Download Phys102 Final-123 Zero Version Coordinator: xyz Monday, July 29

Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 1
Q1.
A sinusoidal wave with a speed of 22.4 m/s is travelling on a string of linear density
15.0 g/m. If the maximum transverse speed of a particle of the string is 7.05 m/s, then
the average power transmitted by this wave is:
A)
B)
C)
D)
E)
8.35 W
4.62 W
5.21 W
1.30 W
9.62 W
Ans:
1
1
Pav = μω2 vym 2 = μv(ωym )2
2
2
1
= × 15.0 × 10−3 × 22.4 × (7.05)2 = 8.35 W
2
Q2.
A car, travelling with a speed of 30.0 m/s, follows a police car travelling at 50.0 m/s.
The siren of the police car emits sound with a frequency of 500 Hz. What frequency is
heard by the driver of the car? The speed of sound is 340 m/s.
A)
B)
C)
D)
E)
474 Hz
527 Hz
397 Hz
534 Hz
468 Hz
Ans:
Speed of car = x ; speed of police car = y
f ′ = f0 .
v + v0
v+x
340 + 30
= f0 .
= (500).
= 474 Hz
v + vs
v+y
340 + 50
Q3.
For an ideal gas, which of the following statements is CORRECT?
A)
B)
C)
D)
E)
Ans:
A
In an isothermal expansion: W = 𝑄.  𝛥𝐸𝑖𝑛𝑡 = 0 for isothermal
In an isothermal expansion: W =ΔEint. 
In an isothermal expansion: Q =ΔEint. 
In an isochoric process: W =ΔEint.  W = 0 for isochoric
In an isochoric process: W = Q. 
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 2
Q4.
Two moles of an ideal gas are in a 6.0 L container at a pressure of 5.0 × 105
Pa. Find the average translational kinetic energy of a single molecule.
A)
B)
C)
D)
E)
3.7 × 10-21 J
1.2 × 10-21 J
7.5 × 10-21 J
1.9 × 10-21 J
9.3 × 10-21 J
Ans:
3
3k pV
kT =
.
2
2 nR
K avg =
3
5.0 × 105 × 6.0 × 10−3
−23
= × 1.38 × 10
×
= 3.7 × 10−21 J
2
2 × 8.31
Q5.
Five moles of an ideal monatomic gas are allowed to expand isobarically
from 40.0 cm3 to 100 cm3. What is the change of entropy of the gas in this
process?
A)
B)
C)
D)
E)
+ 95.2 J/K
+ 57.1 J/K
– 95.2 J/K
– 57.1 J/K
+ 19.1 J/K
Ans:
ΔS = ∫
nCp dT
dQ
Tf
Vf
=∫
= n. Cp . ln ( ) = n. Cp . ln ( )
T
T
Ti
Vi
5
100
= 5 × × 8.31 × ln (
) = +95.2 J/K
2
40.0
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c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 3
Q6.
Two positive point charges are fixed on the x-axis as follows: q1 = 12 µC at
the origin, and q2 = 3.0 µC at x = + 3.0 m. At what coordinate, on the x-axis,
is the electric field due to q1 and q2 zero?
A)
B)
C)
D)
E)
+ 2.0 m
– 1.5 m
+ 3.0 m
+ 1.0 m
– 1.0 m
Ans:
The requested point is between the two charges, a distance x from q1.
magnitude: E1 = E2
kq1
kq 2
=
(3 − x)2
x2
3−x 2
q2
3
1
(
) =
=
=
x
q1
12
4
3−x
1 3
1
= ⇒ − 1=
x
2 x
2
3
3
⇒ =
⇒ x = +2.0 m
x
2
Q7.
In Figure 1, what is the ratio of the electric flux that penetrates surface S1 to the flux
that penetrates surface S2?
A)
B)
C)
D)
E)
3
0
2
1
1/3
Figure 1
Ans:
Φ1 =
q enc
3q + 2q + 4q 9q
=
=
ε0
ε0
ε0
Φ2 =
q enc
3q − 2q + 2q 3q
=
=
ε0
ε0
ε0
Φ1
9q
=
=3
Φ2
3q
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Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 4
Q8.
