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Water flows in a horizontal tube as shown in figure.The pressure of water changes by $600N/m^2$ between A and B where the areas of cross section are $30cm^2$ and $15cm^2$ respectively.Find the rate of flow of water through the tube

$\begin{array}{1 1}(A)\;1880cm^3s^{-1}\\(B)\;1890cm^3s^{-1}\\(C)\;2200cm^3s^{-1}\\(D)\;2300cm^3s^{-1}\end{array} $

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1 Answer

Let velocity at A & B are $V_A$ & $V_B$ respectively
By equation of continuity
$\large\frac{V_B}{V_A}=\frac{30}{15}$$=2$
By Bernoulli's equation
$P_A+\large\frac{1}{2}$$\rho V_A^2=P_B+\large\frac{1}{2}$$\rho V_B^2$
$P_A-P_B=\large\frac{1}{2}$$\rho (2V_A)^2-\large\frac{1}{2}$$\rho V_A^2=\large\frac{3}{2}$$\rho V_A^2$
$\Rightarrow 600=\large\frac{3}{2}$$(1000kgm^{-3})V_A^2$
$V_A=0.63ms^{-1}$
$\therefore$ Rate of flow =$(30cm^2)(0.63ms^{-1})$
$\Rightarrow 1890cm^3s^{-1}$
Hence (B) is the correct answer.
answered May 9, 2014 by sreemathi.v
 

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