68.
(a) We consider a point D on the surface of the liquid in the container, in the same tube of flow with
points A, B and C. Applying Bernoulli’s equation to points D and C, we obtain
p
D
+
1
2
ρv
2
D
+ ρgh
D
= p
C
+
1
2
ρv
2
C
+ ρgh
C
which leads to
v
C
=
2(p
D
− p
C
)
ρ
+ 2g(h
D
− h
C
) + v
2
D
≈
2g (d + h
2
)
where in the last step we set p
D
= p
C
= p
air
and v
D
/v
C
≈ 0.
(b) We now consider points B and C:
p
B
+
1
2
ρv
2
B
+ ρgh
B
= p
C
+
1
2
ρv
2
C
+ ρgh
C
.
Since v
B
= v
C
by equation of continuity, and p
C
= p
air
, Bernoulli’s equation becomes
p
B
= p
C
+ ρg(h
C
− h
B
) = p
air
− ρg(h
1
+ h
2
+ d) .
(c) Since p
B
≥ 0, we must let p
air
− ρg(h
1
+ d + h
2
)
≥ 0, which yields
h
1
≤ h
1,max
=
p
air
ρ
− d − h
2
≤
p
air
ρ
= 10.3 m .