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Problem 1.1 The value of is 3.14159265. . . If C is
the circumference of a circle and r is its radius, determine the value of r/C to four significant digits.

Problem 1.2 The base of natural logarithms is e D
2.718281828 . . .

Solution: C D 2 r )

r
1
D
D 0.159154943.
C
2

To four significant digits we have

r
D 0.1592
C

Solution: The value of e is: e D 2.718281828
(a)(a)
(b)
(c)

Express e to five significant digits.
Determine the value of e2 to five significant digits.
Use the value of e you obtained in part (a) to determine the value of e2 to five significant digits.

To five significant figures e D 2.7183

(b)

e2 to five significant figures is e2 D 7.3891

(c)

Using the value from part (a) we find e2 D 7.3892 which is
not correct in the fifth digit.[Part (c) demonstrates the hazard of using rounded-off
values in calculations.]

Problem 1.3 A machinist drills a circular hole in a
panel with a nominal radius r D 5 mm. The actual radius
of the hole is in the range r D 5 š 0.01 mm. (a) To what
number of significant digits can you express the radius?
(b) To what number of significant digits can you express
the area of the hole?

Solution:a)

The radius is in the range r1 D 4.99 mm to r2 D 5.01 mm. These
numbers are not equal at the level of three significant digits, but
they are equal if they are rounded off to two significant digits.
Two: r D 5.0 mm

b)

The area of the hole is in the range from A1 D r1 2 D 78.226 m2
to A2 D r2 2 D 78.854 m2 . These numbers are equal only if rounded
to one significant digit:
One: A D 80mm2

Problem 1.4 The opening in the soccer goal is 24 ft
wide and 8 ft high, so its area is 24 ft ð 8 ft D 192 ft2 .
What is its area in m2 to three significant digits?
Solution:
A D 192 ft2

1m
3.281 ft

2

D 17.8 m2

A D 17.8 m2

Problem 1.5 The Burj Dubai, scheduled for completion in 2008, will be the world’s tallest building with a
height of 705 m. The area of its groundfootprint will be
8000 m2 . Convert its height and footprint area to U.S.
customary units to three significant digits.

Solution:
h D 705 m
A D 8000 m2

3.281 ft
1m

D 2.31 ð 103 ft

3.218 ft
1m

h D 2.31 ð 103 ft,

2

D 8.61 ð 104 ft2

A D 8.61 ð 104 ft2

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1

Problem 1.6 Suppose that you have just purchased
a Ferrari F355 coupe and you want to know whether
you can use your set of SAE (U.S. Customary Units)
wrenches to work on it. You have wrenches with widths
w D 1/4 in, 1/2 in, 3/4in, and 1 in, and the car has nuts
with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,
and 25 mm. Defining a wrench to fit if w is no more
than 2% larger than n, which of your wrenches can you
use?

Solution: Convert the metric size n to inches, and compute the
percentage difference between the metric sized nut and the SAE
wrench. The results are:
5 mm

1 inch
25.4 mm

D 0.19685.. in,0.19685 0.25
0.19685
D

100

27.0%

10 mm

1 inch
25.4 mm

D 0.3937.. in,

0.3937 0.5
0.3937

100 D

15 mm

1 inch
25.4 mm

D 0.5905.. in,

0.5905 0.5
0.5905

100 D C15.3%

20 mm

1 inch
25.4 mm

D 0.7874.. in,

0.7874 0.75
0.7874

100 D C4.7%

25 mm

1 inch
25.4 mm

D 0.9843.. in,

0.9843 1.0
0.9843

27.0%

n

100 D

1.6%

A negativepercentage implies that the metric nut is smaller than the
SAE wrench; a positive percentage means that the nut is larger then
the wrench. Thus within the definition of the 2% fit, the 1 in wrench
will fit the 25 mm nut. The other wrenches cannot be used.

Problem 1.7 Suppose that the height of Mt. Everest is
known to be between 29,032 ft and 29,034 ft. Based on
this information, to how many...