Question

Water flows through a horizontal tube as shown in the figure. If the difference of heights of water column in the vertical tubes is $h=0.02m$, and the areas of cross section at A and B are $4\times {10}^{-4}{m}^2$ and $2\times {10}^{-4}{m}^2$, respectively, then the rate of flow of water across any section is

• A. $130\times {10}^{-6}{m}^3/s$
• B. $146\times {10}^{-6}{m}^3/s$
• C. $160\times {10}^{-6}{m}^3/s$
• D. $170\times {10}^{-6}{m}^3/s$

Answer 1

The correct option is B $146\times {10}^{-6}{m}^3/s$

Using the continuity equation we have $v_A{a}_A=v_B{a}_B$ or
$v_A\times 4=v_B\times 2$
or
$v_B=2v_A$

Now applying Bernoulli's equation we get

$\frac{1}{2}\rho v_A^2+\rho g{h}_A+p_A=\frac{1}{2}\rho v_B^2+\rho g{h}_B+p_B$

As the height of both the points is same we get equal potential energies. Thus

$\frac{1}{2}\rho v_A^2+p_A=\frac{1}{2}\rho v_B^2+p_B$
or
$p_A-p_B=\frac{1}{2}\rho (v_B^2-v_A^2)=\frac{1}{2}\times 1\times (4v_A^2-v_A^2)$

$p_A-p_B=0.02m$ of water column $=2\times 1\times 1000dyne/c{m}^2$

$\therefore v_A=\sqrt{\frac{4000}{3}}=36.51cm/s$

So, rate of flow $=V_a{a}_A=36.51\times 4=146c{m}^3/s$.