Three charges are arranged as shown in Figure 2. Calculate the work required by an
external agent to remove the negative charge (– e) on the right end to infinity.
A)
B)
C)
D)
E)
3.8 × 10-19 J
7.7 × 10-19 J
1.5 × 10-18 J
1.9 × 10-19 J
1.3 × 10-19 J
Figure 2
Ans:
ke2 ke2 ke2
2ke2 ke2
3ke2
Ui = −
+
−
= −
+
= −
a
2a
a
a
2a
2a
Uf = −
Wext
ke2
a
ke2 3ke2 ke2
= ∆U = Uf − Ui = −
+
=
= + 3.8 × 10−19 J
a
2a
2a
Q9.
A particle (m = 2.0 × 10-9 kg, q = – 5.0 µC) has a speed of 30 m/s at point A
and moves, under the effect of an electric field opposite to its velocity, to
point B where its speed becomes 80 m/s. What is the potential difference VB
– VA?
A)
B)
C)
D)
E)
+ 1.1 V
+ 3.5 V
+ 6.3 V
– 2.4 V
zero
Ans:
Ui + K i = Uf + K f
1
1
qVA + mvA 2 = qVB + mvB 2
2
2
m
(v 2 − vB 2 )
⇒ VB − VA =
2q A
B
A
⃗
E
⃗V
a⃗
2.0 × 10−9
VB − VA =
. (900 − 6400)
−1.0 × 10−5
= +1.1 V
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 5
Q10.
The space between the plates of an isolated parallel-plate capacitor is filled with
plastic of dielectric constant κ1 = 2.10. The potential difference between the plates is
600 V. The plastic is replaced with glass whose dielectric constant is κ2 = 3.40. What
is the potential difference between the plates after inserting the glass?
A)
B)
C)
D)
E)
371 V
247 V
325 V
199 V
462 V
Ans:
Qi = C1 . V1 = (k1 . C0 )(600) = 1260 C0
Qf = C2 . V2 = (k1 . C0 )(Vf ) = 3.40 C0 Vf
isolated: Qi = Qf ⇒ 1260C0 = 3.40C0 Vf ⇒ Vf = 371 V
Q11.
Two resistors, A and B, are made of the same material of cylindrical wires having the
same length. When a potential difference of 110 V is applied to each wire, the powers
dissipated in A and B are 400 W and 100 W, respectively. Ignoring the variation of
resistance with temperature, the ratio of the diameter of A to that of B is:
A)
B)
C)
D)
E)
2
1
1/2
1/4
4
Ans:
R=
ρL
ρL
ρL
4ρL
=
=
=
2
2
A
πr
π(D/2)
πD2
P=
V2
V2
⇒R=
R
P
4ρL V 2
4ρLP
=
⇒ D2 =
2
πD
P
πV 2
⇒ D = K. √P (K = constant)
⇒
DA
PA
400
=√ = √
=2
DB
PB
100
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 6
Q12.
A light bulb has a tungsten filament with a resistance of 18.0 Ω at 20.0 oC. It is
connected to a 120 V source. When operational, it dissipates a power of 100 W. What
is the operational temperature of the filament? Assume that the dimensions of the
filament do not change. The temperature coefficient of resistivity of tungsten is
4.50 × 10-3 (oC)-1.
A)
B)
C)
D)
E)
1850 K
2880 K
1110 K
1280 K
2930 K
Ans:
(120)2
V2
=
= 144 Ω
P
100
∆R
144 − 18
∆R = α. R 0 . ∆T ⇒ ∆T =
=
= 1555.56 K
α. R 0
4.5 × 10−3 × 18
Rf =
Tf = To + ∆T = 293.15 + 1555.56 = 1848.71 → 1850 K
Q13.
A 45 Ω resistor is connected across the terminals of a 10 V battery. If a current of
0.20 A flows through the 45 Ω resistor, what is the internal resistance of the battery?
A)
B)
C)
D)
E)
5Ω
3Ω
9Ω
8Ω
7Ω
i
r
R
ε
Ans:
i=
ε
ε
10
⇒ r+R= =
= 50 Ω
r+R
i
0.20
r = 50 − R = 50 − 45 = 5 Ω
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 7
Q14.
