Water flows in a horizontal tube (see figure). The pressure of water changes by \(700\; \mathrm{Nm}^{-2}\) between \(\mathrm{A}\) and \(\mathrm{B}\) where the area of cross section are \(40\; \mathrm{cm}^{2}\) and \(20\; \mathrm{cm}^{2},\) respectively. Find the rate of flow of water through the tube. ........ \(\mathrm{cm}^{3} / \mathrm{s}\) (density of water \(=1000\; \mathrm{kgm}^{-3}\) )

\((40) \mathrm{V}_{\mathrm{A}}=(20) \mathrm{V}_{\mathrm{B}}\)

\(\mathrm{V}_{\mathrm{B}}=2 \mathrm{V}_{\mathrm{A}}\)

Using Bernoulli's theorem

\(\mathrm{P}_{\mathrm{A}}+\frac{1}{2} \rho \mathrm{V}_{\mathrm{A}}^{2}=\mathrm{P}_{\mathrm{B}}+\frac{1}{2} \rho \mathrm{V}_{\mathrm{B}}^{2}\)

\(\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}=\frac{1}{2} \rho\left(\mathrm{V}_{\mathrm{B}}^{2}-\mathrm{V}_{\mathrm{A}}^{2}\right)\)

\(700=\frac{1}{2} \times 1000\left(4 \mathrm{V}_{\mathrm{A}}^{2}-\mathrm{V}_{\mathrm{A}}^{2}\right)\)

\(\mathrm{V}_{\mathrm{A}}=0.68 \mathrm{m} / \mathrm{s}=68 \mathrm{cm} / \mathrm{s}\)

Rate of flow \(=\mathrm{A}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}\)

\(=(40)(68)=2720\; \mathrm{cm}^{3} / \mathrm{s}\)