BW #7 Ch 13: CQ 33-38, P. 60
Q13.33. Reason: Since the question talks about the “extra pressure” we will ignore the air pressure above the water;
you have an equal amount of air pressure on the inside. The 7 N quoted is the “increased pressure.” Therefore the equation
we need is Equation 13.5 without the po term,
p =  gd
where we want to solve for d . We will also use p  F /A and round g to 10 m/s 2 for one significant figure accuracy.
d
p
F /A
7 N/(7 105 m2)


 10 m
 g  g (1000 kg/m3 )(10 m/s 2 )
The correct answer is D.
Assess: This one significant figure calculation can easily be done in the head without a calculator. The answer seems
plausible, and the other choices seem too small. The units cancel appropriately to leave the answer in m.
Q13.34. Reason: We can determine what will happen to the ball by determining its density and comparing it to the
density of water. The density of the ball may be determined by

M
M
(8 lb)(1 N / 0.2248 lb)


 840 kg / m3
V (4 r 3 / 3) (4 )[(8.5 in)(2.54 cm/in)(m / 100 cm)]3
Since the density of the ball is less than the density of water, it will float in water.
Assess: From the above calculation we not only know that the ball will float but also that it will float with 84% of its
volume below the surface.
Q13.35. Reason: We’ll assume the ball is not being forcibly held under the water (because then the net force would be
zero). A free-body diagram shows only two forces on the ball: the downward force of gravity (w  mb g ), and the upward
buoyant force, whose magnitude equals the weight of the displaced fluid (the weight of 8.0 103 m3 of water).
The downward gravitational force has a magnitude of w  mb g  (0.50 kg)(9.8 m/s 2 )  4.9 N.
The upward buoyant force has a magnitude of w  mw g  ( wVw) g  (1000 kg/m3)(8.0 103 m3)(9.8m/s 2)  78.4N.
Subtracting gives the magnitude of the net force, 78.4 N  4.9 N  73.5 N  74 N.
The correct answer is B.
Assess: This is enough force to cause the basketball to accelerate upward fairly impressively, as you have probably seen
in the swimming pool (if you haven’t, have your teacher assign you a field trip to the pool to try this out).
Q13.36. Reason: Since the object is floating, the buoyant force equals the weight of the water displaced. The volume of
water displaced is 75% of the volume of the object, so  gV  water g ((0.75)V ). So   (0.75) water  750 kg/m3. The correct
choice is B.
Assess: This makes sense. The density of the object must be less than the density of water in order for the object to float.
Q13.37. Reason: We’ll use the equation of continuity, Equation 13.12, and make all the assumptions inherent in that
equation (no leakage, etc.). We’ll also use A   r 2. Call the plunger point 1 and the nozzle point 2, and solve for v1.
v1 
v2 A2 v2 r 22 v2r 22 (10 m/s)(1.0 103 m) 2

 2 
 0.10 m/s
A1
 r 21
r1
(1.0 102 m) 2
The correct answer is B.
Assess: The answer seems to be a reasonable slow and steady pace. The ratio of the speeds is the square of the ratio of the
radii. The last step (the computation) can easily be done without a calculator.
Q13.38. Reason: The question says to ignore viscosity, so we do not need Poiseuille’s equation; Bernoulli’s equation
should suffice. And because the pipe is horizontal we can drop the  gy terms (because they will be the same on both
sides).
p  p2  p1 

1
 v12  v22
2

We are given r1  0.040 m, r2  0.02 m, and v1  1.3 m/s.
We will also use the equation of continuity to solve for v2.
v2 
A1
r2
(0.040 m) 2
v1  12 v1 
(1.3 m/s)  5.2 m/s
A2
r2
(0.020 m) 2
Putting it all together
p  p2  p1 


1
1
 v12  v22  (1000 kg/m3 )[(1.3 m/s) 2  (5.2 m/s) 2]  12,700 Pa
2
2
The magnitude of this is 12,700 Pa, so the correct answer is D.
Assess: Since Pa  N/m2 the units work out.
P13.60. Prepare: Treat the water as an ideal fluid obeying Bernoulli’s equation. There is a streamline connecting point
1 in the wider pipe with point 2 in the narrower pipe.
Solve: Bernoulli’s equation, Equation 13.14, relates the pressure, water speed, and heights at points 1 and 2.
p1 
1 2
1
 v1   gy1  p2   v 22   gy2
2
2
For no height change y1  y2. The flow rate is given as 5.0 L/s, which means
Q  v1 A1  v2 A2  5.0  103 m3/s  v1 
5.0  103 m3/s
 0.6366 m/s
 (0.05 m) 2
2
 0.050 m 
A
 v2  v1 1   0.6366 m/s  
  2.546 m/s
A2
 0.025 m 
Bernoulli’s equation now simplifies to
1 2
1
 v1  p2   v 22
2
2
1
1
 p1  p2   v 22  v12  50  103 Pa  (1000 kg/m3 )[(2.546 m/s) 2  (0.6366 m/s) 2 ]  53 kPa
2
2
Assess: Reducing the pipe size reduces the pressure because it makes v2 > v1.
p1 

