PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HOMEWORK PROBLEMS Chapter 9:
SOLIDS AND FLUIDS
HW-9
PART-A: Hand in your answers in class on scantron on Monday …. November-2010. Write your
name, class (1401) and HW # 9 on the scantron.
1. What is the mass of a solid gold rectangular bar that has dimensions of 4.50 cm x 11.0 cm x
26.0 cm?
(a) 24.8 kg
(b) 45.6 kg
(c) 11.4 kg
(d) 33.2 kg
(e) 19.5 kg

m  gold V  19.3  103 kg m 3
  4.50  10 2 m  11.0  10 2 m   26.0  10 2 m   24.8 kg
2. A 66.0-kg man lies on his back on a bed of nails, with 1 208 of the nails in contact with his
body. The end of each nail has area 1.00 x 10−6 m2. What average pressure is exerted by one nail
on the man’s body?
(a) 2.21 x 105 Pa
(b) 3.09 x 105 Pa
(c) 1.65 x 106 Pa
(d) 5.35 x 105 Pa
4
(e) 4.11 x 10 Pa
On average, the support force each nail exerts on the body is
 66.0 kg  9.80 m s2
mg
F1 

 0.535 N
1 208
1 208
so the average pressure exerted on the body by each nail is
F1
0.535 N
Pav 

 5.35  10 5 Pa
Anail
1.00  10 6 m 2


end
3. A hydraulic jack has an input piston of area 0.050 m2 and an output piston of area 0.70 m2.
How much force on the input piston is required to lift a car weighing 1.2 x 103 N?
(a) 42 N
(b) 68 N
(c) 86 N
(d) 110 N
(e) 130 N
From Pascal’s principle, F1 A1  F2 A2 , so if the output force is to be F2  1.2  103 N , the required
input force is
A 
 0.050 m 2 
F1   1  F2  
1.2  103 N  86 N
 0.70 m 2 
 A2 


4. A lead bullet is placed in a pool of mercury. What fractional part of the volume of the bullet is
submerged?
(a) 0.455
(b) 0.622
(c) 0.714
(d) 0.831
(e) 0.930
According to Archimedes’s principle, the buoyant force exerted on the bullet by the mercury is equal to the
weight of a volume of mercury that is the same as the submerged volume of the bullet. If the bullet is to
float, this buoyant force must equal the total weight of the bullet. Thus, for a floating bullet,
mercury Vsubmerged g  lead Vbullet g
and
Vsubmerged
Vbullet

lead
mercury

11.3  103 kg m3
 0.831
13.6  103 kg m3
1
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-9
5. What is the pressure at the bottom of Loch Ness, which is as much as 754 ft deep? (The
surface of the lake is only 15.8 m above sea level; hence, the pressure there can be taken to be
1.013 x 105 Pa.)
(a) 1.52 X 105 Pa
(b) 2.74 X 105 Pa
(c) 2.35 X 106 Pa
(d) 7.01 X 105 Pa
5
(e) 3.15 X 10 Pa
The absolute pressure at depth h below the surface of a liquid with density  , and with pressure P0 at its
surface, is P  P0   gh . Thus, at a depth of 754 ft in the waters of Loch Ness,

P  1.013  105 Pa  1.00  103 kg m3
  9.80

 1m 
m s2   754 ft  
 2.35  106 Pa
 3.281 ft  


6. Hurricane winds of 95 mi/h are blowing over the flat roof of a well-sealed house. What is the
difference in air pressure between the inside and outside of the house?
(a) 1.2 X 103 Pa
(b) 2.4 X 104 Pa
(c) 3.4 X 103 Pa
(d) 4.0 X 103 Pa
(e) 5.3 X 104 Pa
We assume that the air inside the well-sealed house has essentially zero speed and the thickness of the roof
is negligible so the air just above the roof and that just below the roof is at the same altitude. Then,
Bernoulli’s equation gives the difference in pressure just below and just above the roof (with the pressure
below being the greatest) as
1
P1  P2  air v 22  v12  air g  y2  y1 
2
or
2




1
1


3
P 
1.29 kg m    95 mi h  

0
  0  1.2  103 Pa


2
2.237
mi
h











7. A horizontal pipe narrows from a radius of 0.250 m to 0.100 m. If the speed of the water in the
pipe is 1.00 m/s in the larger-radius pipe, what is the speed in the smaller pipe?
(a) 4.50 m/s (b) 2.50 m/s (c) 3.75 m/s (d) 6.25 m/s (e) 5.13 m/s
From the equation of continuity, A1v1  A2 v 2 , the speed of the water in the smaller pipe is
   0.250 m  2
 A 
v 2   1  v1  
   0.100 m  2
 A2 


