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Fluid Mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsFluid Mechanics2 Marks
Question
Water flows through a horizontal tube as shown in If the difference of heights of water column in the vertical tubes is 2cm, and the areas of cross-section at A and B are $4cm^2$ and $2cm^2$ respectively, find the rate of flow of water across any section.
✓

Answer

The height difference $=2\text{cm}=0.02\text{m}$

The difference of pressure between A and B $=\text{p}_\text{a}-\text{p}_\beta=\rho\text{gh}$

$=\frac{1}{2}\rho \text{Va}^2=\text{p}_\beta+\frac{1}{2}\rho \text{V}\beta^2$

$=1000\times10\times0.02=200\text{N}/\text{m}^2$

Let the rate of flow of water be Qcc/s i.e. $=\text{Q}\times10^{-6}\text{m}^3/\text{s}$

Area of the cross-section at $\text{A}=4\text{cm}^2=\frac{4}{10000}\text{m}^2$

The speed at $\text{A}=\text{v}_\text{a}=\frac{\text{Q}\times10^{-6}}{\Big(\frac{4}{10000}\Big )}\text{m}/\text{s}=\frac{\text{Q}}{400}\text{m}/\text{s}$

Area of the cross-section at $\text{B}=2\text{cm}^2=\frac{2}{10000}\text{m}^2$

The speed at $\text{B}=\text{v}_\beta=\frac{\text{Q}\times10^{-6}}{\Big(\frac{2}{10000}\Big)}\text{m}/\text{s}=\frac{\text{Q}}{200}\text{m}/\text{s}$

Since the height of both the points are the same, from Bernoulli's theorem,

$\text{p}_\text{a}+\frac{1}{2}\rho\text{V}\text{a}^2=\text{p}\beta+\frac{1}{2}\rho\text{V}\beta^2$

$\Rightarrow\text{p}_\text{a}-\text{p}_\beta=\frac{1}{2}\rho\big(\text{V}\beta^2-\text{V}\text{a}^2\big)$

$=\frac{1}{2}\times1000\times\text{Q}^2\Big\{\Big(\frac{1}{2000}\Big)^2-\Big(\frac{1}{400}\Big)^2\Big\}$

$\Rightarrow\text{p}_\text{a}-\text{p}_\beta=\Big(\frac{1}{20}\Big)\times\text{Q}^2\Big\{\frac{1}{4}-\frac{1}{16}\Big\}=\frac{3\text{q}^2}{320}$

$\Rightarrow200=\frac{3\text{Q}^2}{320}$

$\Rightarrow\text{Q}^2=200\times\frac{320}{3}=21333$

$\Rightarrow\text{Q}=146\text{cc}/\text{s}$

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