A manometer connected to a closed tap reads 4 | vedclass

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(D) According to Bernoulli's principle for a horizontal pipe,$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$. When the tap is closed,the

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$10$

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(D) According to Bernoulli's principle for a horizontal pipe,$P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$. When the tap is closed,the velocity $v_1 = 0$,so $P_1 = 4.5 	imes 10^5 \, Pa$. When the tap is opened,the velocity $v_2 = v$ and the pressure $P_2 = 4 	imes 10^5 \, Pa$. Using the equation $P_1 - P_2 = \frac{1}{2} ho v^2$,where $ho = 10^3 \, kg/m^3$ is the density of water. Substituting the values: $(4.5 	imes 10^5 - 4 	imes 10^5) = \frac{1}{2} 	imes 10^3 	imes v^2$. $0.5 	imes 10^5 = 500 	imes v^2$. $50000 = 500 	imes v^2$. $v^2 = 100$. $v = 10 \, m/s$.

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

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