In a horizontal tube, the water pressure chan | vedclass

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(C) Given that, the pressure difference between points $A$ and $B$ is $p_A - p_B = 1500 	ext{ N m}^{-2}$.By using Bernoulli's equation for a horizont

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$4000 	ext{ cm}^3 	ext{ s}^{-1}$

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(C) Given that, the pressure difference between points $A$ and $B$ is $p_A - p_B = 1500 	ext{ N m}^{-2}$.By using Bernoulli's equation for a horizontal tube $(h_A = h_B)$:$p_A + \frac{1}{2} ho v_A^2 = p_B + \frac{1}{2} ho v_B^2$$p_A - p_B = \frac{1}{2} ho (v_B^2 - v_A^2) \quad \dots (i)$Density of water, $ho = 10^3 	ext{ kg m}^{-3}$.Cross-sectional areas at points $A$ and $B$ are:$a_A = 40 	ext{ cm}^2 = 40 	imes 10^{-4} 	ext{ m}^2$$a_B = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2$By the equation of continuity, the rate of flow of water is constant:$a_A v_A = a_B v_B \Rightarrow v_B = v_A \left( \frac{a_A}{a_B} ight) = v_A \left( \frac{40}{20} ight) = 2v_A \quad \dots (ii)$Substituting Eq. $(ii)$ into Eq. $(i)$:$1500 = \frac{1}{2} 	imes 10^3 	imes ((2v_A)^2 - v_A^2)$$1500 = 500 	imes (4v_A^2 - v_A^2)$$3 = 3v_A^2 \Rightarrow v_A^2 = 1 \Rightarrow v_A = 1 	ext{ m s}^{-1}$Therefore, the rate of flow of water is:$Q = a_A v_A = 40 	imes 10^{-4} 	ext{ m}^2 	imes 1 	ext{ m s}^{-1} = 40 	imes 10^{-4} 	ext{ m}^3 	ext{ s}^{-1} = 4000 	ext{ cm}^3 	ext{ s}^{-1}$.

Column - $I$	Column - $II$
$(a)$ Velocity head	$(i)$ $\frac{P}{\rho g}$
$(b)$ Pressure head	$(ii)$ $h$
$(c)$ Potential head	$(iii)$ $\frac{v^2}{2g}$

In a horizontal tube, the water pressure changes by $1500 \text{ N m}^{-2}$ between points $A$ and $B$ as shown in the figure below. The cross-sectional areas at $A$ and $B$ of the tube are $40 \text{ cm}^2$ and $20 \text{ cm}^2$, respectively. Find the rate of flow of water through the tube.

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Different heads are in Column - $I$ and their formulas are given in Column - $II$. Match them appropriately.
Column - $I$ Column - $II$
$(a)$ Velocity head $(i)$ $\frac{P}{\rho g}$
$(b)$ Pressure head $(ii)$ $h$
$(c)$ Potential head $(iii)$ $\frac{v^2}{2g}$

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