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← JEE Advanced 2023 · Paper 2
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JEE Advanced 2023 · Paper 2, Physics question 16

Passage

PARAGRAPH II

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

[Given: The acceleration due to gravity g=10 m s−2g = 10\ \mathrm{m\,s^{-2}} and π=3.14\pi = 3.14]

A cylindrical furnace of height H and diameter D (containing air at rho, Pa, T) sits on the ground; outside air (rho_a, Pa, Ta) enters through its open base. A thin vertical chimney of diameter d and height h rises from the top of the furnace, with an arrow showing air exiting at the top.

Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ gm s−1\mathrm{gm\,s^{-1}}.

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped (marks to all) in the official key. Below is the standard idealised working, which gives about 49.6 g/s. The organisers did not publish why the question was dropped.

Density of hot air. At equal pressure,
ρ=ρaTaT=1.2×300360=1.0 kg/m3.\rho=\rho_a\frac{T_a}{T}=1.2\times\frac{300}{360}=1.0\ \text{kg/m}^3.

Exit speed. Apply Bernoulli along a streamline from the still air outside the furnace inlet to the chimney exit. The outside air column of height H+hH+h supplies the pressure difference (ρa−ρ)g(H+h)(\rho_a-\rho)g(H+h):
v=2(ρa−ρ)g(H+h)ρ=2×0.2×10×101.0≈6.32 m/s.v=\sqrt{\frac{2(\rho_a-\rho)g(H+h)}{\rho}}=\sqrt{\frac{2\times0.2\times10\times10}{1.0}}\approx6.32\ \text{m/s}.
The furnace cross-section is 100 times the chimney's, so the speed inside the furnace is negligible.

Mass flow rate.
m˙=ρ πd24 v=1.0×3.14×0.0025×6.32≈0.0496 kg/s≈49.6 g/s.\dot m=\rho\,\frac{\pi d^2}{4}\,v=1.0\times3.14\times0.0025\times6.32\approx0.0496\ \text{kg/s}\approx49.6\ \text{g/s}.

Why the answer is not unique. It depends on modelling choices the paragraph leaves open:

  • whether the furnace height HH counts in the driving column (using only hh gives about 47 g/s);
  • how the inlet is treated.

This is a plausible reason no single answer could be keyed.