Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 N/m² between x and y where the area of cross-sections are 3cm² and 1.5cm² respectively. If the rate of flow of water through the tube is Px10⁺² cm³/s, then P is (Round off to nearest integer)

Let the velocity at x = v_x and that at y = v_y. 
By the equation of continuity, $\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}=\frac{3{ \mathrm{cm}}^2}{1.5{ \mathrm{cm}}^2}=2$

By Bernoulli's equation,
$$\begin{gathered} \begin{array}{l}{\mathrm{P}}_{\mathrm{x}}+\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{x}}^2={\mathrm{P}}_{\mathrm{y}}+\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{y}}^2 \mathrm{or}, {\mathrm{P}}_{\mathrm{x}}-{\mathrm{P}}_{\mathrm{y}}=\frac{1}{2}\mathrm{\rho }{\left(2{\mathrm{v}}_{\mathrm{y}}\right)}^2-\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{y}}^2=\frac{3}{2}{\mathrm{\rho v}}_{\mathrm{y}}^2\\ \mathrm{or}, 600\frac{\mathrm{N}}{{\mathrm{m}}^2}=\frac{3}{2}\left(1000\frac{\mathrm{kg}}{{\mathrm{m}}^3}\right){\mathrm{v}}_{\mathrm{x}}^2\end{array} \\ \mathrm{or}, {\mathrm{v}}_{\mathrm{x}}=\sqrt{0.4{\mathrm{m}}^2/{\mathrm{s}}^2}=0.63\mathrm{m}/\mathrm{s}\mathrm{. } \end{gathered}$$
The rate of flow $=(3 cm²)(0.63 m/s)=189 cm³/s. =1.89 x 10² cm³/s