56 Various FE Exam Solutions

SOLUTIONS

FOR

THE

EIT SIMULATED EXAM EfooEtfFgfEgff__fsg--gefgf--3_

1.

K

_

e-

[c] [n)2 [A]3

is the equilibrium constant for which of the reactions listed below? a. C + D2 b. C + 3D e. A3

A3

c. 3A

A

d.

3A

C + D3 C + 2D

C + D2

Answer (c) 2. The hydroxyl ion concentration of a solution with a pH of 4 is: a. 1 x 10 moles/L b. 1 x 10-10 moles/L c. 1 x 10 10 moles/L

d. 4 moles/L e. 4 x 10-10 moles/L

Answer (b) 3.

The atomic weight for Ag = 107.88, for N = 14.0, for H = 1, for 0 = 16, and for S = 32. The reaction of 1 lb-mole of AgN03 will produce: a. 311.76 lbs Ag2S04 b. 155.88 lbs Ag2S04 c. 1 Troy Oz Silver

d.73.94lb-mole Ba(N03h e.64.11 lbs barium

Answer(b) The sum of the molecular weights of AgS0 4 is divided by 2 because 1 lb-mole AgN03 will produce 0.5 lb-mole of Ag2S04.

4. The combination of an alkyl radical with a hydroxyl group forms: a. an acid b. an aldehyde

d. an alcohol e. a carboxyl

c. a ketone

Answer(d) 5. You are in charge of parts and must place the orders. You need 2000 parts per year and each part costs $25, and each time you order it costs you $50. Interest is 5%. How many times per year should you order parts? a. 1.25 b. 1.0 c. 11.0

d. 7 e. 3.5

Answer (d)

AM

SESSION

13-25

13-26

i#f--zE-ff-afeE_-ggp_s__f-#sg---d_g-Esffef_ FUNDAMENTALS OF ENGINEERING EXAM

Let x be the EOQ, then the total cost is: C = 2000(25)

+ 50

( 2 000)

+ .05

(25x +50)

X

Take the first derivative and equate to 0: 0 = _ 100,000+ 1.25x x2

x = 282 parts per order 2000/282 = 7,06 orders per year 6. Let S be the accumulated sum, P the principal, invested, i the effective interest rate per compounding period, and n the number of compounding periods. Which of the following formulas correctly relate these quantities? d. S = P(l +in)n-1 e. S = P(l +i)n-1

a. S = P(l+in) b. S = P(1+i)n c. S = P (I + n)i

Answer (b) 7. Equipment is purchased for $6000. In 5 years the Salvage is expected to be $500. What is the book value of the equipment at the end of 3 years using SLD? a. $1400 b. $2700 c. $4200

d.$3300 e. $5400

6ooo - [C6ooo -

2100

Answer (b) 8. With interest at 3.0% quarterly, $1000 will compound to how much after 5 years? a. $1203 b. $2048 c. $3698

d.$1806 e. $1492

From the Interest tables for a single payment and given P = 1000 to find F for n = 20 and i = 3, read 1.8061.

F is 1000 x 1.8061 = 1806.

Answer(d) 9. A facility costs $300,000. What will be the monthly payment for principal and interest only for each $100,000 borrowed at 6.0% for 5 years? a. $2760 b. $3800 e. $8867

d.$1930 e. $1330

Convert the interest rate to months: 6112 = .5% Calculate the periods as 5 x 12 = 60

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WORKBOOK

SOLUTIONS

FOR

AM THEt---gaggEEEafEE-_-dsg__fgfg_#gfEg_fEfe EIT SIMULATED EXAM

Given P = 100,000, find A. From the interest tables the capital recovery factor AlP is .0193 . .0193 x 100000 = $1930/mo Answer(d) 10. All of the following statements about work done on a process are true except: a. The work W depends on the process b. There can be an infinite number of specific heats between state points. c. If dQ amount of heat is transferred to W lb of a substance, the temperature change is dT. d. The change in internal energy is Q - W + LlPV e. LlU depends only on LlT. Answer(d) 11. For any ideal gas, the volume is doubled while the absolute temperature is halved. The pressure will be: a. doubled b. quadrupled c. quartered d. halved e. no change Since V2- 2VI Pz

P1 v1 2Vl

P1 2

Pz

p 1 T1 _ _2_ T1

pl

andT2=T1/2

=

Combining results in P2 = P1/4

2

Answer(c) 12. The net entropy change for an ideal gas that undergoes no temperature change while tripling its volume will be: a. LlS increases by a factor of 1.732 b. LlS decreases by a factor of 1.732 c. LlS increases by a factor of 3.0 d. LlS decreases by a factor of 3.0 e. LlS is constant Answer(c)

SESSION

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-#-g_Eff_f--f-f_fff-Eg_-ff__gcsE#-E___f-E--##th__-ff_apnf# FUNDAMENTALS OF ENGINEERING EXAM REVIEW

13-28

WORKBOOK

13. One of the following statements is true of the theoretical diesel cycle: a. It has two isentropic, one isobaric and one isochoric process. b. It has one polytropic, two isobaric and one adiabatic process. c. It is totally polytropic d. It has no net change on the entropy of the universe e. It burns gasohol.