  1.00 m s   6.25 m s


.
8. Bernoulli’s equation can be used to explain, in part, which of the following phenomena?
(a) the lift on an airplane wing in flight
(b) the curve of a spinning baseball
(c) vascular flutter
(d) reduction in pressure of moving fluids
(e) all these answers
All of these phenomena are the result of a difference in pressure on opposite sides of an object due to a
fluid moving at different speeds on the two sides. Thus, the correct response to this question is choice (e).
Bernoulli’s equation can be used in the discussion of each of these phenomena.
2
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-9
9. A boat develops a leak and, after its passengers are rescued, eventually sinks to the bottom of
a lake. When the boat is at the bottom, is the normal force on the boat
(a) greater than the weight of the boat,
(b) equal to the weight of the boat,
(c) less than the weight of the boat,
(d) equal to the weight of the displaced water, or
(e) equal to the buoyant force on the boat?
The boat, even after it sinks, experiences a buoyant force, B, equal to the weight of whatever water it is
displacing. This force will support part of the weight, w, of the boat. The normal force exerted on the boat
by the bottom of the lake will be n  w  B  w will support the balance of the boat’s weight. The
correct response is (c).
10. Three vessels of different shapes are filled to the same level with water as in Figure
MCQ9.10. The area of the base is the same for all three vessels. Which of the following
statements is valid?
(a) The pressure at the top surface of vessel A is greatest because it has the largest surface area.
(b) The pressure at the bottom of vessel A is greatest because it contains the most water.
(c) The pressure at the bottom of each vessel is the same.
(d) The force on the bottom of each vessel is not the same.
(e) At a given depth below the surface of each vessel, the pressure on the side of vessel A is
greatest because of its slope.
The absolute pressure at depth h below the surface of a fluid having density  is, P  P0   gh where
P0 is the pressure at the upper surface of that fluid. The fluid in each of the three vessels has density
  water , the top of each vessel is open to the atmosphere so that P0  Patmo in each case, and the
bottom is at the same depth h below the upper surface for the three vessels. Thus, the pressure P at the
bottom of each vessel is the same and (c) is the correct choice.
11. An ideal fluid flows through a horizontal pipe having a diameter that varies along its length.
Does the sum of the pressure and kinetic energy per unit volume at different sections of the pipe
(a) decrease as the pipe diameter increases, (b) increase as the pipe diameter increases,
(c) increase as the pipe diameter decreases, (d) decrease as the pipe diameter decreases, or
(e) remain the same as the pipe diameter changes.
Since the pipe is horizontal, each part of it is at the same vertical level or has the same y-coordinate. Thus,
from Bernoulli’s equation (P  1 v 2   gy  constant) , we see that the sum of the pressure and the
2
kinetic energy per unit volume (P 
1
2
v 2 ) must also be constant throughout the pipe, making (e) the
correct choice.
3
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-9
12. A hose is pointed straight up, with water flowing from it at a steady volume flow rate and
reaching a maximum height of h. Neglecting air resistance, which of the following adjustments
to the nozzle will result in the water reaching a height of 4h?
(a) Decrease the area of the opening by a factor of 16.
(b) Decrease the area by a factor of 8.
(c) Decrease the area by a factor of 4.
(d) Decrease the area by a factor of 2.
(e) Give up because the water cannot reach a height of 4h.
Once the water droplets leave the nozzle, they are projectiles with initial speed v0y  vi and having
speed v f  v y  0 at their maximum altitude, h. From the kinematics equation v 2y  v 02 y  2ay (y) the
maximum height reached is h  v i2 2 g . Thus, if we want to quadruple the maximum height (h  4h)
we need to double the speed of the water leaving the nozzle (vi  2vi ) . Using the equation of continuity,
Av i  Avi , it is seen that if I = 2i, it is necessary to have Ai  (vi vi) A  A 2 This says that the
area needs to be decreased by a factor of 2, and the correct choice is (d).
13. When water freezes, it expands about 9.00%. What would be the pressure increase inside
your automobile engine block if the water in it froze? The bulk modulus of ice is 2.00 X 109
N/m2.
a. 1.65 x 107 Pa
b. 1.65 x 108Pa
c. 1.65 x 105 Pa
d. 1.65 x 105 Pa
As a liquid, the water occupied some volume Vl . As ice, the water would occupy volume 1.090Vl if it were
not compressed and forced to occupy the original volume. Consider the pressure change required to
squeeze ice back into volume Vl . Then, V0  1.09Vl and V   0.090Vl , so
 V 
N   0.090 Vl 

