A manometer connected to a closed tap reads 4 | vedclass

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(D) According to Bernoulli's principle for horizontal flow,the total pressure $P_1$ (static pressure when the tap is closed) is equal to the sum of st

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$10$

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(D) According to Bernoulli's principle for horizontal flow,the total pressure $P_1$ (static pressure when the tap is closed) is equal to the sum of static pressure $P_2$ and dynamic pressure $\frac{1}{2}ho v^2$ when the tap is open.$P_1 = P_2 + \frac{1}{2}ho v^2$Rearranging for velocity $v$:$v = \sqrt{\frac{2(P_1 - P_2)}{ho}}$Given $P_1 = 4.5 	imes 10^5 	ext{ Pa}$,$P_2 = 4 	imes 10^5 	ext{ Pa}$,and density of water $ho = 10^3 	ext{ kg/m}^3$.$v = \sqrt{\frac{2(4.5 	imes 10^5 - 4 	imes 10^5)}{10^3}}$$v = \sqrt{\frac{2(0.5 	imes 10^5)}{10^3}} = \sqrt{\frac{10^5}{10^3}} = \sqrt{100} = 10 	ext{ m/s}$.

$A$ manometer connected to a closed tap reads $4.5 \times 10^5 \text{ Pa}$. When the tap is opened,the reading of the manometer falls to $4 \times 10^5 \text{ Pa}$. The velocity of the flow of water is ........ $m/s$.

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