Download key

ID CODE: D
Physics 202 Final Exam
December 22, 2010
Name: ...............KEY.................................
Student ID: .........................
Section: .............................................
TA (please circle):
Matt Elbert
Josh Isaacs
Jianjia Fei
Hyungjun Lim
Frank McNally
Yongyan Rao
Xiao Wang
Michael Wood
Zhen Liu
Instructions:
1. Don’t forget to write down your name, student ID#, and section number. You need do this
on (this page of) your test book and on your Scantron sheet as well.
2. Answer all multiple choice questions in this test book by indicating the best answer among
choices. You must do this both on your test book and on your Scantron sheet. Follow
instructions on the Scantron sheet on how to mark valid answers. The scantron is the final
base for your grade.
3. When you finish, you need to turn in both this test book and the Scantron sheet.
4: Use the blank side of question pages as additional draft spaces. An extra blank sheet is
provided at the end of the test book.
5: Only one answer is allowed per problem/question. All problems have equal weight.
Constants:
ε0= 8.85x10-12 C2/(Nm2), μ0= 4πx10-7 T m/A, c=3.0x108 m/s, ke=8.988x109 Nm2/C2
Please be very careful with the first question even though the answer will not count towards
your grade:
1. ENTER THE ID CODE ABOVE IN THE UPPER RIGHT CORNER
A. ID Code A
B. ID Code B
C. ID Code C
D. ID Code D
E. ID Code E
Sign Conventions for Mirrors and Lenses
d0
>0
concave mirrors
converging lens
object side
<0
convex mirrors
diverging lens
the other side
dI
M=-dI/do
image side (real)
upright
the other side (virtual)
inverted
f
Object Side
Image Side
mirrors
lenses
front
front
front
behind
refraction
surface
opposite to
observer’s side
observer’s side
This is the sign convention we used in class; you are responsible for its proper application
2. In the figure, if Q = 30 μC, q = 5.0 μC, and d = 30 cm, what is the magnitude of the electrostatic
force on q?
F=ke(Qq/d2 – 2Qq/(2d)2) = keQq/d2/2
= 7.5N
A.
B.
C.
D.
E.
15 N
23 N
zero
7.5 N
38 N
3. A long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m3)
distributed throughout its column. Determine the magnitude of the electric field 5.0 cm from the
axis of the cylinder.
A. 25 N/C
Draw a co-axial cylindrical Gaussian surface of r=5cm along the axis of
B. 20 N/C
the cylinder. Take an arbitrary height h.
C. 14 N/C
Gauss’s Law (with symmetry argument):
D. 31 N/C
E. 34 N/C
E(2πrh) =Qenclosed/ε0 = ρ(πr2h)/ε0
Æ E= ρr/(2ε0)= 14.1 N/C
Yes, this is a more difficult test problem. (but it is not too bad, isn’t it?)
4. A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the
particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the
electric potential difference, VB − VA.
A. −50 V
ΔKE = - ΔU =- qΔV
B. −40 V
C. −30 V
Æ VB-VA=ΔV = -ΔKE/q = -30V
D. +30 V
This was in midterm 1, remember?
E. +15 V
5. Determine the energy stored in C1 when C1 = 10 μF, C2 = 12 μF, C3 = 15 μF, and V0 = 70 V.
A.
B.
C.
D.
E.
6.5 mJ
5.1 mJ
3.9 mJ
8.0 mJ
9.8 mJ
C=(1/C1+1/C2+1/C3) = 4μF
Q = CV= 0.28 mC
U1=1/2 Q12/C1 = 3.92mJ
(note Q1=Q)
6. Light bulb A is rated at 60 W and light bulb B is rated at 100 W. The both are designed to operate
at 110 V. When the two light bulbs are connected in series together with a power source, (That is,
the two bulbs and the power source forms a closed circuit in series), which statement is true
A. The 60 W bulb has a greater resistance and consumes more energy than the 100 W bulb.
B. The 60 W bulb has a greater resistance and consumes less energy than the 100 W bulb.
C. The 60 W bulb has a smaller resistance and consumes more energy than the 100 W bulb.
D. The 60 W bulb has a smaller resistance and consumes less energy than the 100 W bulb.
E. We need to know the resistivities of the filaments to answer this question.
at rated voltage: Prated=ΔV2/R Æ R60W > R100W
in circuit P=I2R with same I when in series. Æ larger R consumes more energy
6.
7. In the circuit below, potential difference Vb − Va is measured to be -8.0V , what is the battery
voltage ε1?
