The reading of a pressure meter attached to a | vedclass

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(A) According to Bernoulli's principle for horizontal flow,the total pressure (static pressure + dynamic pressure) remains constant.$P_1 = P_2 + \frac

Vedclass · Accepted Answer

$50$

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(A) According to Bernoulli's principle for horizontal flow,the total pressure (static pressure + dynamic pressure) remains constant.$P_1 = P_2 + \frac{1}{2} ho v^2$Where $P_1$ is the initial pressure when the water is at rest $(4.5 	imes 10^4 \ N/m^2)$,$P_2$ is the pressure when the water is flowing $(2.0 	imes 10^4 \ N/m^2)$,and $ho$ is the density of water $(10^3 \ kg/m^3)$.$P_1 - P_2 = \frac{1}{2} ho v^2$$(4.5 	imes 10^4) - (2.0 	imes 10^4) = \frac{1}{2} 	imes 10^3 	imes v^2$$2.5 	imes 10^4 = 0.5 	imes 10^3 	imes v^2$$v^2 = \frac{2.5 	imes 10^4}{0.5 	imes 10^3} = 5 	imes 10 = 50$$v = \sqrt{50} \ m/s$Given that the velocity is $\sqrt{V} \ m/s$,we have $\sqrt{V} = \sqrt{50}$.Therefore,$V = 50$.

The reading of a pressure meter attached to a closed pipe is $4.5 \times 10^4 \ N/m^2$. On opening the valve,water starts flowing and the reading of the pressure meter falls to $2.0 \times 10^4 \ N/m^2$. The velocity of water is found to be $\sqrt{V} \ m/s$. The value of $V$ is . . . . . .

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