QJEEAdv23Ph62af: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph62af Solution: Considering the air flow to be streamline, the steady mass flow …

We know, $\begin{aligned}& \mathrm{P_a}=\frac{\rho R \mathrm{T}}{M}\end{aligned}$

For a gas R and M are constant. So, $\rho T$ = Constant (for constant pressure).

The density of hot air inside the furnace is = $\rho$

The air gets heated inside the furnace at constant pressure P_a.

$\therefore$ $\rho_{\mathrm{a}} \mathrm{T}_{\mathrm{a}}=\rho \mathrm{T}$

$\Rightarrow$ 1.2 $\times$ 300 = $\rho$ $\times$ 360

$\Rightarrow \rho=1 \mathrm{~kg} / \mathrm{m}^3$

Buoyant force applied on the hot air = $\rho_{\mathrm{a}}$ Vg (Upward direction)

Weight of the hot air = $\rho$ Vg (Downward direction)

$\therefore$ Net force on the hot air = $\rho_{\mathrm{a}}$ Vg - $\rho$ Vg

Let acceleration of the hot air in the upward direction = $a$

and mass of the hot air = $\rho$ V

$\therefore$ $\rho$ V $a$ = $\rho_{\mathrm{a}}$ Vg - $\rho$ Vg

$\Rightarrow$ $a$ = ${{{\rho _a}Vg - \rho Vg} \over {\rho V}}$

= ${{{\rho _a}g - \rho g} \over \rho }$

= ${{1.2 \times 10 - 1 \times 10} \over 1}$

= 2 m/s²

$\therefore$ Velocity(v) of the hot air when exiting the chimney using formula ${v^2} = {u^2} + 2ah$ ,

${v^2} = 0 + 2 \times 2 \times 9$

$\Rightarrow$ v = 6 m/s

Mass flow rate = ${{dm} \over {dt}}$ = $\rho$ Av

= $\rho \times \frac{\pi \mathrm{d}^2}{4} \times\mathrm{v}=1 \times \frac{\pi}{4} \times 10^{-2} \times 6$

= ${{471} \over {10000}}$ kg/s = ${{471} \over {10000}} \times 1000$ gm/s = 47.1