Water is falling down at the rate of 1.8 kg per minute from a vertical tube. The radius at the bottom of the tube is 0.004 m and pressure around it is 0.76m of Hg. The diameter of the tube at a height of 0.5 m from the bottom is 0.005 m. Find the pressure at that point.

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0.724 m of Hg

0.824 m of Hg

0.924 m of Hg

0.624 m of Hg

answer is A.

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Detailed Solution

$\frac{m}{t} = Avρ = {πr}^{2} ρv$
$\begin{array}{l} ∴ v_{1} = \frac{(1 .8 / 60)}{(3 .14) {[\frac{5}{2} \times 10^{- 3}]}^{2} \times 10^{3}} = 1 .5 {ms}^{- 1} \\ A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = {(\frac{r_{1}}{r_{2}})}^{2} v_{1} \\ p_{1} + \frac{1}{2} {ρv}_{1}^{2} + ρgh = p_{2} + \frac{1}{2} {ρv}_{2}^{2} \\ p_{1} = p_{2} + \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) - ρgh \\ = p_{2} + \frac{1}{2} {ρv}_{1}^{2} [{(\frac{r_{1}}{r_{2}})}^{2} - 1] - ρgh \end{array}$
$\begin{array}{l} = [1 .013 \times 10^{5}] + \frac{1}{2} \times 10^{3} \times 2 .25 \times \\ [{(\frac{0 .005 / 2}{0 .004})}^{2} - 1] - 10^{3} \times 9 .8 \times 0 .5 \\ = 0 .72 m of Hg \end{array}$

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