If liquids of different densities filled in a container then the velocity of efflux is
(someone please derive the formula of velocity of efflux)
Answers
Answer: 4 ms^ -1
Explanation:ANSWER
Since area of a hole is very small in comparison to base area A of the cylinder, velocity of liquid inside the cylinder is negligible. Let velocity of efflux be v and atmospheric pressure be P
0
.
Consider two points A (inside the cylinder) and B (must outside the hole) in the same horizontal line as shown in the figure in question.
Pressure at A: P
A
=P
0
+hρ
2
g+(h−y)ρ
1
g
Pressure at A: P
B
=P
0
Applying Bernoulli's theorem at points A and B
P
A
+
2
1
ρ
2
v
2
2
+(ρ
2
gh+ρ
1
g(h−y))=P
B
+
2
1
ρ
1
v
1
2
+(ρ
2
gh+ρ
1
g(h−y)) .....(1)
Substituting P
A
and P
B
in eqn(1) we get,
P
0
+hρ
2
g+(h−y)ρ
1
g+
2
1
ρ
2
v
2
2
+(ρ
2
gh+ρ
1
g(h−y))=P
0
+
2
1
ρ
1
v
1
2
+(ρ
2
gh+ρ
1
g(h−y)) ....(2)
Given,
A=0.5m
2
a=5cm
2
=25×10
−4
m
2
From continuity equation,
Av
2
=av
1
∴
v
1
v
2
=
A
a
=
0.5
25×10
−4
=50×10
−4
Given h=60cm=0.6m
y=20cm=0.2m
ρ
1
=900kgm
−3
ρ
2
=600kgm
−3
P
0
+hρ
2
g+(h−y)ρ
1
g+
2
1
ρ
2
v
2
2
=P
0
+
2
1
ρ
1
v
1
2
As given in the problem, we ignore v
2
since a is very small.
∴P
0
+hρ
2
g+(h−y)ρ
1
g=P
0
+
2
1
ρ
1
v
1
2
∴hρ
2
g+(h−y)ρ
1
g=
2
1
ρ
1
v
1
2
Substituting the values, 0.6×600×10+(0.6−0.2)×900×10=
2
1
×900×v
1
2
⇒v
1
=4ms
−1