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Test Index
JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 3 Questions
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Question : 8
Total: 100
Glycerin of density
1.25
×
10
3
kg
m
−
3
is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are
10
cm
2
and
5
cm
2
and pressure drop across its length is
3
Nm
−
2
. The rate of flow of glycerin through the pipe is
x
×
10
−
5
m
3
s
−
1
. The value of
x
is _______.
[12-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
A
1
=
10
cm
2
∆
P
=
P
1
−
P
2
=
3
N
∕
m
2
(given)
By continuity equation
A
1
v
1
=
A
2
v
2
∴
v
1
=
A
2
A
1
v
2
. . . . . (1)
By bernoulli's equation
P
1
+
1
2
ρ
v
1
2
=
P
2
+
1
2
ρ
v
2
2
P
1
−
P
2
=
1
2
ρ
(
v
2
2
−
v
1
2
)
∆
P
=
1
2
ρ
(
v
2
2
−
A
2
2
A
1
2
v
2
2
)
∆
P
=
1
2
ρ
[
1
−
(
A
2
A
1
)
2
]
v
2
2
3
=
1
2
×
1.25
×
10
3
(
1
−
(
5
10
)
2
)
v
2
2
3
=
1
2
×
1.25
×
10
3
[
1
−
1
4
]
v
2
2
3
=
1
2
×
1.25
×
10
3
×
3
4
v
2
2
∴
v
2
=
8
×
10
−
2
m
∕
s
So discharge rate
=
A
2
V
2
=
5
×
10
−
4
×
8
×
10
−
2
=
4
×
10
−
5
m
3
∕
s
© examsnet.com
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