QJEEAdv23Phc3cf: JEE Advanced 2023 Physics - Class 11 JEEAdv23Phc3cf Solution: When the chimney is closed using a cap at the top, a pressure…

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

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

After chimney is closed,

Pressure at the bottom surface, P₁ = P_a - $\rho$ gh = P_a - (1)(10)(9)

Pressure at the bottom surface, P₂ = P_a - $\rho$ _ag(h + H) = P_a - (1.2)(10)(9 + 1)

$\therefore$ Pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap

= P₁ - P₂

= P_a - (1)(10)(9) - P_a + (1.2)(10)(9 + 1)

= 120 - 90 = 30