This preview shows pages 1–20. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 222, Summer 2010
FINAL EXAM Friday, August 6, 2010 Name (Printed):
Section Number: Recitation Instructor: INSTRUCTIONS: 1. This is a two hour exam consisting of 30 multiple~choice questions. Questions 101
through 130 are each worth 1 point. Questions 121 through 130 are based on your work
done during the laboratory periods. However, you can answer these questions also by
using physics information from the lectures. I will count all your correct answers for a total
of up to 30 points. Use your previous bubble sheet answering questions # 101 through #
130 on the back of your bubble sheet from the first three exams.. 2. Use a number 2 pencil when marking your bubble sheet. Do not use ink. Ask for a pencil
if you did not bring one. Fill in the appropriate bubble completely. If you need to change any
entry, you must completely erase your previous entry. Also, circle your answer on the
exam. 3. Carefully read each problem and its five possible answers before beginning to work on
the problem. Select only one answer for each problem. Choose the answer that is closest to the correct one. 4. Before handing in your exam, be sure that your answers on your bubble sheet are what
you intend them to be. You should copy down your answers on a piece of a scratch paper
for comparison with the answer key to be posted later. 5. When you are finished with the exam, place all exams materials, including the bubble
sheet, the exam itself, and scratch paper that you used for the exam, in your folder and return the folder. Enjoyed working with you this summer and wish you all the best in your future
endeavors. ' Good Luck!  Art Meyers A resistanccicss' LC circuit consists of a capacitor and an inductor, as drawn in ﬁgure 101. When the
current ﬁttwing through an indoclor changeaa voitage {emf} gets created. in the inﬁuctor that opposes
the Chang: in the cment. So‘ inductors ham: an "inertia" with restJed to changes in am. just like masses have inertia with respcct to changes in velocity. In an LC circuit, charge osciilates been}: and forth bcm'een the two capacitor plates. just like a pooduium’e
position oscillates back and forth between the two enopoims of its Swing. For instance‘ in ﬁgure 101
suppose the bottom plate initially carries negative charge Excess eicctmns flow of? the bottom pirate.
:hzough the inductor. and onto the top plate So. the top plate now eateries tho negative charge. But then
the negative charge “sioshes‘ back onto the bottom piate. In this manner! the charge oscillates. At any given moment. the energy stored in the capacitoris Lie = in; %. where Q denotes the net charge
on the top piate. and C denotes the capacitance. The energy stored in the inductor is UL =~*= LP, wherei
denotes the cuntm through. 2h: ciztuit, and L is a. CEIHSIEBI calied the inductance. Since the circuit is
essentially rcsistanceless. no heat dissipates as current occiliabcs hack cm! forth. .1}! this. circuit. C = 0.10 fonds and L = 2.0 item’s. [Fmds and henms are both SI units.) Initial i}: at
time t '= [L the bottom piote has charge — 1.0 cottiombs. and no current is ﬂowing. 101. Woich graph best rcpicscots the charge on the top piatc of the capacitor as a function of time? 'L L
L C it”? pint: A. B:
gr ' Thatth 35:": ' I . :— 0 .
Figure 101 73, 1 “mt
E
..i
D.
