between these holes are equivalent to,
$\Delta h=51cm$
Let the velocity of water, flowing through A be ${{v}_{a}}$ and ${{v}_{b}}$ that through point B, and also discharging rate through A
${{Q}_{a}}={{S}_{{{v}_{a}}}}$
And similarly through B
${{Q}_{b}}={{S}_{{{v}_{b}}}}$$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.
Note:
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.
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