7. The spring of the pressure gauge shown in Figure P14.7 has a force constant of
1250 N/m, and the piston has a diameter of 1.20 cm. As the gauge is lowered into
water in a lake, what change in depth causes the piston to move in by 0.750 cm?
Figure P14.7
SOL. Assuming the spring obeys Hooke’s law, the increase in force on the piston
required to compress the spring an additional amount Δx is
ΔF = F − F0 = (P − P0)A = k(Δx)
The gauge pressure at depth h beneath the surface of a fluid is
P − P0 = ρgh
so we have
ρghA = k (Δx)
or the required depth is
h = k (Δx) ρgA
If k = 1250 N/m, A = πd2/4, d = 1.20 × 10−2 m, and the fluid is water (ρ =
1.00 × 103 kg/m3), the depth required to compress the spring an additional
Δx = 0.750 × 10−2 m is h = 8.46 m
h=8.46 m
22. Mercury is poured into a U-tube as shown in Figure P14.22a. The left arm of the
tube has cross-sectional area A1 of 10.0 cm2, and the right arm has a cross-sectional
area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm
as shown in Figure P14.22b. (a) Determine the length of the water column in the
right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what
distance h does the mercury rise in the left arm?
Figure P14.22
SOL. (a)
Using the definition of density, we have
h 

mwater
A2  water
100 g
 5.00 cm2 1.00 g/cm3 
 20.0 cm
ANS. FIG. P14.
(b)
ANS. FIG. P14.22 (b) represents the situation after the water is
added. A volume (A2h2) of mercury has been displaced by water in
the right tube. The additional volume of mercury now in the left
tube is A1h. Since the total volume of mercury has not changed,
A2 h2  A1h or h2 
A1
h
A2
[1]
At the level of the mercury–water interface in the right tube, we
may write the absolute pressure as:
P = P0 + ρwaterghω
The pressure at this same level in the left tube is given by
P = P0 + ρHgg(h + h2) = P0 + ρwaterghω
which, using equation [1] above, reduces to

 Hg h 1 

or
h
A1 
   water h
A2 
 water h
 Hg (1  A1 / A2 )
.
Thus, the level of mercury has risen a distance of
1.00 g/cm   20.0 cm 
h
13.6 g/cm  1  10.0 / 5.00 
3
3
h  0490 cm above the original level.
26. The gravitational force exerted on a solid object is 5.00 N. When the object is
suspended from a spring scale and submerged in water, the scale reads 3.50 N (Fig.
P14.26). Find the density of the object.
Figure P14.26
SOL. Refer to Figure P14.26. We observe from the left-hand diagram,
 Fy  0  T1  Fg  mobject g  object gVobject
and from the right-hand diagram,
 Fy  0  T2  B  Fg  T2  B  T1
which gives
T2 −T1 = B
where the buoyant force is
B = mwater g = ρwVobject g
Now the density of the object is
object 
object 
mobject
Vobject
T1
T1  T2

T1 g T1
B   g  B
1000 kg

m 3   5.00N 
1.50 N
 3.33 103 kg m 3
36. A hydrometer is an instrument used to determine liquid density. A simple one is
sketched in Figure P14.36. The bulb of a syringe is squeezed and released to let the
atmosphere lift a sample of the liquid of interest into a tube containing a calibrated
rod of known density. The rod, of length L and average density ρ0, floats partially
immersed in the liquid of density ρ. A length h of the rod protrudes above the
surface of the liquid. Show that the density of the liquid is given by

