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