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JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 2 Questions
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Question : 40
Total: 100
Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density
d
. The area of the base of both vessels is
S
but the height of liquid in one vessel is
x
1
and in the other,
x
2
. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:
[4 Sep. 2020 (II)]
g
d
S
(
x
2
2
+
x
1
2
)
g
d
S
(
x
2
+
x
1
)
2
3
4
g
d
S
(
x
2
−
x
1
)
2
1
4
g
d
S
(
x
2
−
x
1
)
2
Validate
Solution:
Initial potential energy,
U
1
=
(
ρ
S
x
1
)
g
.
x
1
2
+
(
ρ
S
x
2
)
g
.
x
2
2
Final potential energy,
U
f
=
(
ρ
S
x
f
)
g
.
x
f
2
×
2
By volume conservation,
S
x
1
+
S
x
2
=
S
(
2
x
f
)
x
f
=
x
1
+
x
2
2
When valve is opened loss in potentail energy occur till water level become same.
Δ
U
=
U
i
−
U
f
Δ
U
=
ρ
S
g
[
(
x
1
2
2
+
x
2
2
2
)
−
x
f
2
]
=
ρ
S
g
[
x
1
2
2
+
x
2
2
2
−
(
x
1
+
x
2
2
)
2
]
=
ρ
S
g
2
[
x
1
2
2
+
x
2
2
2
−
x
1
x
2
]
=
ρ
S
g
4
(
x
1
−
x
2
)
2
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