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Answer, Key – Homework 2 – David McIntyre
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Chapter 15 problems.
001 (part 1 of 2) 5 points
A simple U-tube that is open at both ends
is partially filled with a liquid of density
553 kg/m3 . Water is then poured into one
arm of the tube, forming a column 11 cm in
height, as shown in the following diagram.
The density of water is 1000 kg/m3 .
h1
l
water
liquid
What is the difference, h1 , in the heights of
the two liquid surfaces?
Correct answer: 8.8915 cm.
Explanation:
Basic Concepts: gauge pressure, variation of pressure with depth Because the liquid
in the U-tube is static, the pressure exerted
by the water column of height l in the right
branch of the tube must balance the pressure
exerted by the liquid of height h1 + l in the
left branch. Therefore,
P0 + (l + h1 ) ρ` g = P0 + l ρw g
Solving for h1 ,
¶
µ
ρw
−1
h1 = l
ρ`
µ
¶
1000 kg/m3
= 11 cm
−1
553 kg/m3
= 8.8915 cm.
002 (part 2 of 2) 5 points
A simple U-tube that is open at both ends is
1
partially filled with water. A liquid of density
553 kg/m3 is then poured into one arm of the
tube, forming a column 11 cm in height, as
shown in the following diagram.
h2
l
liquid
water
What is the difference, h2 , in the heights of
the two liquid surfaces?
Correct answer: 4.917 cm.
Explanation:
Because the liquid in the U-tube is static,
the pressure exerted by the water column of
height l − h2 in the left branch of the tube
must balance the pressure exerted by the liquid of height l poured into the right branch.
Therefore,
P0 + (l − h2 ) ρw g = P0 + l ρ` g
Solving for h2 ,
µ
¶
ρ`
h2 = l 1 −
ρw
¶
µ
553 kg/m3
= 11 cm 1 −
1000 kg/m3
= 4.917 cm
003 (part 1 of 1) 0 points
Consider an object that floats in water but
sinks in oil. When the object floats in a glass
of water, half of the object is submerged. We
now slowly pour oil into the glass so it completely covers the object.
oil
Answer, Key – Homework 2 – David McIntyre
Note: The second figure does not necessarily
show the correct changes, only the experimental setup.
Compared to the water level, the object
1. moves up correct
2. moves down
3. stays in the same place
Explanation:
Consider the point where we have poured
in an amount of oil exactly enough to cover
the object:
We can infer that it does not matter how
much more oil we pour into the glass after this
point, since it will exert an evenly distributed
pressure on both water and object. Before
this point, however, the oil added will be on
top of the water only. Therefore, the oil will
push the water down, while not pushing the
object down. The extra pressure from the
oil will be transferred to the bottom of the
object, so it must push the object up.
004 (part 1 of 3) 4 points
A cube of wood whose edge is 12 mm is
floating in a liquid in a glass with one of
its faces parallel to the liquid surface. The
density of wood is 812 kg/m3 , that of liquid is
1296 kg/m3 .
How far below the liquid surface is the bottom face of the cube?
Correct answer: 7.51852 mm.
Explanation:
Given:
s = 12 mm
ρw = 812 kg/m3
ρ` = 1296 kg/m3
h
2
Wood
air
liquid
s
According to Archimedes’ principle, we
have
ρ w s3 g = ρ ` h s 2 g ,
and solving for h, we have
h=s
ρw
ρ`
(812 kg/m3 )
(1296 kg/m3 )
= 7.51852 mm .
= (12 mm)
005 (part 2 of 3) 3 points
A light oil is gently poured onto the immiscible liquid surface to form a 4 mm thick layer
above the liquid. The density of the oil is
606 kg/m3 .
When the wood cube attains hydrostatic
equilibrium again, what will be the distance
from the top of the liquid to the bottom face
of the cube?
Correct answer: 5.64815 mm.
Explanation:
Given:
ha = 4 mm
ρo = 606 kg/m3 .
air
Wood
hw
liquid
alcohol
ha
We assume that the top of the cube is still
above the oil’s surface. If ha is the width of
the oil layer, we have that the buoyant force
is
B = (ρo s2 ha + ρ` s2 hw ) g .
Once again, the buoyant force must be equal
to the weight of the cube, so
ρw s3 g = (ρo s2 ha + ρ` s2 hw ) g .
Answer, Key – Homework 2 – David McIntyre
3
Solving for hw , we get
ρw s = ρ o h a + ρ ` h w
ρw s ρo h a
−
ρ`
ρ`
(812 kg/m3 ) (12 mm)
=
(1296 kg/m3 )
(606 kg/m3 ) (4 mm)
−
(1296 kg/m3 )
= 5.64815 mm .
hw =
006 (part 3 of 3) 3 points
Additional light oil is poured onto the liquid
surface until the top surface of the oil coincides with the top surface of the wood cube
(in hydrostatic equilibrium).