an
5° 0
4.; “mi: [3“; Home: of the. above graphs represent the charge in the top piate of the capacitor as a function
of time 102. An ion of mass m and of charge +4: is in circular orbit around a ﬁxed
pom: charge Q. with charge 8.0 11C. The radius of the orbit is 0.20 m. and
the speed of the ion in the orbit is 1.2xto‘ tan/5.1% uniform external magnetic v :12 x 1 05”“
ﬁeld, perpendicular to the plane of the orbit. is present. The magnetic force I
on the ion is equal to the electric force in magnitude and in direction at all 4—)@
points of the orbit. In the ﬁgure. the mass m of the ion is closest to:
A) 61:10'zﬁ kg 
B) 51:10 '1‘ kg
C) 43:10 '2‘ kg
D) 81:10‘m kg
E) 77:10 '2‘ kg If an electron is accelerated through a potential of 300 V and then enters, from the left, a region with an
electric field of 12.00 x 106 We pointing down {see figure below]: 103. What is the election speed leaving the accelerating potential? [in m/s] A) 1.05 x 10 1“ a) 5.25 x 1013 C) 10.27 x 105 0) 1.05 x 10" E) Need more information to solve the problem. 104. What magnetic ﬁeld pointing into the pa per will allow the electron to travel to the right without
being deflected up or down? [See ﬁgure above] A) 0.987 T B) 1.05 T C) 1.168T D) Not enough information given in Problem #103. E) Impossible since ﬁelds are in the wrong direction for there to be no deflection. 105. You have a long solenoid which has 1500 turns in 20 cm with an AC current of 3.5 amps flowing through the coil. This coil has a diameter of 4 cm and a resistance of 25 ohms. What is the
magnetic field inside the center of the solenoid? [ in Tesla] A) 6.60 x 10‘3
a) 0.3299 C) 3.299 x 10'2
0) 4.67 x 10'2
E) 1.649 106. If a coi] of 300 turns with a diameter of 1.00 inch was placed in the above solenoid (assume i3.rms = 10 x 10" T) what will be the Ems voltage reading on an AC voltmeter or on a digital scope
(in volts) if the frequency = 5000 Hz? [in Volts] A) 15
B) 2.5
c) 21
0) 23.5
E) 30 107. Consider a series R1. circuit consisdng ofa battery a switch an inductor and a resistor alt
connected in series. The switch is closed and the current and the voltage change with time are
measured. noted and plozteti in graph :4! and graph #2. The graphing persons forgot to label the
axes. As a physicist, you wit! dc: this for them. These are graphs ofctttrent and voltage versus
time. Yin: comet axes labels are as follows: A C a E Q [2
A) time current time voltage
3) current time voltage time
C) time voltage time current
D) voltage time current time E) Home of the afﬁne combinations‘ 108 _ in the ﬁgure. an insulated wire is bent into a circular 100p of radius 6.0 cm
and has two long straight sections. The loop is in the xy plane, with the center
at the origin. The straight sections are parallel to the zaxis. The wire carries a current of 8 A, The magnitude of the magnetic field at the origin, in ttT. is
closest to: A} 90
B) 80
C) 110
D) 100
E) 70 109. 110. 111. 112. An eiectron has the same de Brogiie wavelength as an 800 nm photon. The speed of the
electron is closest to: [in m/s] A) 500
B) 700
C) 300
o) 900
E) 1000 An object 8 m high is located 125 m in front ofa concave mirror which has a focal length of
200 m. Which of the following set of positions, size, and character of the image is correct? Image Position ﬂ Character of Image
A) 500 m +32 m erect virtual
B) 333 m +213 m erect virtual
C) +333 m 21.3 m inverted real
D) +500 m 32 rn inverted real
E) None of the above. Determine the focal length of a converging lens which will project the image of a lamp magnified
4 diameters, upon a screen 10 m from the iamp (in m). A) 1.25 B) 1.6 C) 2.0 D) 2.5 E) Not enough information. A narrow beam of light strikes a glass plate (n = 1.60) at an angle of 530 to the normal. if the plate is 20 mm thick, what will be the lateral displacement of the beam after it emerges from
the plate (in mm)? A) 26.5
a) 15.0
C) 11.5
o) 9.0 E) Not enough information. 113. 114. 115. 116. Two identical bea kers are ﬁlled to thesame level, one filled with water (n = 1.351) and the other
with mineral oil, (n = 1.47) they are both viewed from directly above. Which beaker appears to
contain the greater depth of liquid? A) same depth 3) water depth greater
C) mineral oil depth greater
D) Not enough information. An amateur lens grinder wants to grind a converging lens of crown glass (n = 1.52) with the
same curvature on both sides and a focal length of 25 cm. What radius of curvature must he
grind on each face? [in cm] A) 13
B) 25
C) 26
D) 50
E) Not enough information. The interference pattern of two identical slits separated by a distance d = 0.25 mm is observed
on a screen at a distance of 1 m from the plane of the slits. The slits are illuminated by
