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JEE Advanced 2023 Paper 2 Solutions

Paragraph for next two questions

A cylindrical furnace has height

(\mathrm{H})

and diameter

(\mathrm{D})

both

1\ \text{m}

. It is maintained at temperature

360\ \text{K}

. The air gets heated inside the furnace at constant pressure

P_a

and its temperature becomes

T=360\ \text{K}

. The hot air with density

\rho

rises up a vertical chimney of diameter

d=0.1\ \text{m}

and height

h=9\ \text{m}

above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density

\rho_a=1.2\ \text{kg}\ \text{m}^{-3}

, pressure

P_{a}

and temperature

T_{a}=300\ \text{K}

enters the furnace. Assume air as an ideal gas, neglect the variations in

\rho

and

T

inside the chimney and the furnace. Also ignore the viscous effects.

[Given: The acceleration due to gravity

g=10\ \text{m}\ \text{s}^{-2}

and

\pi=3.14

]

Section: Physics

Question : 16 of 51

Marks: +1, -0

\mathrm{gm}\ \mathrm{s}^{-1}

Solution:

\because P M = \rho R T

And

P

inside furnace is constant

\therefore \rho R T =

constant

\rho T =

constant

\therefore \rho_a T_a = \rho T

1.2(300) = \rho (360)

\rho = 1\ \text{kg/m}^3

Now, applying Bernoulli's theorem at the bottom and the top of chimney

P_a + \frac{1}{2}\rho (0)^2 + 0 = (P_a - \rho_a g h) + \frac{1}{2}\rho (v^2) + \rho g h

v = \sqrt{\frac{2 (\rho_a - \rho) g h}{\rho}}

\Rightarrow v = 6\ \text{m/s}

\therefore \frac{dm}{dt}

at exit

= \rho v \left( \frac{\pi d^2}{4} \right)

= \frac{1 \times 6 \times 3.14 \times (0.1)^2}{4}

= 0.0471\ \text{kg/s}

= 47.10\ \text{g/s}

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