JEE Advanced 2023 — Paper 2 — PHY-IV solution

JEE Advanced2023Paper 2PHY-IV

PARAGRAPH-II

A cylindrical furnace has height ( $H$ ) and diameter ( $D$ ) both $1$ m. It is maintained at temperature $360$ K. The air gets heated inside the furnace at constant pressure $P_{a}$ and its temperature becomes $T = 360$ K. The hot air with density $ρ$ rises up a vertical chimney of diameter $d = 0.1$ m and height $h = 9$ m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density $ρ_{a} = 1.2$ kg m $^{- 3}$ , pressure $P_{a}$ and temperature $T_{a} = 300$ K enters the furnace. Assume air as an ideal gas, neglect the variations in $ρ$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects.

[Given: The acceleration due to gravity $g = 10$ ms $^{- 2}$ and $π = 3.14$ ]

Q2. When the chimney is closed using a cap at the top, a pressure difference $Δ P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of $Δ P$ is ______ N m $^{- 2}$ .

Your answer

Solution

Density of hot air (unchanged): $ρ_{a} T_{a} = ρT \Rightarrow 1.2 \times 300 = ρ \times 360 \Rightarrow ρ = 1 kg m^{- 3}$ .

With the cap in place the air is static everywhere. Take $P_{a}$ at the base (ground level) outside the furnace.

Pressure on the bottom face of the cap (inside, hot air): The hot air column inside the chimney has height $h$ (from the top of the furnace to the cap). Since the furnace base is at $P_{a}$ ,

$P_{in} = P_{a} - ρ g h$ .

Pressure on the top face of the cap (outside, cold atmosphere): The atmospheric column of height $H + h$ above the ground gives

$P_{out} = P_{a} - ρ_{a} g (H + h)$ .

Pressure difference:

$Δ P = P_{in} - P_{out} = ρ_{a} g (H + h) - ρ g h = 1.2 \times 10 \times 10 - 1 \times 10 \times 9 = 120 - 90 = 30 N m^{- 2}$ .

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