monochromatic light of wavelength 589.3 nm [sodium D) traveling perpendicular to the plane of
the slits. Bright bands are observed on each side of the central maximum. Caiculate the
separation between adjacent bright bands (in mm). A) 1.08
B) 2.36
C) 4.72
D) 5.80
E) None of the above. The headlights of a distant automobile are 1.4 m apart. if the diameter of the pupil of the eye is
3 mm, what is the maximum distance at which two headlights can be resolved? {in m} Consider
the headlights as point sources of wavelength 500 mm. A)  2030
B) 5700
c) 5100
o) 5900 E) No enough information. 117. 118. 119. 120. A grating having 15,000 lines per inchproduces spectra of a mercury arc. The green line of the
mercury spectrum has a wavelength of 546.1 nm. What is the angular separation between the
1st order green line and the 2nd order green line? A) 13.3”
a) 21.40
c} 40.20
D) 59.00
E) None of the above. In an important experiment in 1927, a beam of electerons was scattered off a crystal of nickel.
The intensity of the scattered beam varied with the angle of scattering, and the analysis of
these results lead to the confirmation of A) the particle nature of light. B) the Bohr model of the atom. C) the wave nature of light. D) the Rutherford model of the nucleus.
E) the quantization of energy levels. You have a blackbodv radiator with a maximum intensity at a wavelength of 414.3 nm, what is
the Temperature of the blackbodv radiator? [in Kelvin] A) 6000
B) 5500
c) 7000
D) 7500
E) Not enough information. The work functions for potassium and cesium are 2.25 and 2.14 eV, respectively. Will the
photoelectric effect occur for either or both of these elements with an incident light of
wavelength 565 nm? A) neither cesium nor potassium
B) cesium only C) potassium only D) both E) Not enough information. PHYSICS 222 Summer 2010  Laboratory Final Exam 121. When an electrophorus is charged according to the procedure explained in prelab and lab
2222, the charge accumulated on the metal plate of the electrophorus comes from: The handle of the electrophorus The plastic plate that was rubbed before placing the metal plate on top of it.
The air around the system. The ground, via the table under the plastic plate. The ground, via your finger when you touch the metal plate. 921.0272» 122. The voltage across a resistor is measured to be 4.5 i 0.2 V and the current through it is
found to be 1.2 i 0.3 A. What is the range of acceptable values of the resistance that we can
calculate from our data? Between 2.9 Q and 5.2 Q.
Between 2.9 Q and 4.3 9.
Between 3.1 Q and 4.8 Q.
Between 3.3 Q and 4.8 Q.
Between 3.3 Q and 4.3 Q. 9999‘!» 123. Incandescent buibs are an example of nonohmic system. Which of the following graphs is
the best representation of the current versus voltage graph for such a system? I I >V 9V >V V 124. The graph illustrates the decrease with time of an exponentially decaying voltage. 20 IL. 15 H
D U1 Potential difference (V) Time (5) What is the time constant of this decay? a. 4.1 s
b. 5.6 s
c. 7.2 s
d. 9.4 s
e. 12 s 125. A laser beam strikes a semicircular transparent block as shown below. The ine0ming beam
is aligned with the 0° position and hits the block right at the center of the ﬂat surface. Semicircular plexiglass
block Laser beam 220°
Polar graph paper When the flat surface of the block is oriented along the 100°—280° line, the refracted ray
comes out at an angle of 182°. What is the index of refraction of block? a. 1.2
b. 1.5
c. 2.3
d. 2.9
e. 5.0 126. In the electron tube used in experiment 2226 Electron beam in a magnetic ﬁeld, electrons
were produced in a hot ﬁlament, accelerated towards an anode, and some of them went
through" a slit and into a glass tube placed between the coils of a Helmholtz pair that
produced a magnetic field in the tube. To increase the number of electrons in the beam (i.e.,
to make the beam more intense), we should increase: The current through the ﬁlament. The current through the anode. The current in the coils. The number of turns in the coils. None of the above, the intensity of the beam is ﬁxed by the ﬁlament material only. 9999‘?» 127. In experiment 2228, the test coil in the ﬁgure below was placed inside the solenoid to
observe the effects of magnetic induction. Oscilloscope Function Rd 51 0 Test coil
3.;
3:
generator . Bk Solenoid : (15 cm long] BNC cable Assume that for certain conditions, the oscilloscope produces the following graph: IIIIIIIII
ﬂl‘ﬂlﬂl “KIWI! Ground P Illlil If we keep all conditions the same but double the number ofturns in the test coil, which
of the following graphs is the closest to what the oscillosc0pe will read? 12?”  “WWIIIll .Illlllﬂllﬂ IllIIIIIIlIlII'I Illllllllll‘ll I‘IIIiIFIIIlIF IMIIMIL. IIIIIIII
B l
7 ‘ . all“!