P   B 
   2.00  109
 1.65  108 Pa  1600 atm


m2   1.09 Vl 
 V0 
14. A stainless-steel orthodontic wire is applied to a tooth, as in Figure P9.6. The wire has an
unstretched length of 3.1 cm and a diameter of 0.22 mm. If the wire is stretched 0.10 mm,
find the magnitude and direction of the force on the tooth. Disregard the width of the tooth
and assume Young’s modulus for stainless steel is 18 X 1010 Pa.
a.
b.
c.
d.
14 N
18 N
22 N
26 N
From Y  F L0 A ( L ) the tension needed to stretch the wire by 0.10 mm is
4
PHYS-1401: College Physics-I
F 

Y A  L 
L0

CRN 55178

Y  d2
Khalid Bukhari
HW-9
  L 
4 L0
18  1010 Pa   0.22  10 3 m 2  0.10  10 3 m 
4  3.1  10 2 m 
 22 N
The tension in the wire exerts a force of magnitude F on the tooth in each direction along the length of the wire
as shown in the above sketch. The resultant force exerted on the tooth has an x-component of
Rx  Fx   F cos 30 F cos 30 0 , and a y-component of
Ry  Fy  F sin 30º  F sin 30º  F  22 N .
Thus, the resultant force is
R  22 N directed down the page in the diagram .
15. A large man sits on a four-legged chair with his feet off the floor. The combined mass of
the man and chair is 95.0 kg. If the chair legs are circular and have a radius of 0.500 cm at
the bottom, what pressure does each leg exert on the floor?
a. 2.96 x 106 Pa
b. 3.96 x 106 Pa
c. 2.46 x 106 Pa
d. 2.16 x 106 Pa
We shall assume that each chair leg supports one-fourth of the total weight so the normal force each leg
exerts on the floor is n  mg 4 . The pressure of each leg on the floor is then
Pleg 


 95.0 kg  9.80 m s2
n
mg 4


 2.96  106 Pa
2
Aleg
r2

2
4 0.500  10 m


16. A hypodermic syringe contains a medicine with the density of water (Fig. P9.47). The
barrel of the syringe has a cross-sectional area of 2.50 X 10−5 m2. In the absence of a force on
the plunger, the pressure everywhere is 1.00 atm. A force of magnitude 2.00 N is exerted
on the plunger, making medicine squirt from the needle. Determine the medicine’s flow
speed through the needle. Assume the pressure in the needle remains equal to 1.00 atm and
that the syringe is horizontal.
a.
b.
c.
d.
11.6 m/s
12.6 m/s
13.6 m/s
14.6 m/s
From Bernoulli’s equation, choosing at y = 0 the level of the syringe and needle, so the flow speed in the
needle, P2  21 v 22  P1  21 v12 is
5
PHYS-1401: College Physics-I
v2 
v12 
CRN 55178
Khalid Bukhari
HW-9
2  P1  P2 

In this situation,
P1  P2  P1  Patmo   P1  gauge 
Thus, assuming   0,
v2 
0

2 8.00  10 4 Pa
1.00 
103

kg m 3
F
2 .00 N

 8.00  10 4 Pa
A1
2 .50  10 5 m 2
 12.6 m s
17. On October 21, 2001, Ian Ashpole of the United Kingdom achieved a record altitude of
3.35 km (11 000 ft) powered by 600 toy balloons filled with helium. Each filled balloon had
a radius of about 0.50 m and an estimated mass of 0.30 kg. Estimate the net upward force on
all 600 balloons.
a. 1.5 N
b. 1.8 N
c. 2.2 N
d. 2.7 N
9.37
(a)

 4 3  
 600  air gVballoon   600  air g 
r 
 3

balloon

4

 600  1.29 kg m 3 9.80 m s 2
 0.50 m 3   4.0  10 3 N  4.0 kN
3


3
Fy  Btotal  mtotal g  4.0  10 N  600  0.30 kg  9.8 m s 2  2.2  10 3 N  2.2 kN
Btotal  600  Bsingle

(b)



6
