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 .