JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Fluid Mechanics

An ideal gas of density

ρ = 0.2 k g m^{- 3}

enters a chimney of height h at the rate of

α = 0.8 k g s^{- 1}

from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is

A_{1} = 0.1 m^{2}

and the upper end is

A_{2} = 0.4 m^{2}

. The pressure and the temperature of the gas at the lower end are 600 Pa and 300 K, respectively, while its temperature at the upper end is

150 K

. The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take

g = 10 m s^{- 2}

and the ratio of specific heats of the gas

γ = 2

. Ignore atmospheric pressure.

Which of the following statement(s) is(are) correct?

(A)The pressure of the gas at the upper end of the chimney is $300 P a$ .
(B)The velocity of the gas at the lower end of the chimney is $40 m s^{- 1}$ and at the upper end is $20 m s^{- 1}$ .✓ Correct
(C)The height of the chimney is $590 m$ .
(D)The density of the gas at the upper end is $0.05 k g m^{- 3}$ .

Answer: (B)

Solution

From continuity:

ρ_{1} A_{1} v_{1} = 0.8

v_{1} = \frac{0.8}{0.2 \times 0.1} = 40 m / s

For adiabatic process with

γ = 2

\frac{P _{2}}{P _{1}} = (\frac{T _{2}}{T _{1}})^{γ / (γ - 1)} = (\frac{150}{300})^{2} = \frac{1}{4}

P_{2} = 150 P a

From ideal gas law:

ρ \propto P / T

ρ_{2} = ρ_{1} \times \frac{P _{2}}{P _{1}} \times \frac{T _{1}}{T _{2}} = 0.2 \times \frac{1}{4} \times 2 = 0.1 k g / m^{3}

From continuity:

v_{2} = \frac{0.8}{ρ _{2} A _{2}} = \frac{0.8}{0.1 \times 0.4} = 20 m / s

Using energy conservation:

2 P_{1} / ρ_{1} - 2 P_{2} / ρ_{2} = \frac{v _{2}^{2} - v _{1}^{2}}{2} + g h

6000 - 3000 = \frac{400 - 1600}{2} + 10 h

h = 360 m

Only option (B) is correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options