Siphon tube

Taking the mouth of the nozzle as the reference for zero gravitational potential and applying Bernoulli's theorem for the mouth of nozzle and the water in the tank, we get

$\begin{aligned} & P_0 + H \rho g + 0 = P_0 + 0 + \dfrac 12 \rho v^2 \\ \implies & v = \sqrt{2gH} \end{aligned}$

Via equation of continuity we have

$\begin{aligned} & AV = av \\ \implies & \dfrac{\pi D^2}{4} V = \dfrac{\pi d^2}{4} \cdot \sqrt{2gH} \\ \implies & V = \dfrac{d^2}{D^2} \cdot \sqrt{2gH} \end{aligned}$

Again taking mouth of nozzle as the reference for zero gravitational potential and applying Bernoulli's theorem for the mouth of nozzle and point C, we get

$\begin{aligned} P_C + (H+h) \rho g + \dfrac 12 \rho V^2 &=& P_0 + 0 + \dfrac 12 \rho v^2 \\ \implies P_C &=& P_0 - (H+h) \rho g + \dfrac 12 \rho \left( v^2 - V^2 \right) \\ &=& P_0 - (H+h) \rho g + \dfrac 12 \rho \left( 1 - \dfrac{d^4}{D^4} \right) 2gH \\ &=& P_0 - \rho g \left( h + H \cdot \dfrac{d^4}{D^4} \right) \end{aligned}$

Putting required values, we get

$P_C = 99334.375 \ \text{Pa} = 745.069 \ \text{mm of Hg} \approx \boxed{745}$