Q50 / 51 · JEE Advanced 2023 (Paper 2) · Physics

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 Pa 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 ρa=1.2 kg m-3, pressure Pa and temperature Ta=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 m s-2 and π=3.14]

Figure


Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ gm s−1.

Correct answer: 7850

Solution

Applying Bernoulli’s theorem between top and bottom points of furnace :

Pa+12ρaV2=Pa+ρgH+12ρV2                ... 1

Also, since PM=ρRT

⇒ ρa×300=ρ×360

⇒ ρ=1 kg m-3             ... 2

From equation(1), we can write

⇒ V220.2=1×10×1

⇒ V=10 m s-1

Required mass flow rate =V×πD24×ρ

=10×3.144×12×1 kg s-1

=7.85 kg s-1

=7850 gm s-1

Hence, the correct answer is 7850.

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