The absolute value of the potential difference between points a and b in the Figure 6
is:
Figure 6
A) 14 V
B) 21 V
C) 41 V
D) 11 V
E) 19 V
Ans:
R eq = 2 kΩ
εnet = 12 + 12 − 4 = 20 V
εnet
20
=
= 1 × 10−2 A
R eq 2 × 103
Va + 12 − (1 × 103 × 1 × 10−2 ) + 12 = Vb ⇒ Va − Vb = −24 + 10 = −14 V
i=
|Va − Vb | = 14 V
Q15.
Three resistors of resistances 2 Ω, 6 Ω and 12 Ω are connected to a battery, as shown
in Figure 7. If the total current through the circuit is I = 5 A, what is the current
through the 12 Ω resistor?
Figure 7
2Ω
A) 0.6 A
B) 0.3 A
C) 0.1 A
6Ω
D) 0.2 A
E) 0.9 A
12 Ω
Ix
I
Ans:
1
1 1 1
6+2+1
9
= + +
=
=
R eq
2 6 12
12
12
Type equation here.
Ω
V
12
4
⇒ R eq =
= Ω
9
3
5 × 4 20
ε = I. R eq =
=
V
3
3
Now, consider the loop containing 12 Ω and ε:
+ε − 12 Ix = 0 ⇒ Ix =
ε
1
20
20
=
×
=
= 0.555 A ⇒ 0.6 A
12
12
3
36
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 8
Q16.
In the multi-loop circuit shown in Figure 8, the current through the 2.0 kΩ resistor is:
Figure 8
A)
B)
C)
D)
E)
1.2 mA
3.1 mA
1.9 mA
0.7 mA
2.5 mA
𝑖2 6.0 kΩ
3.0 kΩ 𝑖1
A
i
2.0 kΩ
6.0 V
6.0 V
Ans:
3.0 kΩ
Take the big outer loop:
+ 6.0 − 600i2 + 6000i1 − 6.0 = 0 ⟹ i1 = i2
At junction A: i = i1 + i2 = 2i2 ⟹ i2 =
Now, take the left loop:
6 − 6000 i2 − 2000 i = 0
6 − 3000 i − 2000 i = 0
⇒i=
i
2
6
= 1.2 × 10−3 A
5000
Q17.
In the circuits shown in Figure 9, all the capacitors have the same
capacitance and initial charge, and all the resistors have the same resistance.
At time t = 0, switch S is closed such that each capacitor discharges through
its resistors network. Rank the time taken for discharging the capacitor to
have its initial charge, shortest time first.
Figure 9
A)
B)
C)
D)
E)
B, C, A
C, B, A
A, C, B
C, A, B
B, A, C
Ans:
τ = RC ⟶ shortest τ ⇒ smallest R
A: R = 2R
R
B: R =
2 ∴ B has the shortest time then C then A
3R
C: R =
2}
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 9
Q18.
A particle (mass = 6.0 ×10-6 kg) moves in the xy plane with a speed of
km/s in a direction that makes an angle of 37o above the positive x axis.
the instant it enters a magnetic field of (5.0 î ) mT, it experiences
acceleration of (8.0 k̂ ) m/s2. What is the charge of the particle? Ignore
gravitational force on the particle.
A)
B)
C)
D)
E)
4.0
At
an
the
– 4.0 µC
– 4.8 µC
+ 4.0 µC
+ 4.8 µC
– 5.0 µC
Ans:
⃗F = ma⃗ = 6.0 × 10−6 × 8.0 k̂ = 48 × 10−6 k̂ (N)
⃗ = −(4.0 × 103 × 5.0 × 10−3 × sin37°)k̂ = −12 k̂ (N)
⃗v × B
⃗ = q (v
⃗ ): thus q = (48× 10−6 ) / (– 12) = – 4.0 C
F
⃗ × B
Q19.
A region of space contains perpendicular uniform electric and
magnetic fields, with the electric field pointing in the positive y
direction. In this region, a proton travels in the positive x direction
with constant velocity. What is the direction of the magnetic field?
A)
B)
C)
D)
E)
+ z direction
– z direction
+ y direction
– y direction
– x direction
⃗e
F
+
⃗
v
Ans:
A
⃗FB
Q20.