Answer (a) 14. A liquid has begun to freeze. A salt is added to the liquid. When the salt is dissolved: a. The rate of solidification increases b. The rate of solidification becomes zero c. The liquid warms up d. Sublimation begins e. The salt settles to the bottom of the tank Answer(b) 15. What is the enthalpy of 5 lbs of a fluid occupying 10 ft3, with an internal energy of 1000 Btu!lb at a pressure of 1.5 atm? a. 5004.81 Btu

d. 2750.4 Btu e. 5000.28 Btu

b. 5040.81 Btu c. 36752 Btu u = 5 lbs x 1000 pv =

(1. 5 h

atm) (14. 7

ll..t.JJ.. = SOOOBtu lb

2)

lbs. atm 1n

in2) (

(144 ft 2

Btu ) (10 ft3) 778 ft-lbs

= u + pv = 5040.81 Btu Answer(b)

16. Air expands isentropically such that its pressure is increased by 50%. The initial temperature is 150 OF. What is the final temperature.

a. 168.42 OF b. 224.91 OF c. 179 OF

d. 610 OR e. 672.98 OR

k - 1

T2 = T1 (:: )-k(150 +460)

1.4- 1

4_ __ 1._

=

(684.91 -

460)

224.91.F

Answer (b)

40.81Btu

SOLUTIONS

FOR

THE

s-g-ET-fdgessg-t-f_gg-cf_df.EE#EEgg EIT SIMULATED EXAM

17. What is the equation that must be satisfied by the coordinates of every point (x,y) that is equidistant from the origin and the point (1,1)? a. 2x + 2y = 1 b. X - y = 1 X

c.

d. 2x + 2y 0 e. X + y = 1

+ y2 - 1

2

The locus of points (x,y) forms the perpendicular bisector of the point drawn between (0,0) and ( 1,1 ). This line must pass

G

through (x,y) and

1

y

l.

-1

rn

2

( 0' 0)

X

-1,

m;

1

X + y

1

Answer(e) 18. An elementary game is played by rolling a die and drawing a ball from a bag containing 3 white and 7 black balls. The player wins by rolling a number less than 4 and drawing a black ball from the bag. What is the probability of winning on the first try? a. 13/20 b. 112 c. 7/20

d. 12/20 e. 7/1

The probability of rolling a number less than 4 is that of rolling a 1, 2, or 3.

This is 1/2. The probability of drawing a black ball is 7/(7+3) or 7/10. The probability of the occurance of two independent events is the product of the probability of the two events or: 1 7 x10 2

=

7 20 Answer(c)

AM

SESSION

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t#fg-f_f#Tsc#_f_TE_ffg_f-_ FUNDAMENTALS OF ENGINEERING

13-30

EXAM

-3--3-3--333 REVIEW WORKBOOK

19. Find the general solution to the following differential equation: ydx-y2dy=dy

a. (x + y) 2 b.

= 4y

d. 2x

+ C

+ y = 2xy + C

X

c. 2x = y

2

+

lny2

e. y2

= yekt = x3 +

+ C C

+ C

Separate variables and integrate y dx =

x

( 1 + y2 ) dy

y +

+ c 2 2x = ln y2 + y2 + C = ln

Answer(c) 20. Let A = 2i - 4j + k, B = i +j - 3k, and C = -i +2j + 2k. (A x B) · (C x A) is ? a. -5 b. 1li + 7j + 6k c. 145

A

X

d. -13 e. 15

i 2 1

B

j -4 1

(12 i 11 i

C

+ 7j

X

A

10 i X

B)

-3

+ 2k)

-

(i - 6j - 4k)

+ 6k

i -1

j 2

2

-4

(2i + 4j + 4k)

(A

+ j

k 1

· (C

x

A)

k 2 1

(-81 -j +4k)

+ 5j

= (11

X

10) + (7

X

5) + (6

X

0)

145

Answer (c)

SOLUTIONS

21.

