Given:
The pressure in the lower pipe is,
[tex]P_1=120\text{ kPa}[/tex]
The speed of water is,
[tex]v_1=1\text{ m/s}[/tex]
The radius of the lower pipe is
[tex]\begin{gathered} r_1=12\text{ cm} \\ =0.12\text{ m} \end{gathered}[/tex]
the radius of the upper pipe is
[tex]\begin{gathered} r_2=6\text{ cm} \\ =0.06\text{ m} \end{gathered}[/tex]
The height of the upper pipe is,
[tex]h_2=2\text{ m}[/tex]
The density of water is,
[tex]\rho=1000kg/m^3[/tex]
Using the continuity equation,
[tex]\begin{gathered} \pi r^2_{1^{}}v_1=\pir_2^{}^2_{}v_2 \\ v_2=\frac{r^2_{1^{}}v_1}{r^2_{2^{}}} \\ v_2=\frac{0.12\times0.12\times1}{0.06\times0.06} \\ v_2=4\text{ m/s} \end{gathered}[/tex]
Hence the speed of water is 4 m/s.
Applying Bernoulli's principle we get,
[tex]\begin{gathered} P_1+\frac{1}{2}\rho(v_1)^2+\rho g\times0_{}=P_2+\frac{1}{2}\rho(v_2)^2+\rhogh_2 \\ 120\times10^3+\frac{1}{2}\times1000\times1^2+0=P_2+\frac{1}{2}\times1000\times4^2+1000\times9.8\times2 \\ P_2=120500-27600 \\ P_2=92.9\text{ kPa} \end{gathered}[/tex]
Hence, the pressure is 92.9 kPa.
