The Pitot tube shown in the figure is used to | vedclass

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(B) Let $v$ be the velocity of the gas in the pipe.Applying Bernoulli's theorem at the opening of the Pitot tube (stagnation point $A$) and at a point

Vedclass · Accepted Answer

$Q = S\sqrt{\frac{2\Delta h \rho_0 g}{\rho}}$

Vedclass · Answer

(B) Let $v$ be the velocity of the gas in the pipe.Applying Bernoulli's theorem at the opening of the Pitot tube (stagnation point $A$) and at a point $B$ in the free stream flow:$P_A = P_B + \frac{1}{2} \rho v^2$Here,$P_A$ is the stagnation pressure and $P_B$ is the static pressure of the gas.The pressure difference is measured by the liquid column of height $\Delta h$:$P_A - P_B = \Delta h \rho_0 g$Substituting this into the Bernoulli equation:$\frac{1}{2} \rho v^2 = \Delta h \rho_0 g$$v^2 = \frac{2 \Delta h \rho_0 g}{\rho}$$v = \sqrt{\frac{2 \Delta h \rho_0 g}{\rho}}$The volume flow rate $Q$ is given by the product of the cross-sectional area $S$ and the velocity $v$:$Q = S v = S \sqrt{\frac{2 \Delta h \rho_0 g}{\rho}}$Thus,the correct option is $B$.

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