A proton moves around a circular path (radius = 2.0 mm) in a uniform 0.25 T
magnetic field. What total distance does this proton travel during a 1.0-s time
interval?
A)
B)
C)
D)
E)
48 km
82 km
71 km
59 km
7.5 km
Ans:
mv 2
qBR
⇒v=
R
m
qBRt
1.6 × 10−19 × 0.25 × 2.0 × 10−3 × 1.0
S = v. t =
=
= 48 km
m
1.67 × 10−2
qvB =
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 10
Q21.
A straight 20-cm wire is bent at its midpoint so as to form an angle of 90o, as
shown in Figure 10, and carries a current of 10 A. It lies in the xy plane in a
region where the magnetic field is in the positive z direction and has a
constant magnitude of 3.0 mT. What is the magnitude of the total magnetic
force on this wire?
Figure 10
A)
B)
C)
D)
E)
4.2 × 10-3 N
3.2 × 10-3 N
5.3 × 10-3 N
2.1 × 10-3 N
6.0 × 10-3 N
Ans:
Portion on the x − axis:
⃗F1 = (iL1 B)ĵ = 10 × 0.1 × 3.0 × 10−3 ĵ = 3.0 × 10−3 ĵ (N)
Portion on the y − axis:
⃗F2 = (iL2 B)î = 3.0 × 10−3 î (N)
⃗Fnet = ⃗F1 + ⃗F2 = (î + ĵ) × 3.0 × 10−3 (N)
Fnet = √1 + 1 × 3.0 × 10−3 = 4.2 × 10−3 N
Q22.
A loop of radius r = 5.00 cm, carrying a current I = 0.200 A, is placed inside
a magnetic field B = 0.300 ˆi (T). The normal to the loop is parallel to a unit
vector n̂  0.600 ˆi  0.800ˆj . Calculate the magnitude of the torque on the
loop.
A)
B)
C)
D)
E)
3.78 × 10-4 N.m
1.13 × 10-4 N.m
0.600 × 10-4 N.m
4.72 × 10-4 N.m
Zero
Ans:
⃗ = N iAn̂ = N iπr 2 n̂ = 1 × 0.200 × π × 25 × 10−4 × (−0.6î − 0.80ĵ)
μ
= −9.42 × 10−4 î − 1.26 × 10−3 ĵ (A. m2 )
τ⃗ = μ
⃗ × ⃗B = +3.78 × 10−4 k̂ (N. m)
King Fahd University of Petroleum and Minerals
Physics Department
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Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 11
Q23.
Figure 11 shows three circuits consisting of straight radial lengths and
concentric circular arcs (either half- or quarter-circles of radii r or 2r). The
circuits carry the same current in the indicated direction. Rank the circuits
according to the magnitude of the magnetic field produced at the center of
curvature (the dot), greatest first.
Figure 11
A)
B)
C)
D)
E)
B2, B1, B3
B1, B2, B3
B3, B2, B1
B2, B3, B1
B1, B3, B2
i
i
i
1
2
3
Ans:
1: Contribution from the big circle only
2: Contribution from both circles
3. Inner circle cancels some of the outer
Q24.
A wire is bent as shown in FIGURE 12, and carries a current i = 15 mA along the
indicated directions. What is the magnitude of the magnetic field at the center of the
loop, C, if the radius of the loop is R = 5.0 cm?
Figure 12
A)
B)
C)
D)
E)
Ans:
1.3 × 10−7 T
3.3 × 10−6 T
1.1 × 10−8 T
3.1 × 10−7 T
5.3 × 10−8 T
2
μ0 i
4π × 10−7 × 15 × 10−3
Straight: B1 =
=
= 6.0 × 10−8 T
2πR
2π × 5.0 × 10−2
Out of page
−7
−3
μ0 iϕ
4π × 10 × 15 × 10 × 2π
Circle: B2 =
=
= 18.8 × 10−8 T
4πR
4π × 5.0 × 10−2
Into page
Bc = B2 − B1 = 12.8 × 10
−8
T → into the page
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 12
Q25.