How thick is the whole layer of the light
oil?
Correct answer: 8.41739 mm.
Explanation:
air
Wood
hw
liquid
alcohol
007 (part 1 of 1) 0 points
A styrofoam slab has a thickness of 8.4 cm
and a density of 247 kg/m3 . What is the area
of the slab if it floats just awash (top of slab
is even with the water surface) in fresh water
when a 98.3 kg swimmer is aboard?
Correct answer: 1.5541 m2 .
Explanation:
Since the slab floats awash, the buoyant force
is B = ρw V g = ρw dAg. This force is equal in
magnitude to the weight of the slab and the
swimmer,
ρw dAg = ρs dAg + M g
Solving for A,
A=
M
= 1.5541 m2 m
d(ρw − ρs )
008 (part 1 of 2) 0 points
Calculate the flow rate of blood (ρ =
0.837 g/cm3 ) in an aorta with a crosssectional area of 0.902 cm2 if the flow speed is
46.9 cm/s.
Correct answer: 35.4083 g/s.
Explanation:
ha
Flow Rate = ρ A v
If the width of the oil layer is ha , the buoy= (0.837 g/cm3 ) (0.902 cm2 ) (46.9 cm/s)
ant force is now
= (0.837 g/cm3 ) (42.3038 cm3 /s)
h
i
= 35.4083 g/s .
B = ρo s2 ha + ρ` s2 (s − ha ) g .
Equating the buoyant force with the weight
of the wood cube, we get
h
i
ρw s3 g = ρo s2 ha + ρ` s2 (s − ha ) g
ρw s = ρo ha + ρ` (s − ha )
ha (ρ` − ρo ) = s (ρ` − ρw )
ρ` − ρ w
ρ` − ρ o
= (12 mm)
¸
·
(1296 kg/m3 ) − (812 kg/m3 )
×
(1296 kg/m3 ) − (606 kg/m3 )
= 8.41739 mm .
ha = s
009 (part 2 of 2) 0 points
Assume: The aorta branches to form a large
number of capillaries with a combined crosssectional area of 4550 cm2 .
What is the flow speed in the capillaries?
Correct answer: 0.00929754 cm/s.
Explanation:
From the equation of continuity, we have
A1 v1
A
µ 2
¶
0.902 cm2
=
(46.9 cm/s)
4550 cm2
= 0.00929754 cm/s .
v2 =
Answer, Key – Homework 2 – David McIntyre
010 (part 1 of 1) 0 points
You use a fan to blow air across the surface
of the exposed mercury in a barometer. The
.
level of mercury in the tube will
1. drop correct
2. rise
3. be unaffected
Explanation:
Solution:
Bernoulli’s principle says that forcing the air
to move across the barometer will cause a
decrease in the air’s pressure. Since this pressure is holding the mercury up in the barometer tube, the level of mercury in the tube will
drop.
011 (part 1 of 1) 10 points
A tall water cooler tank is standing on the
floor. Some fool punched two small holes
in the tank’s wall, one hole at a height of
45 cm above the floor and the other hole
47 cm directly above the first hole and 92 cm
above the floor. Each hole produces a jet of
water that emerges in a horizontal direction
but eventually hits the floor at some distance
from the tank.
If the two water jets (emerging from each
hole) hit the floor at exactly the same spot,
how high H is the water level in the tank
(relative to the room’s floor)?
Correct answer: 137 cm.
Explanation:
Apply Bernoulli’s equation to the stream of
water jetting from the first hole. This stream
ultimately begins within the tank, at the surface level H where the pressure is equal to the
atmospheric pressure and the speed of water
flow is essentially zero. When this stream
emerges from the hole, the pressure is again
equal to the atmospheric pressure but the
height h1 is lower than H and the stream’s
velocity is v1 6= 0. Applying Bernoulli’s equation
1
P + ρ g h + ρ v 2 = const
2
4
to this situation, we obtain
p
v1 = 2 g (H − h1 ) .
This formula is known as Torricelli’s theorem.
