MCQ
Water is flowing through a horizontal tube according to the figure. Its diameter at two points are $0.3\,m$ and $0.1\,m$  respectively. Pressure difference between these two points is equal to $0.8\,m$ of water column. Find rate of flow of water in the tube ..... $ltr/s$
  • $32$
  • B
    $72$
  • C
    $198$
  • D
    $49.5$

Answer

Correct option: A.
$32$
a
According to Bernoulli's thereom

$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}$

$\Rightarrow P_{1}-P_{2}=\frac{1}{2} \rho\left[v_{2}^{2}-v_{1}^{2}\right]$

Let rate of flow $Q=A_{1} v_{1}=A_{2} v_{2}$

$\mathrm{h} \rho \mathrm{g}=\frac{1}{2} \rho\left[\frac{\mathrm{Q}^{2}}{\mathrm{A}_{2}^{2}}-\frac{\mathrm{Q}^{2}}{\mathrm{A}_{1}^{2}}\right]$

$\Rightarrow 2 g h=Q^{2}\left[\frac{A_{1}^{2}-A_{2}^{2}}{A_{1}^{2} A_{2}^{2}}\right]$

$\Rightarrow Q=\sqrt{\frac{2 g h A_{1}^{2} A_{2}^{2}}{A_{1}^{2}-A_{2}^{2}}}$

$Q=A_{1} A_{2} \sqrt{\frac{2 g h}{A_{1}^{2}-A_{2}^{2}}}$

$=\pi r_{1}^{2} \times \pi r_{2}^{2} \sqrt{\frac{2 \times 10 \times 0.8}{\left(\pi r_{1}^{2}\right)^{2}-\left(\pi r_{2}^{2}\right)^{2}}}$

$=\frac{\pi^{2} r_{1}^{2} r_{2}^{2}}{\pi} \sqrt{\frac{16}{\left(225 \times 10^{-4}\right)^{2}-\left(25 \times 10^{-4}\right)^{2}}}$

$=\frac{10 \times 225 \times 10^{-4} \times 25 \times 10^{-4} \times 4}{\pi \times 10^{-4} \sqrt{250 \times 200}}$

$=\frac{225 \times 25 \times 4 \times 10^{-3}}{\pi \times 10^{2} \sqrt{5}}$

$=\frac{45 \sqrt{5}}{\pi} \times 10^{-5} \mathrm{m}^{5} / \mathrm{s}=32 \mathrm{ltr} / \mathrm{s}$

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