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JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 1 Questions
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© examsnet.com
Question : 22
Total: 100
In the diagram shown, the difference in the two tubes of the manometer is
5
c
m
, the cross section of the tube at
A
and
B
is
6
m
m
2
and
10
m
m
2
respectively. The rate at which water flows through the tube is
(
g
=
10
m
s
−
2
)
[Online April 19, 2014]
7.5 cc/s
8.0 cc/s
10.0 cc/s
12.5 cc/s
Validate
Solution:
According to Bernoulli's theorem,
P
1
+
1
2
ρ
v
1
2
=
P
2
+
1
2
ρ
v
2
2
∴
v
2
2
−
v
1
2
=
2
g
h
.......(1)
According to the equation of continuty
A
1
v
1
=
A
2
v
2
......(2)
A
1
A
2
=
6
m
m
2
10
m
m
2
From equation(2)
,
A
1
A
2
=
v
2
v
1
=
6
10
or,
v
2
=
6
10
v
1
Putting this value of
v
2
in equation (1)
(
6
10
v
1
)
2
−
(
v
1
)
2
=
2
×
10
3
×
5
[
∵
g
=
10
m
∕
s
2
=
10
3
c
m
∕
s
2
and
h
=
5
c
m
]
Solving we get
v
1
=
10
8
Therefore the rate at which water flows through the tube
=
A
1
v
1
=
A
2
v
2
=
6
×
10
8
=
7.5
c
c
∕
s
© examsnet.com
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