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Test Index
JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 3 Questions
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© examsnet.com
Question : 1
Total: 100
An air bubble of volume
1
cm
3
rises from the bottom of a lake
40
m
deep to the surface at a temperature of
12
∘
C
. The atmospheric pressure is
1
×
10
5
Pa
, the density of water is
1000
kg
∕
m
3
and
g
=
10
m
∕
s
2
. There is no difference of the temperature of water at the depth of
40
m
and on the surface. The volume of air bubble when it reaches the surface will be:
[8-Apr-2023 shift 1]
3
cm
3
4
cm
3
2
cm
3
5
cm
3
Validate
Solution:
👈: Video Solution
Pressure at surface
=
P
atm
=
1
×
10
5
Pa
v
surface
=
?
Pressure at
h
=
40
m
depth
P
=
P
atm
+
ρ
gh
P
=
10
5
+
10
3
×
10
×
40
P
=
5
×
10
5
Pa
V
=
1
cm
3
Temp. is constant
P
1
V
1
=
P
2
V
2
10
5
×
v
=
5
×
10
5
×
1
v
=
5
cm
3
© examsnet.com
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