FOR

THE

EIT SIMULATED EXAM t-Eg_-E-ggesgfsg--E_E-Egsdg_---f_E-g_

X + 2

Iff(x) =

C x- 2

C

and C = constant What is f(x) ? a. X+ C b. 3C c.

d. X-C e. 1

c

Expand the determinant: f(x)

[ (x + 2)

•

(C)]

-

-2

[ (Cx

Cx + 2C + C

Differentiate f(x)

=C Answer (c)

22. Which of the following is not true? a. sin (x + y) =sin x cosy+ cos x sin y b. sin x I sin y = 2 sin 112 (x + y) I cos 112 (x - y) c. sin x +sin y = 2 sin 112 (x + y) cos 112 (x- y) d. sin x - sin y = 2 cos 1/2 (x + y) cos 1/2 (x- y) e. sin (x - y) = sin x cos y - cos x sin y Answer(b) 23. Ifthe function: y = 3x

2

is integrated between the limits of x = -3 and x = +3, the result is: a. 0 b. 108 c. 18

d. 54 e. 27

By the power rule for integration:

y =

3x313

6

-3

27 - (- 27) = 54

Answer(d)

AM

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13-32

-E-Egffpf-fg__f_fg#f_f-s__fEg#--hfffE_f3g__fFUNDAMENTALS OF ENGINEERING EXAM

24. The Laplace Transform of a step function of magnitude a is: a. I (s +a) b. a/s c. (s +a)/ a

d. s/a e. s +a

Answer (b) 25. For the circuit below, what is the voltage across the 50Q resistor?

20Q 20Q

lOOV

a. 28.6V b. 83.3V c. 63.9V

d. 75.4V e. 46.8V

The circuit simplifies to two resistors in series, the 50Q and the 2 parallel 20Q which are equivalent to lOQ. V = 100 = 50 I + 10 1: I = 1.166A Vso = IR50

= 1.166 x 50= 83.3V Answer (b)

26. A parallel plate capacitor with area A, separation distance D, and separator permitivity e has a capacitance that is: a. directly proportional to A and e, and inversely proportional to D b. directly proportional to D and e, and inversely proportional to A c. inversely proportional to A and D, but directly proportional to e d. inversely proportional to e and D, but directly proportional to A e. inversely proportional to A and e, but directly proportional to D Answer (a)

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WORKBOOK

SOLUTIONS

FOR

THE

sfdfdf-fg_ftff_T.gg#EgfEggEs_f3gfEIT SIMULATED EXAM

27. For the op amp circuit below, the voltage is:

c R

a.

Rt Vr

Vo

Ri

b.

Rt Vr _Rt v2

Vo

R1

c.

R2

Vo

jroL Vi Ri

Vo

j

d

roL Vi R (1 + jroRC

e. vo

_L

RC

f Vidt Answer (e)

28. For the circuit shown below, the average power in watts dissipated in the balanced 3 phase load is:

z = 10 + j5 = 120 V

VAC

a. 2507 watts b. 3456 watts c. 5272 watts Sph= tan

-1

d. 978 watts e. 1728 watts

_5_

10

120 120 Iph= !1o + j s!

26.6.

Vph=

10.73

Cos Gph 10.73 X .8942 = 3456w

PL = 3 Vph lph

=3 X

120

X

Answer (b)

AM

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s-Ef_E--gf#E--__#gff_#-Ecff__#Eg_---#eg_-EffE-Ef-nOF ENGINEERING EXAM REVIEW FUNDAMENTALS

13-34

29. The inductance of a coil of wire is calculated using one of the formulae below: a.

d L

N

b.

L

N

112

L

N2

N

A2

L

Jl A2 1

e.

1

c.

L

Jl A 1

N2

Jl A2 1

Jl A 1

Answer (c)

30. The switch in the circuit shown below has been closed for a long time. The expression to calculate L is:

d.

a. i

Io

e-RL/t

Io

eR/Lt

Io

e-t/RL

Io

e- (R/L) t

i

Io

et/'t

e.

b. i

i

c. i

Answer(d)

3l.The power factor for the circuit below is most nearly:

4Q

j3

a. 0.2 b. 0.4 c. 0.6

d. 0.8 e. 0.9

z

= 4 + j3 = 36.9° pf = cos

e

e

5 L36.9° cos (36.9°)

.79968

Answer(d)

