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EAMCET Engineering 2003 Solved Paper
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Section:
Physics
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© examsnet.com
Question : 1
Total: 160
'A horizontal uniform glass tube of
100
cm
, length sealed at both ends contains
10
cm
mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively
31
∘
C
and
76
cm
of mercury. If the air column at one end is kept at
0
∘
C
and the other end at
273
∘
C
, then the pressure of air which is at
0
∘
C
is : ( in cm of
Hg
)
76
88.2
102.4
122
Validate
Solution:
On keeping the temperature of the ends of tube at
0
∘
C
and
273
∘
C
Applying ideal gas equation
p
1
V
1
T
1
=
p
2
V
2
T
2
=
p
3
V
3
T
3
76
×
45
(
273
+
31
)
=
p
2
×
l
(
273
+
0
)
=
p
3
(
90
−
l
)
273
+
273
76
×
45
304
=
p
2
×
l
273
=
p
3
(
90
−
l
)
546
From Part II and III,
p
2
×
l
273
=
p
3
(
90
−
l
)
546
(Mercury column is at rest, so pressure difference,
p
2
−
p
3
=
0
⇒
p
2
=
p
3
∴
p
2
×
l
273
=
p
2
(
90
−
l
)
546
⇒
2
l
=
90
−
l
⇒
l
=
30
cm
From I and II,
⇒
76
×
45
304
=
p
2
×
30
273
p
2
=
76
×
45
×
273
30
×
304
p
2
=
102.4
© examsnet.com
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