I'llI‘
IIIVIIII‘II .‘
a
l JIIIIH'
I _L‘_"III‘!
_Il 128. 129. 130. A digital AC voltmeter is connected to a function generator that produces a sinusoidal
voltage with an amplitude of 5.0 V. What does the voltmeter read? 09.0579 2.5 V
3.5 V
5.0V
7.1 V
10V When you hold a lens with focal length f = 5 cm at about 30 cm from your eye and look
at an object through this lens (say, a lamp), what do you sec? 99.02?!» An inverted, larger image of the lamp.
An upright, larger image of the lamp. An inverted, smaller image ofthe lamp.
An upright, smaller image of the lamp.
Nothing, the image is completely blurred. When white light shines on a thin ﬁlm of oil ﬂoating on a puddle of water, beautiful
rainbow patterns appear. Why does this happen? 3.. b. Because different ratios of the amounts of oil and water mixed in different places
produce a different reﬂection spectrum. Because different ratios of the amounts of oil and water mixed in different places
prdduce a different absortion Spectrum. Because different wavelengths produce constructive interference for different
thicknesses ofthe oil film. Because different thicknesses of the oil film result in different indices of refraction.
Because different wavelengths are reﬂected in different directions. You may record your answers on this page and take Ft with you after the exam to compare to the posted solutions. 109—"* 118
F— 119
ET— 1 120
112
T 121
W 114 I 115 123 116 124
i125 126 117 L 127 128 129 136 Formula Sheet  Physics 222  Summer 2010 Vectors and math lgl: (Afmjmzz 3§=ABcosf9=AxBx+AyBy+Asz ﬁxfr = AB sine 2x3 =(AyB: AAEBy)f+[AzBJ —AXB:)}‘+(AIBy —AyBx) w[Jifo 4ac >1 3‘2") ax2+bx+c=0 :> x: 2:: d n H d _ d . ~ femto (i) ——x =nx —smx=cosx —cosx=smx _ aix dx dx plCO (p)
Geometry nano (n) perimeter circle: 21:}? area sphere: 4x322 micro— (0) area circle: ER: volume Sphere; gm}? milli (m) 1 revolution = 2:: radians = 360° centi (C) Conversion factors 11m3 = 1000 liters 1 atm = 1.01x105 Pa =760 mm Hg Physical constants g 2931ng e=1.6><10‘19 C _ Wb _ 42 C2 _ s
#024NXIOTE 80—8.85X10 Nlmz 6—3.00x10 mils
Fluids
F
13:; 13:12—ng §£=Av=constant P+pgy+épvg :constant
Gravitation
—' Mm = E = _ Mm. U. . : G34,
IFNewton r2 g G r2 U G r cucular mint 1'! r
T 2m?” F x 13 = constant g =9.81 00st G=6.67><10'” Nmzi’kg2 ME =5.97x1024 kg RE =6.38x106 m Electrostatics . A ﬂ F __ 4 a #
FCoulomh=keq;_gzr E:q— E=ﬁXE U2—15‘E
0
(I) _ E. d.. I qcncloscd _. q _. “ O a
5—! ‘“ Ezk—F E=k—P E=i—x 212:2; 6” “r2 e r 260
(«Q dQ dQ
2— VEvoiume 0':— AEarea x1=— fElen th
’0 (IV ( ) dA ( ) d; ( g )
VA—VB 2]:Edf E=_€w W=—AU=—quV (2’:qu
_ 9 _
V—kev— AVFiEa'
r
1 N 2 2
kg = =8.99><109 ‘1‘ e0 =8.85><10"3 C 921.60x10‘” c
41:60 0 Nm2
me :9.11><10'31 kg mp 21.67x10'2? kg
Capacitors
C Q C —C C C "L—L+L+ l +
_? sq — I+ 2+ 3+... Cu] C] C2 C3
A 27:6 L ab
0d Ida/b) C :r 0 _a C 43:60}?
