Let air absolute pressure is P, then apply Bernaulli’s theorem just on surface of water and just outside the nozzle.

$\mathrm{P}+0+0={\mathrm{P}}_0+\frac{{\mathrm{\rho v}}^2}{2}+\mathrm{\rho gh}$

Where h = 0.5 m and level of water is taken as reference level for potential energy head.

$\Rightarrow \mathrm{P}-{\mathrm{P}}_0=\frac{{\mathrm{\rho v}}^2}{2}+\mathrm{\rho gh}$

$\frac{{10}^3\times {30}^2}{2}+{10}^3\times 10\times 0.5=4.55\times {10}^5 \mathrm{N}/{\mathrm{m}}^2$

P – P₀ is representing the required gauge pressure.