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JEE Advanced 2023 Paper 2 Solutions
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© examsnet.com
Question : 11
Total: 52
An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius
0.1
mm
is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume
V
0
to
100
101
V
0
with the movable piston. Considering air as an ideal gas, the height
(
h
)
of the liquid column in the capillary above the liquid level in
cm
is _________________
[Given: Surface tension of the liquid is
0.075
N
m
−
1
, atmospheric pressure is
10
5
N
m
−
2
, acceleration due to gravity
(
g
)
is
10
m
s
−
2
, density of the liquid is
10
3
kg
m
−
3
and contact angle of capillary surface with the liquid is zero]
Your Answer:
Validate
Solution:
Let
P
f
be the air pressure
P
0
v
0
=
P
f
v
f
P
0
v
0
=
P
f
(
100
101
)
v
0
P
f
=
101
×
10
3
Pa
(
∵
P
0
=
10
5
Nm
−
2
)
Now, consider the 4 points shown in diagram
P
d
−
P
c
=
2
T
R
(
∵
P
d
=
P
0
)
∴
P
c
=
P
0
−
2
T
R
Now,
P
a
=
P
b
(also,
P
a
=
P
f
)
P
f
=
ρ
g
h
+
P
c
101
×
10
3
=
(
10
3
×
10
×
h
)
+
(
10
5
−
2
×
0.075
0.1
×
10
−
3
)
© examsnet.com
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