0 L
Lh
Figure P14.36
SOL. Let A represent the horizontal cross-sectional area of the rod, which we presume
to be constant. The rod is in equilibrium:
 Fy  0 :  mg  B  0  0Vwhole rod g  fluidVimmersal g
0 ALg   A  L  h  g
The density of the liquid is  
0 L
Lh
42. Water moves through a constricted pipe in steady, ideal flow. At the lower point
shown in Figure P14.42, the pressure is P1 = 1.75 × 104 Pa and the pipe diameter is
6.00 cm. At another point y = 0.250 m higher, the pressure is P2 = 1.20 × 104 Pa and
the pipe diameter is 3.00 cm. Find the speed of flow (a) in the lower section and (b)
in the upper section. (c) Find the volume flow rate through the pipe.
Figure P14.42
SOL. (a)
The mass flow rate and the volume flow rate are constant:
 A11   A22 121   r222
Substituting,
(3.00 cm)2υ1 = (1.50 cm)2 υ2 → υ2 = 4υ1
For ideal flow,
1
1
P1   gy1  12  P2   gy2  22
2
2
1
2
1.75 104 Pa + 0 + 1000 kg m3  1 
2
 1.20 104 Pa + 1000  9.8  0.250  Pa

Solving for υ1 gives 1 
(b)
From part (a), we have
(c)
The volume flow rate is
1
2
1000 kg m3   41 

2
3050 Pa
 0.638 m s
7 500 kg m3
υ2 = 4υ1 = 2.55m/s
 r121    0.030 m   0.638m s   1.80 103 m3 s
2
49. The Venturi tube discussed in Example 14.8 and shown in Figure P14.49 may be
used as a fluid flowmeter. Suppose the device is used at a service station to measure
the flow rate of gasoline (ρ = 7.00 × 102 kg/m3) through a hose having an outlet
radius of 1.20 cm. If the difference in pressure is measured to be P1 − P2 = 1.20 k Pa
and the radius of the inlet tube to the meter is 2.40 cm, find (a) the speed of the
gasoline as it leaves the hose and (b) the fluid flow rate in cubic meters per second.
Figure P14.49
SOL. (a)
Since the tube is horizontal, y1 = y2 and the gravity terms in Bernoulli’s
equation cancel, leaving
1
1
P1  12  P2  22
2
2
or
  
2
2
2
1
2  P1  P2 


2 1.20 103 Pa 
7.00 102 kg m3
and
1
22  12  3.43m2 s2
ANS. FIG. P14.49
From the continuity equation, A1υ1 = A2υ2, we find
2
2
A 
r 
 2.40 cm 
2   1 1   1  1  
 1
 1.20 cm 
 A2 
 r1 
or
υ2 = 4υ1
[2]
Substituting equation [2] into [1] yields 1512  3.43m2 s2 and
υ1 = 0.478 m/s
Then, equation [2] gives
υ2 = 4(0.478 m/s) = 1.91 m/s
(b)
The volume flow rate is
A11  A22   r22 2   1.20 102 m  1.91m s 
2
 8.64 104 m3 s
53. A siphon is used to drain water from a tank as illustrated in Figure P14.53. Assume
steady flow without friction. (a) If h = 1.00 m, find the speed of outflow at the end
of the siphon. (b) What If? What is the limitation on the height of the top of the
siphon above the end of the siphon? Note: For the flow of the liquid to be
continuous, its pressure must not drop below its vapor pressure. Assume the water
is at 20.0°C, at which the vapor pressure is 2.3 k Pa.
Figure P14.53
SOL. (a)
We use Bernoulli’s equation,
1
P0   gh  0  P0  0  32
2
which gives. 3  2gh .
If h = 1.00 m, then υ3 = 4.43 m/s.
ANS. FIG. P14.53
(b)
Again, from Bernoulli’s equation,
1
1
P   gy  22  P0  0  32
2
2
Since υ2 = v3,
P = P0 − ρgy
Since P ≥ 2.3 kPa, the greatest possible siphon height is given by
y
P0  P 1.013105 Pa  2.30 103 Pa

