Question

Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. the separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem.

Answer 1

Given:
Separation between the cross sections at A and B = 15/16 cm
Speed of water at A, v_A = 25 cm/s
Speed of water at B, v_B = 50 cm/s
(c) By Bernoulli's equation, we get:
$$\begin{gathered} \frac{1}{2}p{v}_{\mathrm{A}}^2+\mathrm{\rho }g{h}_{\mathrm{A}}+P_{\mathrm{A}}=\frac{1}{2}p{v}_{\mathrm{B}}^2+\mathrm{\rho }g{h}_{\mathrm{B}}+P_{\mathrm{B}} \\ \Rightarrow P_{\mathrm{A}}\mathit{-}{P}_{\mathit{B}}=\left(\frac{1}{2}\right)p\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right)+\rho g\left(h_{\mathrm{B}}-h_{\mathrm{A}}\right)[h_{\mathrm{A}}-h_{\mathrm{B}}=\left(\frac{15}{16}\right)\mathrm{cm}] \\ \Rightarrow P_{\mathit{A}}\mathit{-}{P}_{\mathrm{B}}=\frac{1}{2}(2500-625)-1000\times \frac{15}{16} \\ =0 \end{gathered}$$