Water flows in a horizontal tube (see figure) | vedclass

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(C) The rate of flow of water is constant,so $A_{A} V_{A} = A_{B} V_{B}$.Given $A_{A} = 40 \; cm^{2}$ and $A_{B} = 20 \; cm^{2}$,we have $40 V_{A} = 2

Vedclass · Accepted Answer

$2720$

Vedclass · Answer

(C) The rate of flow of water is constant,so $A_{A} V_{A} = A_{B} V_{B}$.Given $A_{A} = 40 \; cm^{2}$ and $A_{B} = 20 \; cm^{2}$,we have $40 V_{A} = 20 V_{B}$,which implies $V_{B} = 2 V_{A}$.Using Bernoulli's theorem for a horizontal tube: $P_{A} + \frac{1}{2} ho V_{A}^{2} = P_{B} + \frac{1}{2} ho V_{B}^{2}$.Rearranging gives $P_{A} - P_{B} = \frac{1}{2} ho (V_{B}^{2} - V_{A}^{2})$.Substituting the given values ($P_{A} - P_{B} = 700 \; Pa$,$ho = 1000 \; kg/m^{3}$):$700 = \frac{1}{2} 	imes 1000 	imes ((2 V_{A})^{2} - V_{A}^{2})$$700 = 500 	imes (4 V_{A}^{2} - V_{A}^{2})$$700 = 500 	imes 3 V_{A}^{2}$$V_{A}^{2} = \frac{700}{1500} = \frac{7}{15} \; m^{2}/s^{2}$$V_{A} = \sqrt{\frac{7}{15}} \approx 0.683 \; m/s = 68.3 \; cm/s$.The rate of flow $Q = A_{A} V_{A} = 40 \; cm^{2} 	imes 68.3 \; cm/s \approx 2732 \; cm^{3}/s$. Given the options,the closest value is $2720 \; cm^{3}/s$.

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