I3= (va-vb)/R3 = 0.4 A
consider loop R2ε2R3R2 : I2R2+ε2-I3R3=0 Æ I2= -2.2A
at junction a: 0=I1+I2+I3 Æ I1= 1.8A
now consider loop R1R3ε1R1: -ε1 +I1R1 –I3R3=0
Æ ε1= 10V
A.
B.
C.
D.
E.
5.0 V
10.0 V
15.0V
18.0 V
28.0 V
8. A positively charged particle has a velocity in the negative z direction at point P. The magnetic
force on the particle at this point is in the negative y direction. Which one of the following
statements about the magnetic field at point P can be determined from this data?
A. Bx is positive.
B. Bz is positive.
C. By is negative.
FB=qvxB = q(-vk) x (Bxi+Byj+Bzk) = -qv(Bxj-Byi)
D. By is positive.
FB in -j ÆBy=0 and Bx>0
E. Bx is negative.
9. Two long (and thin) parallel wires separated by 4.0 mm each carry a current of 24 A. These two
currents are in the same direction. What is the magnitude of the magnetic field at a point that is
between the two wires and 1.0 mm from one of the two wires?
A. 4.8 mT
B. 6.4 mT
B1=μ0I/(2πr1) and B1=μ0I/(2πr1)
r1=1.0mm and r2=3.0 mm
C. 3.2 mT
D. 9.6 mT
Use right hand rule to determine that B1 and B2 must be opposite in dir.
E. 5.3 mT
B=|B1-B2| =3.2 mT
10. A planar loop consisting of four turns of wire, each of which encloses 200 cm2, is oriented
perpendicularly to a magnetic field that increases uniformly in magnitude from 10 mT to 25 mT in
a time of 5.0 ms. What is the resulting induced current in the coil if the resistance of the coil is 5.0
Ω?
A. 60 mA
B. 12 mA
ε=dΦ/dt=NAdB/dt=4x200cm2x(25-10)mT/(5ms) =0.24V
C. 0.24 mA
I=ε/R=0.048A
D. 48 mA
E. 6.0 mA
11. As shown in the figure, a circular circuit loop of area A=10cm2 connected with a resistor
R=100Ω and a capacitor C=1 μF is placed inside an external uniform magnetic field B.
The plane of the loop is normal to the field. At t=0, the magnetic field starts to increase at
a constant rate dB/dt=103 T/s. How long does it take to charge the capacitor to 80% full?
(hint: the time constant of a RC circuit is τ=RC)
A.
B.
C.
D.
E.
0.8x10-4 s
1.0x10-4 s
1.6x10-4 s
2.0x10-4 s
2.5x10-4 s
τ=RC=1.0x10-4 s
q=Qfull(1-e-t/τ)=0.8Qfull Æ -e-t/τ=0.2 Æt=1.6x10-4 s
12. Still in the above setting, when the capacitor is fully charged, how much energy is stored in
the capacitor?
ε=dΦ/dt =AdB/dt = 1V Æ Qfull =Cε = 1μC Æ U=1/2 Cε2 =5x10-7 J
A.
B.
C.
D.
E.
5.0 x10-13 J
1.0 x10-10 J
5.0 x10-7 J
4.0 x10-2 J
2.0 J
13.
A.
B.
C.
D.
Still in the same setting, which side of the capacitor is charged positive?
upper side
bottom side
the capacitor will not be charged
not enough information to determine.
Per Lenz’s law
14.
A.
B.
C.
D.
E.
At what frequency will a 12-μF capacitor have a reactance XC = 300Ω?
44 Hz
88 Hz
XC =1/(2πfC) Æ f=44.2 Hz
180 Hz
350 Hz
280 Hz
15. The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by
in SI units. Find the average power per square meter carried
A.
B.
C.
D.
E.
by the EM wave.
720 W
172 W
500 W
2 × 107 W
86 W
Bmax=1.2x10-6
I=uc =1/2 Bmax2/μ0 c =172 W/m2
16. Figures below show two settings for the “jumping ring” experiment. The two settings are identical
except for that the battery polarizations are opposite. The metal ring is made of aluminum which is
a conducting but non-magnetic material. When the switch is closed, how will the ring move?
setting A
A.
B.
C.
D.
E.
setting B
the ring in setting A jumps up and the ring in setting B moves down.
the ring in setting B jumps up and the ring in setting A moves down.
both up
both down
neither ring moves as aluminum is non-magnetic.
per Lenz’s law (and this was demo’ed and explained in class.)
