Answer
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Hint: We know that Bernoulli's theorem, in fluid dynamics, relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar. The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one point along its flow. Because the Bernoulli equation is equal to a constant at all points along a streamline, we can equate two points on a streamline.
Complete step by step answer
We know that Bernoulli's equation can be viewed as a conservation of energy law for a flowing fluid. We saw that Bernoulli's equation was the result of using the fact that any extra kinetic or potential energy gained by a system of fluid is caused by external work done on the system by another non-viscous fluid.
Velocity of water surface $_{1}=0$
From Bernoulli equation $\mathrm{P}+\dfrac{\rho \mathrm{v}^{2}}{2}+\rho \mathrm{gh}=\mathrm{constant}$
$\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{1}^{2}}{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{2}^{2}}{2}+\rho \mathrm{gh}_{2}$
$\dfrac{\rho \mathrm{v}_{2}^{2}}{2}=\rho \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)$
$\mathrm{v}_{2}=\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Velocity of water is $\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Speed of water from hole is independent of the area. Hence, in steady flow the speed of water leaving the larger hole is the same as the speed of the water leaving the smaller.
So, the correct answer is option D.
Note: We can conclude that Bernoulli's principle is then cited to conclude that since the air moves slower along the bottom of the wing, the air pressure must be higher, pushing the wing up. However, there is no physical principle that requires equal transit time and experimental results show that this assumption is false. According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains constant. He proposed a theorem for the streamline flow of a liquid based on the law of conservation of energy.
Complete step by step answer
We know that Bernoulli's equation can be viewed as a conservation of energy law for a flowing fluid. We saw that Bernoulli's equation was the result of using the fact that any extra kinetic or potential energy gained by a system of fluid is caused by external work done on the system by another non-viscous fluid.
Velocity of water surface $_{1}=0$
From Bernoulli equation $\mathrm{P}+\dfrac{\rho \mathrm{v}^{2}}{2}+\rho \mathrm{gh}=\mathrm{constant}$
$\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{1}^{2}}{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{2}^{2}}{2}+\rho \mathrm{gh}_{2}$
$\dfrac{\rho \mathrm{v}_{2}^{2}}{2}=\rho \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)$
$\mathrm{v}_{2}=\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Velocity of water is $\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Speed of water from hole is independent of the area. Hence, in steady flow the speed of water leaving the larger hole is the same as the speed of the water leaving the smaller.
So, the correct answer is option D.
Note: We can conclude that Bernoulli's principle is then cited to conclude that since the air moves slower along the bottom of the wing, the air pressure must be higher, pushing the wing up. However, there is no physical principle that requires equal transit time and experimental results show that this assumption is false. According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains constant. He proposed a theorem for the streamline flow of a liquid based on the law of conservation of energy.
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