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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 29
Total: 54
An ideal gas of density
ρ
=
0.2
kg
m
−
3
enters a chimney of height
h
at the rate of
α
=
0.8
kg
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
ms
−
2
and the ratio of specific heats of the gas
γ
=
2
. Ignore atmospheric pressure.
The pressure of the gas at the upper end of the chimney is
300
Pa
.
The velocity of the gas at the lower end of the chimney is
40
ms
−
1
and at the upper end is
20
ms
−
1
The height of the chimney is
590
m
.
The density of the gas at the upper end is
0.05
kg
m
−
3
.
Validate
Solution:
dm
dt
=
ρ
1
A
1
v
1
=
0.8
kg
∕
s
A
v
1
=
0.8
0.2
×
0.1
=
40
m
∕
s
g
=
10
m
∕
s
2
γ
=
2
Gas undergoes adiabatic expansion,
p
1
−
γ
T
γ
=
Constant
P
2
P
1
=
(
T
1
T
2
)
r
1
−
γ
P
2
=
(
300
150
)
2
−
1
×
600
P
2
=
600
4
=
150
Pa
Now
ρ
=
PM
RT
⇒
ρ
∝
P
T
ρ
1
ρ
2
=
(
P
1
P
2
)
(
T
1
T
2
)
=
(
150
600
)
(
300
150
)
=
1
2
ρ
2
=
ρ
1
2
=
0.1
kg
∕
m
3
Now
ρ
2
A
2
v
2
=
0.8
⇒
v
2
=
0.8
0.1
×
0.4
=
20
m
∕
s
Now
W
on gas
=
∆
K
+
∆
U
+
(Internal energy)
P
1
A
1
∆
x
1
−
P
2
A
2
∆
x
2
=
1
2
∆
mV
2
2
−
1
2
∆
mV
1
2
+
∆
mgh
+
f
2
(
P
2
∆
V
2
−
P
1
∆
V
1
)
⇒
2
P
1
∆
V
1
∆
m
−
2
P
2
∆
V
2
∆
m
=
V
2
2
−
V
1
2
2
+
gh
⇒
2
×
600
0.2
−
2
×
150
0.1
=
20
2
−
40
2
2
+
10
h
h
=
360
m
© examsnet.com
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