UleVZ =._Q_Z._=1QV 31:15}? E=KEU CZKC
2 2C 2 2 °
Electric current and resistor circuits
dQ ' 1 L
I=— V=IR J=—= E: = _
dz A qnvd JP R pA
p=po(1+a(T—yg)) R=Rﬂ(1+a(T—I;,))
R —R +R +R + ~1——i+i+i+ ’ V2
cq _ I 2 3 '” Rm RI R2 R3 “‘ P :1 R :?
RCcircuits Q(t)=Q(°0)(—e_%] Q(z)=g(0)e% r=RC I(t):1(0)e_% Magnetic ﬁeld _ .. _. _. _. _. .. _. mv
F=q(E+~3xB) szmzxs ®B=JBdA R=lqlB
,u=IA 1:;1XB Uz—gif;
é=ﬁzqﬁ><9 cilﬁfdix; 3:“)! 5J0” c2: 1
4x r2 4:? r 23x L Zn #050
2 q n H
BI — [JG{a 3’2 Bx = B : yon! : #0 enclosed
2(x2 + :22) 2“
Induction
dtD _ __ —~ a ~_ __d(133
8=—N (#3 g_gS[va]dz SEE d3 dt
~ 4 ’ ‘ _ a’CDE
$3 ‘ : #0 (15' + ID )cucloscd ED — g Maxwell’s equations
@E.dﬁ=Qenclosed 80
~ ~ dd)
~  861! 2 ' + E : — ¢ #0 [1C 80 dt Jenclosed
d!
Inductance
gl=_Mdiz “:1 M=NPm=N2fDm 54g hﬂ
d: d: :2 a. d: f
. ~ ’; _i L 1 BE
I=Ioo[l_e ] szoe’ TIE LzﬂonzfA U—ELIZ "22%
1
= cos !+ a) = —
9' Q0 ('5‘) LC iv / 1 R2
q=Qe 3" cos w’t+ w’: ___
0 ( ‘9). LC 4L2 AC current I=I t Ira :—I 1r1'I1's=— Vnm=— Vmis=1nnsZ
J (305.50 V K J5 _ J5
l
VR=IR VIE=in Xi‘sz VC:IXC X —X
V=IZ Z=ﬁ/}i’2+(XL~)(C)2 tan¢=¥ v=Vcos(mt+¢)
: ' l V _ N
p Iv average = COS ¢ : [nnsVrms cost? : IliIISR V? _ —f VII] = V212
EM waves
E(x,t) : Em 005(kxwwt)} 306,!) : Bmax coonc—wtyz Emax = cBmax
l 2 l dp _ S _ EB
5:53 0: u=iaoE2+ 1 82:50E2:B_ EE—Ewmj
Vgoluo 2 2:510 due #06
~ — ~ E B 2 I 7
S : X B I = Savemge = max max : Em“ : l flLillian: : 18008;“
#0 2"”0 28/10 2 o 2
Geometric optics
c . . . n, 2 n2
n:— 191. :6? nlsm 6i =n2 sm 62 sm 19”” =—' I=Imax cos g5 tan (9,, =—
v n. nl
1+L::i~ nz:i:_i f5: hericalmirror :£ i: nin —num L_i
S S f s p 2 f um R, R2
nr,+n_;:_n;,—n,, m=£=_ﬂ M=g
s s R y nbs (9 Interference and diffraction 3, _ 2 ¢ _ rz—rl _ dsiné‘
(Isinﬁzmﬂ. y:mRE I—Imaxcos €35275 A 2?3' A
dsin6=[m+l]/1 2t=mlu 2;:[m+_l_];L”
2 2
