JEE Advanced 2022 — Paper 1 — Question 29

${{\text{P}}^{1-\gamma }}{{\text{T}}^{\gamma }}=$ Const.

$\begin{matrix}{{\text{P}}_{2}}=150\text{ }\!\!~\!\!\text{ }Pa\#\left( 1 \right) \\\end{matrix}$

$\frac{\text{dm}}{\text{dt}}={{\rho }_{1}}\text{ }\!\!~\!\!\text{ }{{\text{A}}_{1}}{{\text{v}}_{1}}$

$\begin{matrix}{{\text{v}}_{1}}=40\text{ }\!\!~\!\!\text{ }m/s\#\left( 2 \right) \\\end{matrix}$

$\rho =\frac{\text{PM}}{\text{RT}}$

$\begin{matrix}{{\rho }_{2}}=0.1Kg/{{\text{m}}^{3}}\#\left( 3 \right) \\\end{matrix}$

${{\rho }_{1}}{{A}_{1}}{{\text{v}}_{1}}={{\rho }_{2}}\text{ }\!\!~\!\!\text{ }{{\text{A}}_{2}}{{\text{v}}_{2}}$

$\begin{matrix}{{\text{v}}_{2}}=20\text{ }\!\!~\!\!\text{ }m/s\#\left( 4 \right) \\\end{matrix}$

From work energy theorem, ${{P}_{1}}{{A}_{1}}{{v}_{1}}dt-{{\rho }_{2}}{{A}_{2}}{{V}_{2}}dt+{{\rho }_{1}}{{A}_{1}}{{v}_{1}}dtg\left( 0 \right)-{{\rho }_{2}}{{A}_{2}}{{v}_{2}}dtg\left( h \right)$ $=\frac{1}{2}{{\rho }_{2}}{{A}_{2}}{{\text{v}}_{2}}\text{dt}_{2}^{2}-\frac{1}{2}{{\rho }_{1}}\text{ }\!\!~\!\!\text{ }{{\text{A}}_{1}}{{\text{v}}_{1}}\text{dtv}_{1}^{2}+\frac{1}{\gamma -1}\left[ {{\text{P}}_{2}}\text{ }\!\!~\!\!\text{ }{{\text{A}}_{2}}{{\text{v}}_{2}}\text{dt}-{{\text{P}}_{1}}\text{ }\!\!~\!\!\text{ }{{\text{A}}_{1}}{{\text{v}}_{1}}\text{dt} \right]$ $\text{h}=360\text{ }\!\!~\!\!\text{ m}$.