WORKBOOK

SOLUTIONS

FOR

THE

SEN.gs#fgf-f_fffpftgffg--Egf-fgE EIT SIMULATED EXAM

32. If the load on a test material is carried beyond the ultimate tensile strength of the material, the material will: a. Return to its normal configuration b. Permanently maintain its deformed condition c. Continue bending until stress failure occurs d. Expand lengthwise. e. Continue bending until strain failure occurs Answer (c) 33. Wrought iron coated with magnesium is left to the elements. Over enough time what will happen? a. The iron and the magnesium will form an irreversible chemical bond b. Magnesium will act as an anode and preferentially corrode c. Magnesium will act as a cathode and preferentially corrode d. Iron will act as an anode and preferentially corrode e. Iron will act as a cathode and preferentially corrode Answer (b) 34. Given the temperature/composition diagram below, at what point will the components of the mixture be totally miscible?

c

B

Weight %

a. A b.B c. c

d.D e. E

Point A is the eutectic point.

Answer (a)

AM

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13-35

-0dg 13-36

FUNDAMENTALS

ENGINEERING EXAM OF-EffEsegf_-gpE8g__-g_ff--ago-tg_

35. Electrical and thermal conductivities for various materials differ relative magnitude and order. This phenomenon is due to: a. Electrons and heat travel at the same speed but encounter resistance differently. b. Heat and electrical energy are both wave forms that counteract each other's motion. c. Electrons and heat are conducted through metals by valence or free electrons and are influenced differently by lattice structure. d. Various materials have different densities. e. When heat and electrons move through a material, the material changes crystalline structure.

Answer (c) 36. A 25,000-lb truck is going up an incline of 4% and must maintain a speed of 40 mph. Neglecting friction, the engine must develop how much horsepower? d. 120hp e. 112hp

a. lOOhp b. 109hp c. 106hp

v

100

The power required is -F·V. As friction is neglected only the weight is considered. The angle between V and W is 90° + tan-1.04 = 92.29°.

- F · V

2500 (-=L) 550

X 5280)) 3600 - F · V = 106.4 hp

( (35

COS

92.29• Answer (c)

37. A vehicle is traveling at 50 mph on a level road and must come to a stop without its cargo sliding.

The coefficient of friction between the vehicle and the cargo is 0.6. What is the minimum amount of time required for vehicle to come to a complete stop. d. 2.8 sec a. 10 sec b. 5.2 sec e. 3.8 sec c. 7.6 sec

REVIEW

WORKBOOK

SOLUTIONS

FOR

EIT SIMULATED EXAM THE iT#-g-f_fE#_fE-g_-Esgg--EgfdffN-g_-

To avoid sliding ma/mg must be ::; 0.6. Thus llmax = 0.6 x 32.2 = 19.37 ft/sec2

V = 50 X 5280 = 73.33 sec 3600 t

73.33 19.32

3.798sec Answer (e)

38. Find the maximum height in meters.

H

vo = 50

X

m/s

a. 90m b. 96 m c. 101m H

20m - - L,-20

T sin2 e 2g

d. 110m e. 106m

50 2 2

X X

. 87 2 9.8

95. 67m Answer(b)

39. The total distance traveled by the particle from t = 4 tot= 8 is:

t

a. 4v b. 8v c. 10v

d.O e. -8v

The distance traveled is the area between the curve and the line v=O. Since the areas of the triangles above the curve between 4 and 6 is the same in magnitude but opposite in sign as the area between 6 and 8, the distance traveled is 0.

Answer(d)

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13-38

--Ef-f__gf--#s_fe--f_EEgfEgf-#osg_--g_--gOF ENGINEERING EXAM REVIEW FUNDAMENTALS

40. For a planet, acceleration due to gravity is g, and the radius of the planet is r.

If we move to a point 3r from the surface of the planet, the strength of the gravitational field will be: d. g/4 e. 16g

a. g b. 4g c. g/16

Total distance from the center of the planet is 4r

Answer(c) 41. When the load is applied as shown, the edge BC of the block moves 0.03mm to the right. Determine the shear modulus.

20000N

a. 50,300 MPa b. 41,700 MPa c. 38,600 MPa G

....EL._ =

A

.012

20000 X .2

d. 32,500 MPa e. 26,200 MPa

X

X

• 15 .00003

41,700 MPa Answer(b)

42. For the beam below the maximum moment occurs at: 10000#

WORKBOOK

SOLUTIONS

tccd.3_gg-f-ff-dffggg.ES#s_e_Etdfg THE EIT SIMULATED EXAM

FOR

a. 12 ft from the left b. 8 ft 8 in. from left c. 8 ft from the left d. 10 ft from the left e. at the center of the beam The maximum moment is when the point of zero shear: 18.67

Answer (b) 43. Determine the maximum shear stress in the shaft.