17. A light ray is incident on the surface of water (n = 1.33) at an angle of 60° relative to the normal to
the surface. The angle of the reflected wave is
A. 80°
B. 40°
C. 20°
θ’=θ
D. 60°
E. 30°
18. A laser beam is incident from air at an angle of 30.00° to the vertical onto a solution of Karo syrup
in water. If the beam is refracted to 19.24° to the vertical, what is the index of refraction of the
syrup solution?
A. 1.331
B. 1.414
n=sinθ1 /sinθ2 =1.517
C. 1.732
D. 1.133
E. 1.517
19. A light ray whose frequency is 6.00 × 1014 Hz in vacuum is incident on water (n = 1.33). The
wavelength of the light after it enters the water is (in nm)
A. 798
B. 500
C. 665
λ=c/f = 500nm λn=λ/n= 376 nm
D. 376
E. 266
20. A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what
angle relative to the normal to the interface of the two liquids is light totally reflected?
A. 78°
B. 88°
C. 68°
D. 49°
E. the critical angle is undefined
sinθcritical = 1.333/1.361 Æ θcritical =78.4o
21. An object is placed 25 cm in front of a converging lens of focal length 20 cm. How far past the lens
is the image?
A. 20 cm
B. 40 cm
do=25cm, f=20 cm
C. 60 cm
1/do+1/di=1/f Ædi=100 cm
D. 80 cm
E. 100 cm
22.
A.
B.
C.
D.
E.
In the same setting above, is the image
reduced, virtue, and upright
enlarged, virtue, and upright
reduced, real, and inverted
enlarged, real, and inverted
enlarged, virtue, and inverted.
di>0 Æreal, M=-di/do=-4 Æenlarged, inverted
23. Still in the same setting, a second lens of focal length 25cm is place 60 cm behind the first lens.
How far past the new lens does the final image form?
A. 20 cm
B. 40 cm
for the 2nd lens:
C. 15 cm
do=60-100=-40cm, f=25 cm
D. 47 cm
E. 67 cm
1/do+1/di=1/f Æ di=15 cm
\
24. Which ray diagram is correct? The three rays in each diagram are distinguished by different types
of lines. Hint: Check the behavior of each ray as it interacts with the lens.
A.
B.
C.
D.
E.
A
B
C
D
E
25. The working principle of a glass prism is illustrated in the figure below. Compare the index of
refraction for three colors in the glass, nred, nyellow, and nviolet .
A.
B.
C.
D.
E.
nred > nyellow > nviolet
nyellow > nred > nviolet
larger n ÅÆ bends more
nviolet > nyellow > nred
nred > nviolet > nyellow
Index of refraction is independent of the color.
26. A laser beam (λ = 694 nm) is incident on two slits 0.100 mm apart. Approximately how far apart
(in m) will the bright interference fringes be on the screen 5.00 m from the double slits?
A. 3.47 × 10−3
B. 3.47 × 10−2
Δy =λ/d L= 3.47x10-2m
C. 3.47 × 10−4
D. 3.47 × 10−6
E. 3.47 × 10−5
27. As shown in the figure, a light beam of λ=600nm is incident at a near normal angle on a
thin oil film floating on water. What is the minimum thickness of the oil film that can
maximize the transmission of light into water?
air
~0o
oil, n=1.5
(ignore all absorption)
A.
B.
C.
D.
E.
100 nm
150 nm
200 nm
250 nm
300 nm
water, n=1.3
consider two beams in water:
Δφ =2π/λn 2t + (0−0) = 2π
Æt=λn/2 =600/1.5/2=200nm
28. Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the
width of the central maximum on a screen 1.0 m from the slit?
A. 2.0 cm
B. 4.2 mm
Δy=2λ/aL= 4.2mm
C. 1.1 cm
D. 2.0 mm
E. 0.70 mm
29. For a multi-slit interference pattern as shown, how many slits are used?
A.
B.
C.
D.
E.
2
3
4
5
6
3+2=5
30. A ruby laser beam (λ = 694.3 nm) is sent outwards from a 2.7-m diameter telescope to the moon,
384,000 km away. What is the smallest possible radius of the big red spot on the moon?
A. 500 m
B. 250 m
C. 120 m
Δ=1.22λ/DL= 120m
D. 1.0 km
E. 2.7 km
31. A spy satellite is equipped with a lens of 2m in diameter. It operates at an altitude of 500km above
the earth’s surface and is sensitive to wavelength about 600nm. What is the minimum distance
between two objects on earth it can resolve?
A. 1 m
B. 2 m
Δ=1.22λ/DL= 0.18m
C. 600 nm
D. 300 nm
E. 0.2 m