 10.1m
g
103 kg m3 9.80 m s2 
61. Figure P14.61 shows a valve separating a reservoir from a water tank. If this valve
is opened, what is the maximum height above point B attained by the water stream
coming out of the right side of the tank? Assume h = 10.0 m, L = 2.00 m, and θ =
30.0°, and assume the cross-sectional area at A is very large compared with that at
B.
Figure P14.61
SOL. Consider the diagram in ANS. FIG. P14.61 and apply Bernoulli’s equation to
points A and B, taking y = 0 at the level of point B, and recognizing that vA
is approximately zero. This gives:
PA 
1
1
2
 w  0    w g  h  L sin    PB   w B2   w g  0 
2
2
Now, recognize that PA = PB = Patmosphere since both points are open to the
atmosphere (neglecting variation of atmospheric pressure with altitude).
Thus, we obtain
B  2 g  h  L sin  
Now the problem reduces to one of projectile motion with υyi = υB sin θ.
Then using,  yf2   yi2  2a  y  where y = ymax, υyf=0, and a = –g, we find
y 
0   yi2
2a
2
 B2 sin 2   2 g  h  L sin    sin 


2g 
2g
y   h  L sin   sin 2 
y  10.0 m  (2.00 m) sin 30.0 sin 2 30.0
ymax  2.25 m  about the level where the water emerges 
ANS. FIG. P14.61
71.A 1.00-kg beaker containing 2.00 kg of oil (density = 916.0 kg/m3) rests on a scale.
A 2.00-kg block of iron suspended from a spring scale is completely submerged in
the oil as shown in Figure P14.71. Determine the equilibrium readings of both
scales.
Figure P14.71
SOL.
Looking first at the top scale and the iron block, we have
T+B = Fg,iron
where T is the tension in the spring scale, B is the buoyant force, and Fg,iron
is the weight of the iron block. Now if miron is the mass of the iron block,
we have
miron = ρironV
so V 
miron
Then,
B = ρoilVirong
iron
 Vdisplaced oil
Therefore,
T  Fg,iron  oilViron g  miron g   oil
miron
iron
or


T  1  oil
 iron

 miron g


916 kg/m3 
 1 
(2.00 kg)  9.80 m/s 2 
3 
 7860 kg/m 
 17.3 N
Next, we look at the bottom scale which reads n (i.e., exerts an upward
normal force n on the system). Consider the external vertical forces acting
on the beaker-oil-iron combination.
F
y
 0gives
T  n  Fg,beaker  Fg,oil  Fg,iron  0 


n   mbeaker+ moil+ miron  g  T   5.00 kg  9.80 m s  17.3 N
Thus, n = 31.7 N is the lower scale reading.
2
81. A U-tube open at both ends is partially filled with water (Fig. P14.81a). Oil having
a density 750 kg/m3 is then poured into the right arm and forms a column L = 5.00
cm high (Fig. P14.81b). (a) Determine the difference h in the heights of the two
liquid surfaces. (b) The right arm is then shielded from any air motion while air is
blown across the top of the left arm until the surfaces of the two liquids are at the
same height (Fig. P14.81c). Determine the speed of the air being blown across the
left arm. Take the density of air as constant at 1.20 kg/m3.
Figure P14.81
SOL.
(a)
Consider the pressure at points A and B in ANS. FIG. P14.81(b).
Using the left tube: PA=Patm+ρwg(L−h)
Using the right tube: PB=Patm+ρogL
But Pascal’s principle says that PA = PB.
Therefore,Patm+ρwg(L−h)=Patm+ρogL
or ρwh=(ρw−ρo)L, giving
   o 
h w
L

w


 1000 kg m3  750 kg m 3 

  5.00 cm 
3
1000
kg
m


 1.25cm
ANS. FIG. P14.81
(b)
Consider part (c) of the diagram showing the situation when the air
flow over the left tube equalizes the fluid levels in the two tubes.
First, apply Bernoulli’s equation to points A and B (yA=yB, υA=υ,
and υB=0).
This gives:
PA 
1
 a 2   a gy A
2
1
2
 PB   a  0    a gyB
2
and since yA = yB, this reduces to
PB  PA 
1
 a 2
2
1
Now consider points C and D, both at the level of the oil-water
interface in the right tube. Using the variation of pressure with
depth in static fluids, we have
PC = PA + ρagH + ρwgL
and PD = PB + ρagH + ρogL
But Pascal’s principle says that PC = PD. Equating these two gives:
PB+ρagH+ρogL=PA+ρagH+ρwgL
or
PB – PA = (ρw – ρo) gL
[2]
Substitute equation [1] for PB − PA into [2] to obtain
1
a 2    w   o  gL
2
or