T"di?'oa

l>"

..

!oooo# 500 Ft Lb

a. 3160 psi b. 4140 psi c. 7340 psi

d. 6520 psi e. 5730 psi

Due to torque: 't

500

Tc J

X

12

1t X

Due to elongation: (J

1:.

A

1

X

3820 psi

32

10000 12

3180 psi

1t X

Combined: 'tmax

1_

2

Y3180 2 + 4

X

3820 2

4137 psi

Answer (b)

44. Brass cannot be used to reinforce concrete because: a. It is too dense b. It differs in shear capabilities c. It cannot carry enough load d. It does not stick to concrete e. It differs in coefficient of thermal expansion Answer (e)

AM

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ftp.#fpTfcgpgEgEEg_ffzf-Tsg_-cgosEp FUNDAMENTALS OF ENGINEERING EXAM

13-40

45. What increase in temperature ("F) is necessary to cause a 1 in. dia 10ft long steel rod eith fixed ends to buckle. There is no initial stress (a = 6.5 x 1Q-6) a.66 b. 26 c. 56

d.46 e. 16

a 41t 2

X

LlT EA

1t

1

X

X_A_

64

LlT

26.4"F

Answer (b) 46. For a fluid the number that relates inertial force to compressibility force is? a. Reynolds Number b. Mach Number c. Weber Number

d. Drag Number e. Froude Number Answer(b)

Consider the diagram of the pumping system as shown: D

E

c

A

and the following equations: a.

d Q

b.

e. hp out hp supplied

hxg

c. !2_ + Hp = p 2 + Z2

w

w

47. Which equation applies to the fluid as it passes through point D? Answer (a) 48. Which equation applies to the fluid as it passes from point B to point C? Answer (c)

REVIEW

WORKBOOK

SOLUTIONS

EIT SIMULATED EXAM THE t.cc#_-_f-ffspf-fggfEftggg-csz_fEgg

FOR

49. Which equation applies to the fluid as it passes from point D to pointE? Answer(e) 50. Which equation applies to the fluid as it passes from point A to point B? Answer (b) 51. For the system shown below, the pressure Pis? a. 5.46psi b. 6.87psi e . 4.72psi

d. 8.92psi e. 7.50psi

62 . 5 lbs X 1 f t

62 . 5 lbs x 1 ft p

+

144

=

. 2 .!..!1....

f

1 44

fe

X

13. 6

3 . 2 .!..!1....

fe

p = 5.46 psi Answer (a) 52. What is the moment of the force F about the origin 0 in the system below?

All distances are in feet.

(13 , 6, 0)

k (1 , 1 , 0)

i

a. -7k ft lbs b. 7 ft lbs c. 17k ft lbs

d.- 17 ft lbs e. 27k ft lbs

AM

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-E3_fEgggfEgEfg--gfEtg-#gpf3g-_-Ege-Esse-OF ENGINEERING EXAM REVIEW FUNDAMENTALS

13-42

(li

+ 1j + Ok) (12 i

+ 5j + Ok)

1

1

0

12

5

0

i

j

k

-7k ft#

Answer(a) 53. If equilibrium must be maintained, what is the maximum value that F can have?

100#

.........,__

Coefficient of friction for all surfaces is o. 4

F

a. 90 lbs b. 95 lbs e. 100 lbs

F = (100

X

d. 105 lbs e. l10Ibs

.4) + (175

X

.4) = 110 lbs

54. The magnitude of the reaction at B is: 100#

50 #/Ft

r-3' a. 34 lbs b. 44 lbs c. 74 lbs FB

X

6 = (150

FB = 87.5 lbs

d. 88 lbs e. 94 lbs

X

1.5) + (100

X

3)

or 88 Answer(d)

WORKBOOK

SOLUTIONS

FOR

THEi-Ef_-TNggg_B-N-spe__-f_EE_-#--gfESgEIT SIMULATED EXAM AM

55. The y coordinate of the centroid of the area below is? a. 3.4 b. 2.6 c. 4.2

d. 4.6 e. 5.0

y

1

8

6 3

------;,-x 4

y

(24

X

3)

+ ((}

X

4

X

3)

X

X

3)

+

(6x4)+(4x} )

3))

3.4 Answer (a)

56. The force system below is best described as: a. Nonconcurrent, noncoplanar b. Concurrent c. Coplanar d. Concurrent and coplaner e. 2 dimensional 50Jf

X

Since the forces have no common point or plane, they must be Non concurrent, non coplanar.

Answer (a)

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13-43