U-tube

logistic_guy

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Consider a \(\displaystyle \text{U}\)-tube whose arms are open to the atmosphere. Now water is poured into the \(\displaystyle \text{U}\)-tube from one arm, and light oil \(\displaystyle \left(\rho = 790 \ \text{kg/m}^3\right)\) from the other. One arm contains \(\displaystyle 70\)-\(\displaystyle \text{cm}\)-high water, while the other arm contains both fluids with an oil-to-water height ratio of \(\displaystyle 6\). Determine the height of each fluid in that arm.
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\(\displaystyle P_L = P_R\)

\(\displaystyle \rho g h_1 + P_{\text{atm}} = \rho_o g h_o + \rho g h_2 + P_{\text{atm}}\)

\(\displaystyle \rho h_1 = \rho_o h_o + \rho h_2\)

We are given \(\displaystyle \frac{h_o}{h_2} = 6\).

Then,

\(\displaystyle \rho h_1 = \rho_o h_o + \rho h_2 = \rho_o 6h_2 + \rho h_2 = (6\rho_o + \rho)h_2\)

Or

\(\displaystyle h_2 = \frac{\rho h_1}{6\rho_o + \rho} = \frac{1000 \times 0.7}{(6)790 + 1000} = 0.122 \ \text{m} = \textcolor{blue}{12.2 \ \text{cm}}\)


\(\displaystyle h_o = 6h_2 = 6(0.122) = 0.732 \ \text{m} = \textcolor{blue}{73.2 \ \text{cm}}\)
 
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