Now consider what happens to the water after
it emerges from the hole with velocity v1 in the
horizontal direction. The Bernoulli equation
tells us the speed of the water at every point of
its trajectory, but it has nothing to say about
the direction of the flow. Instead, we use the
fact that the water droplets comprising the jet
exert no net force on each other (the pressure
is constant everywhere outside the tank), so
each droplet is essentially in a free fall. A
drop falling to the floor from height h1 with
no initial vertical component of the velocity
(the jet emerges s
horizontally from the hole)
2 h1
to fall, during which
g
time it flies through a horizontal distance
p
L1 = v1 t1 = 4 (H − h1 ) h1 .
takes time t1 =
In exactly the same way, the jet emerging from
the second
p hole has horizontal initial velocity
v2 = 2 g (H − h2 ) and hits the floor at the
distance
p
L2 = 4 (H − h2 ) h2 .
If the two jets hit the floor at the same spot,
it means that L2 = L1 , L22 = L21 or
0 = L22 − L21
= 4 (H − h2 ) h2 − 4 (H − h1 ) h1
= 4 H h2 − 4 h22 − 4 H h1 + 4 h21
= 4 (h2 − h1 ) (H − h2 − h1 ),
and since h2 − h1 6= 0, we should have H −
h2 − h1 = 0 or
H = h1 + h2 = 137 cm.
And given the above result, we immediately
find that the two jets hit the floor at the
distance
p
L1 = L2 = 4 h1 h2 = 128.686 cm
from the tank.
Answer, Key – Homework 2 – David McIntyre
012 (part 1 of 3) 4 points
A siphon of circular cross section is used to
drain water from a tank as shown below. The
siphon has a uniform diameter of 3.09 cm and
the surface of the reservoir is at a height of
h1 = 2.19 m (see figure).
Assume: A steady flow, no friction for the
water, and that the water cannot sustain a
negative pressure.
Note: Density of water is 103 kg/m3 .
h2
5
013 (part 2 of 3) 3 points
Assume: The reservoir is very large so that
h1 barely changes.
Note: 1 liter = 10−3 m3 .
How long would it take to siphon 486 L of
water from the reservoir?
Correct answer: 1.64865 min.
Explanation:
Since the reservoir is assumed to be large,
we assume that the height h1 will not change
appreciably; i.e., v is constant. In a time t,
the volume of water pumped V will be
V = Avt,
where A = cross sectional area of the siphon.
h1
V
d2
v
π
4
4V
=
π d2 v
4 (486 L) (0.001 m3 /L)
=
π [(3.09 cm) (0.01 m/cm)]2
(0.0166667 min/s)
×
(6.55164 m/s)
= 1.64865 min .
t=
What is the speed of the flow at the end of
the siphon?
Correct answer: 6.55164 m/s.
Explanation:
Basic concept: Bernoulli’s equation
P+
1 2
ρ v + ρ g y = constant
2
along a streamline.
Solution: Define the potential energy to
be zero at the outside end of the siphon.
Bernoulli’s equation says that along a streamline,
P+
ρ v2
+ ρ g y = constant .
2
So at the surface of the reservoir and at the
end of the siphon, these sums have the same
value
Patm + 0 + ρ g h1 = Patm +
1 2
ρv + 0.
2
So the velocity at the end of the siphon is
p
v = 2 g h1
q
= 2(9.8 m/s2 )(2.19 m)
= 6.55164 m/s .
014 (part 3 of 3) 3 points
Note: Instead of water, suppose the tank were
filed with a liquid with density 1400 kg/m3 .
What is the maximum value h2 can have in
order for the siphon to still work?
Hint: We note that the velocity v in the
tube is a constant since the cross sectional
area A is a constant.
Correct answer: 5.19338 m.
Explanation:
We note that in the siphon tube liquid flows
at constant speed and Q = v A is conserved.
Now equating the expression for Bernoulli’s
equation at the end of the siphon to that at
the height h2 , we have
Patm +
1
1 2
ρ v + 0 = Pht + ρ v 2 + ρ g (h1 + h2 ) ,
2
2
Answer, Key – Homework 2 – David McIntyre
where Pht is the pressure at height h2 + h1
and Pht ≥ 0 (liquid cannot sustain a negative
pressure).
Note: If a negative pressure occurs on the
discharge side of the siphon, the velocity in
the discharge side of the siphon will increase.
Neglecting viscous considerations, the siphon
will not work. Thus, (h1 + h2 ) (instead of
simply h2 ) must be included on the righthand side of Bernoulli’s equation.
⇒
⇒
Patm − ρ g (h2 + h1 ) = Pht ≥ 0
ρ g (h2 + h1 ) ≤ Patm
Patm
h2 ≤
− h1 .
ρg
The limiting height is
Patm
− h1
ρg
(101300 Pa)
− 2.19 m
=
(1400 kg/m3 )(9.8 m/s2 )
= 5.19338 m .
h2 =
6