Three long wires have currents flowing perpendicular to the page with directions as
indicated in FIGURE 13. Wire 1 is at y = 2.0 m on the y-axis, wire 2 is located at the
origin, and wire 3 is at x = 2.0 m on the x-axis. If I1 = 1.0 A, I2 = 2.0 A, and I3 = 3.0
A, what is the magnitude of the net force per unit length on wire 2 due to the other
two wires?
Figure 13
A)
B)
C)
D)
E)
−7
6.3 × 10 N/m
5.3 × 10−5 N/m
5.3 × 10−9 N/m
3.6 × 10−7 N/m
1.3 × 10−7 N/m
f21
f23
Ans:
f = force per unit lenght
f21
μ0 I1 I2
4π × 10−7 × 1.0 × 2.0
N
=
=
= 2.0 × 10−7
2πd
2π × 2.0
m
μ0 I2 I3
4π × 10−7 × 2.0 × 3.0
N
f23 =
=
= 6.0 × 10−7
2πd
2π × 2.0
m
N
f2,net = √4.0 + 36 × 10−7 = 6.3 × 10−7
m
Q26.
Each of the four wires in Figure 14 carries a 2.0 A current into or out of the page.
 
What is the value of the line integral  B.d s for the indicated path of integration?
Figure 14
A)
B)
C)
D)
E)
−2.5 × 10−6 T.m
+2.5 × 10−6 T.m
−1.5 × 10−7 T.m
+1.5 × 10−7 T.m
+5.5 × 10−6 T.m
Ans:
ienc = +2.0 − 2.0 − 2.0 = −2.0 A
⃗ . ds = μ0 . ienc = −2.5 × 10−6 (T. m)
∮B
King Fahd University of Petroleum and Minerals
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c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
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Final-123
Monday, July 29, 2013
Zero Version
Page: 13
Q27.
A solenoid with N turns carries a current of 12 A and has a length of 43 cm. If the
magnitude of the magnetic field generated at the center of the solenoid is 90 mT,
what is the value of N?
A)
B)
C)
D)
E)
2.6 × 103
1.9 × 103
2.2 × 103
1.1 × 103
1.7 × 103
Ans:
B = μ0 ni =
μ0 Ni
L
BL
90 × 10−3 × 43 × 10−2
=
= 2.6 × 103
μ0 i
4π × 10−7 × 12
⇒N=
Q28.
A flexible loop has a radius of 12.0 cm and is placed with its plane perpendicular to a
uniform magnetic field of magnitude 0.150 T, as shown in Figure 3. The loop is
grasped at points A and B and stretched at constant rate until its area is zero. If it takes
0.300 s to stretch the loop, what is the magnitude of the induced emf during this time
interval?
Figure 3
A)
B)
C)
D)
E)
22.6 mV
6.31 mV
2.76 mV
63.1 mV
16.3 mV
Ans:
εind =
dϕB
d
dA
∆A
(BA) = B
=
= B.
dt
dt
dt
∆t
(B)(πr 2 )
0.150 × π × 144 × 10−4
=
=
= 22.6 mV
∆t
0.300
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Final-123
Monday, July 29, 2013
Zero Version
Page: 14
Q29.
A conducting loop is placed near a long straight wire carrying a constant current I, as
shown in Figure 4. Which of the following actions will induce a counterclockwise
current in the loop?
Figure 4
A) Moving it up away from the
wire 
B) Moving it toward the wire
clockwise
C) Moving it to the right (parallel
to the wire) No change
D) Moving it to the left (parallel to
the wire) No change
E) Rotating it through an axis that coincides with the wire No change
Ans:
A
Q30.
Figure 5 shows a metallic bar being moved to the right on two conducting parallel
rails in a uniform magnetic field of magnitude 2.5 T, directed into the page. If R = 6.0
Ω, and L = 1.2 m, what is the magnitude of the applied force required to move the bar
at a constant speed of 2.0 m/s?
Figure 5
A)
B)
C)
D)
E)
3.0 N
5.0 N
2.0 N
4.0 N
6.0 N
Ans:
εind = BLv = 2.5 × 1.2 × 2.0 = 6.0 V
i=
εind
6.0
=
= 1.0 A
R
6.0
F = iLB = 1.0 × 1.2 × 2.5 = 3.0 N
King Fahd University of Petroleum and Minerals
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