Answer :

Correct answer is : 0.375

Mass flow rate through nozzle remain constant

$\begin{array}{l}\therefore {\left(\rho AV\right)}_{inlet}={\left(\rho AV\right)}_{outlet}\\ 0.1\times V_i=0.02V_O\end{array}$

V_O = 5 V_i

∴ V₂ = 5 V_i

Bernoulli’s equation for nozzle

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}\because \left[z_1=z_2\right]$

$\frac{0.36\times {10}^3}{1.2\times 9.81}+\frac{V_1^2}{2g}-\frac{{\left(5V_1\right)}^2}{2g}=\frac{P_2}{\rho g}$

$\frac{0.36\times {10}^3}{1.2g}-\frac{24V_1^2}{2g}=\frac{P_2}{\rho g}$

P₂ = P_atm = 0 (Gauge)

∴ V₁ = 5 m/s V₂ = 25 m/s

∵ P₂ = P_O = P_atm = 0 (Gauge)

But at point O there will be a pressure due to motion of fluid which will exert a pressure on the plate.

Now,

Bernoulli’s equation between exit point & point O

$\frac{P_2}{\rho g}+\frac{V_2^2}{2g}=\frac{P_3}{\rho g}+\frac{V_3^2}{2g}$

$P_3=\frac{1}{2}\rho V_2^2$

$P_3=\frac{1}{2}\times 1.2\times {\left(25\right)}^2=375Pa$

∴ P₃ = 0.375 kPa