n” . 6 sinigj sirﬁ asiné =m£ ,6: 2:1" asm 1 :1“ ———~—— I : m 9052 2 )1 max ﬂ 2
{a} R=—=Nm 5in61:1‘22— 2dsin6=ml Ch. 38 Photons 71f = c E = hf = ll; = Fin) (n = 2nf h : 6.626X10—34Js = 4.136X10—15 eVs_ h = Ehh = 1.055 x10‘34Js = 6.583 x10—16eVs
' TE
1240 eVnm E hf h 2n: E:—— :—:—=—:fzk kz— CVZK :hf—(D
Mnm) p C ‘3 7L 7L 0 max
hydrogen atom: 51; 2 R (nf = 1,2,...; n; : nf+ 1,nf+ 2, ...) R = 1.097X107fm
line spectra: hf = h—C = Ei—Ef I hydrogen atom: En = u 13:26V (n = 1,2,3, ....) black body radiation: I = 0T4 0 z 5.6? X 10—3 22:4 MT = 2.90 ><10'3 mK
m_.
. 1 2 Jﬁ
Qh.39 Matter Partlcles p = my K = imv = 2m a p = \f2mK
h 211: _, a 1.23 mm W
E=hf=hco =—=hk 2— =hk electrons:?L=—~"—
p 71 i 1’ \f—K(eV)
electron diffraction (normal incidence): K = eVab Maxima: dsine = m (m: 1,2,...)
wave packet (pulse): AxAkx 2 1 AyAky 2 1 A2 Akz 2 1 At Ad) 2 1
particles: AxApx 2 Ft AyApy 2 h AZ‘APZ 2 h At AB 2 35
X2
phase velocity = (Mk group velocity = dmfdk P041, x2) 2 I t1(x)2 dx
X1 Ch.40 112 ﬁt? U2 _ mtx
En = (8m2)n2 : (2m2]n2 (n = 1,2,3,...) IIIHOQ = 8111(T]
Ch. 41
hydrogen atom: EH 2 — 1332‘“! (n = 1,2,3, ....)
L = ’V€(f+1) F: with E = 0,1,2,...,n«1 LZ = mgh with mg: 0,:t1,:2,..., if
_, L eﬁ
llorb = nus ; Mom: ~mwB ME = E = 9.274x 10‘24 FT
electron spin: S = “\ls(s+ Uh with s:% S2 = mg? with m5 = i% S
is = Jung use = —2msuB = 1 1mg shell: ﬁxed n subshell: ﬁxed n, E orbital: ﬁxed n, 3, mg shell designations: K for n 2 l; L for n = 2; M for n = 3, subshell designations: 5 fort? = 0, p forf = i,d forf = 2,ffor£J = 3, g for»? = 4, Order of subshell ﬁlling: Is, 25,2p, 3s, 3p, 45, 3d, 4p, 55, 4d, 5p, 68, 4f, 5d, 6p, 75, 5f, 6d, Uz—E'ﬁ Ifﬁ=B(z)R,then: U=pZB and Fz:—%=nzdliz) _ (13.6 eV)Zeff2 Multielectron atoms: E1 = 112 (n = 1,2,3, ....) ...
View Full
Document
 Fall '05
 Ogilvie

Click to edit the document details