Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
Let velocity at A & B are $V_A$ & $V_B$ respectively
By equation of continuity
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$
$\therefore$ 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App