Question

Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³ s⁻¹. Find the difference of mercury levels in the U-tube.

Answer 1

Given:
Area of cross section of the wide portions of the tube, a_a = 5 cm²
Area of cross section of the narrow portions of the tube, a_b= 2 cm²
Now, let v_a and v_b be the speeds of water at A and B, respectively.
Rate of flow of water through the tube = 500 cm³/s
$$\begin{gathered} \Rightarrow v_{\mathrm{A}}=\left(\frac{500}{5}\right)=100\mathrm{cm}/\mathrm{s} \\ \mathrm{From}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{continuity},\mathrm{we}\mathrm{have}: \\ {v}_{\mathrm{A}}{a}_{\mathrm{A}}=v_{\mathrm{B}}{a}_{\mathrm{B}} \\ \Rightarrow \frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{a_{\mathrm{B}}}{a_{\mathrm{A}}}=\frac{2}{5} \\ \Rightarrow 5v_{\mathrm{A}}=2v_{\mathrm{B}} \\ \Rightarrow v_{\mathrm{B}}=\left(\frac{5}{2}\right)v_{\mathrm{A}}...\left(\mathrm{i}\right) \\ \mathrm{From}\mathrm{the}\mathrm{Bernoulli}\text{'}\mathrm{s}\mathrm{equation},\mathrm{we}\mathrm{have}: \\ \frac{1}{2}\rho v_{\mathrm{A}}^2+\rho g{h}_{\mathrm{A}}+p_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+\rho g{h}_{\mathrm{B}}+p_{\mathrm{B}} \\ \Rightarrow p_{\mathrm{A}}-p_{\mathrm{B}}=\frac{1}{2}\left[p\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right)\right] \end{gathered}$$
Here,
ρ is the density of the fluid.
p_A and p_B are the pressures at A and B.
h is the difference of the mercury levels in the U-tube.
$$\begin{gathered} \mathrm{Now}, \\ h\times 13.6\times 980=\frac{1}{2}\times 1\times \frac{21}{4}(100{)}^2[\mathrm{Using}\left(\mathrm{i}\right)] \\ \Rightarrow h=\frac{21\times (100{)}^2}{2\times 13.6\times 980\times 4}[\because p_{\mathrm{A}}-p_{\mathrm{B}}] \\ =1.969\mathrm{cm} \end{gathered}$$

Therefore, the required difference is 1.969 cm.