2 gL   w  o 
a
 1000 kg m3  750 kg m3 
 2 9.80 m s  0.050 0 m  

3
1.20 kg m



2

  14.3m s
84. A jet of water squirts out horizontally from a hole near the bottom of the tank
shown in Figure P14.84. If the hole has a diameter of 3.50 mm, what is the height h
of the water level in the tank?
Figure P14.84
SOL. First, consider the path from the viewpoint of projectile motion to find the speed
1
at which the water emerges from the tank. From y   yi t  a y t 2 with
2
vyi = 0, Δy = −1.00 m, and ay = −g, we find the time of flight as
t
2  y 
2.00 m

 0.452s
ay
g
From the horizontal motion, the speed of the water coming out of the hole
is
2   xi 
x 0.600 m

 1.33m s
t
0.452s
We now use Bernoulli’s equation, with point 1 at the top of the tank and
point 2 at the level of the hole. With P1 = P1 = Patm and v1 ≈ 0, this gives
1
2
 gy1   gy2  22
or
h  y1  y2 
22
2g
1.33m s 

2g
2
 9.00  102 m = 9.00 cm
85. An ice cube whose edges measure 20.0 mm is floating in a glass of ice-cold water,
and one of the ice cube’s faces is parallel to the water’s surface. (a) How far below
the water surface is the bottom face of the block? (b) Ice-cold ethyl alcohol is gently
poured onto the water surface to form a layer 5.00 mm thick above the water. The
alcohol does not mix with the water. When the ice cube again attains hydrostatic
equilibrium, what is the distance from the top of the water to the bottom face of the
block? (c) Additional cold ethyl alcohol is poured onto the water’s surface until the
top surface of the alcohol coincides with the top surface of the ice cube (in
hydrostatic equilibrium). How thick is the required layer of ethyl alcohol?
SOL. Let s stand for the edge of the cube, h for the depth of immersion, ρice for the
density of the ice, ρw for the density of water, and ρal for the density of the
alcohol.
(a)
According to Archimedes’s principle, at equilibrium we have
B = Fg
，
 W ghs 2  ice gs 3  h  s
ice
W
With ρice = 0.917 × 103 kg/m3,
ρw = 1.00 × 103 kg/m3 and s = 20.0 mm,
we obtain h = 20.0(0.917) = 18.34 mm ≈ 18.3 mm
(b)
We assume that the top of the cube is still above the alcohol
surface. Letting hal stand for the thickness of the alcohol layer, we
have
 ice   al
s

 W   W
al gs 2 hal   W gs 2 hW  ice gs 3 so hW  

 hal .

With ρal = 0.806×103 kg/m3
and hal = 5.00 mm,
we obtain hw = 18.34−0.806(5.00) = 14.31 mm ≈ 14.3 mm.
To check our assumption above, the bottom of the cube is below
the top surface of the alcohol 14.4 mm+5.00 mm = 19.3 mm, so the
top of the cube is above the surface of the alcohol 20.0 mm–19.3
mm = 0.7 mm. The assumption was valid.
  s  hal , so Archimedes’s principle gives
Here, hW
(c)
al gs 2 hal   W gs 2  s  hal   ice gs 3  al hal   w  s  hal   ice s
hal  s
  W  ice   20.0 mm 1.000  0.917 


  W  al 
1.000  0.806 
 8.557  8.56 mm