A manometer connected to a closed tap reads 3 | vedclass

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(B) According to Bernoulli's theorem for a horizontal flow of an incompressible,non-viscous fluid:$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho

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$10$

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(B) According to Bernoulli's theorem for a horizontal flow of an incompressible,non-viscous fluid:$P_1 + \frac{1}{2}ho v_1^2 = P_2 + \frac{1}{2}ho v_2^2$When the tap is closed,the velocity $v_1 = 0$,so the pressure $P_1 = 3.5 	imes 10^5 \ N/m^2$.When the tap is opened,the pressure $P_2 = 3.0 	imes 10^5 \ N/m^2$ and the velocity is $v_2 = v$.Assuming the density of water $ho = 10^3 \ kg/m^3$:$P_1 = P_2 + \frac{1}{2}ho v^2$$\frac{1}{2}ho v^2 = P_1 - P_2$$v^2 = \frac{2(P_1 - P_2)}{ho}$$v^2 = \frac{2(3.5 	imes 10^5 - 3.0 	imes 10^5)}{10^3}$$v^2 = \frac{2(0.5 	imes 10^5)}{10^3} = \frac{1.0 	imes 10^5}{10^3} = 100$$v = \sqrt{100} = 10 \ m/s$.

$A$ manometer connected to a closed tap reads $3.5 \times 10^5 \ N/m^2$. When the valve is opened,the reading of the manometer falls to $3.0 \times 10^5 \ N/m^2$. The velocity of the flow of water is ........ $m/s$.

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