Phys102
Coordinator: xyz
Fi nal-131
Monday, December 30, 2013
Page: 1
Q1.
A transverse sinusoidal wave travels along a stretched string that has a linear mass
density of 5.00 g/m. The amplitude of the wave is 5.00 cm, its frequency is 80.0 Hz,
and the tension in the string is 22.0 N. What is the average power transmitted by this
wave?
A)
B)
C)
D)
E)
105 W
195 W
285 W
254 W
512 W
Ans:
μ = 5.00 × 10−3 kg/m
ym = 5.00 cm
f = 80.0 Hz
τ = 22.0 N
ω = 2πf = 160 π rad/s
τ
v = √ = 66.3 m/s
μ
Pavg =
1
μv(ωym )2
2
= 0.5 × 5 × 10−3 × 66.3 × (160π × 0.05)2 = 105 W
Q2.
The maximum pressure amplitude that the human ear can tolerate is 28 Pa. What is
the corresponding displacement amplitude for sound waves in air at a frequency of 2.0
kHz? [The speed of sound in air is 340 m/s, and density of air is 1.2 kg/m3]
A)
B)
C)
D)
E)
5.5 µm
8.3 µm
2.2 µm
11 µm
4.2 µm
Ans:
∆Pm = ρvωSm
⇒ Sm =
∆Pm
∆Pm
28
=
=
= 5.46 × 10−6 m = 5.5 μm
ρvω
2πρvf
2π × 1.2 × 340 × 2000
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 2
Q3.
How much ice at – 10 oC must be mixed with 0.20 kg of water, initially at 20 oC, in
order for the final equilibrium temperature to be 0 oC with all the ice melted?
[specific heat of ice = 2220 J/kg.K]
A)
B)
C)
D)
E)
47 g
63 g
85 g
75 g
12 g
Ans:
Q1 = mi∙ ci ∙ ∆T = (m)(2220)(10) = 22200 m
Q2 = mi . Lf = 333000 m
Q3 = mw ∙ cw ∙ ∆T = (0.2)(4190)(−20) = −16760 J
Q1 + Q2 + Q3 = 0: 355200 m = 16760 ⇒ m = 0.047 kg = 47 g
Q4.
In an adiabatic process, the temperature of one mole of an ideal monoatomic gas is
decreased from 500 K to 400 K. What is the work done in the process?
A)
B)
C)
D)
E)
1.25 kJ by the gas
1.25 kJ on the gas
2.08 kJ by the gas
2.08 kJ on the gas
zero
Ans:
adiabatic: Q = 0
3R
3
. ∆T = − nR∆T
2
2
= −(−1.5)(1)(8.31)(400 − 500) = +1246.5 J = +1.25 kJ
⇒ W = −∆Eint = −n. Cv . ∆T = −n.
by the gas
Q5.
In which of the following, is the entropy change per cycle negative?
A)
B)
C)
D)
E)
a perfect refrigerator 
a Carnot refrigerator
a real refrigerator
a Carnot heat engine
none of the other answers
Ans:
perfect refrigerator: W = 0
⇒ QH = QL = Q
∆S =
QH
QL
1
1
=
= Q( − ) < 0
TH
TL
TH TL
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 3
Q6.
Two particles of charges q1 = + 3.0 µC and q2 = – 27 µC, are fixed 10 cm apart. How
far from q1 should a third charge be located so that the net electrostatic force on it, due
to q1 and q2, is zero?
A)
B)
C)
D)
E)
5.0 cm
15 cm
20 cm
10 cm
25 cm
Ans:
See the figure:
q3
L
d +3
q1
−27
q2
The 3rd charge (q 3 ) should be placed to the left of q1 .
F31
kq 3 q1
kq 2 q 3
d + L 2 q2
L 2 27
= F32 :
=
⇒(
) =
⇒ (1 + ) =
=9
(d + L)2
d2
d
q1
d
3
⇒1+
L
L
L 10
=3⇒
=2 ⇒d= =
= 5 cm
d
d
2
2
Q7.
A proton with a speed of 4.0×106 m/s moves in a uniform electric field of magnitude
3.8×103 N/C. The field is acting to decelerate the proton. How far does the proton
travel before it is brought momentarily to rest?
A)
B)
C)
D)
E)
22 m
5.6 m
33 m
44 m
11 m
Ans:
vf2 = vi2 − 2ax
0=
vi2
vi2
vi2 m 1.67 × 10−27 × 16 × 1012
− 2ax ⇒ x =
=
.
=
2a
2 eE
2 × 1.6 × 10−19 × 3800
= 21.97 m ≈ 22 m
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 4
Q8.
A solid insulating sphere has a radius of 20 cm, and a uniform charge density of
3.0 nC/m3. What is the magnitude of the electric field at a distance of 40 cm from the
center of the sphere?
A)
B)
C)
D)
E)
5.6 N/C
4.4 N/C
6.3 N/C
7.1 N/C
8.9 N/C
Ans:
Q = ρ. V =
4π 3
ρR
3
Q
1
4πρR3
3.0 × 10−9 × (0.20)3
E=
=
.
=
= 5.6 N/C
4πε0 r 2
4πε0 r 2
3
3 × 8.85 × 10−12 × 0.16
Q9.
A point charge Q = + 1.5 µC is located at the origin. Point A is at x = + 1.0 m, and
point B is at x = + 3.0 m. How much work needs to be done by an external agent to
move a charge
q = – 1.0 µC from A to B?
A)
B)
C)
D)
E)
+ 9.0 mJ
– 9.0 mJ
+ 4.5 mJ
– 4.5 mJ
zero
Ans:
VA =
kQ
9 × 109 × 1.5 × 10−6
=
= 13.5 kV
rA
1.0
kQ
9 × 109 × 1.5 × 10−6
VB =
=
= 4.5 kV
rB
3.0
W = q. ∆V = q(VB − VA ) = −1.0 × 10−6 × (4.5 − 13.5) × 103 = +9.0 × 10−3 J
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 5
Q10.
Two point charges (+q and – q) are fixed on a line, as shown in FIGURE 1. Rank the
electric potentials at points 1, 2, 3; greatest first.
Figure 1
A)
B)
C)
D)
E)
1, 2, 3
(1 and 3) tie, then 2
3, 2, 1
2, then (1 and 3 tie)
2, 3, 1
Ans:
kq i
ri
1 1
2kq
V1 = kq [ − ] =
d 3d
3d
V= ∑
1 1
V2 = kq [ − ] = 0
d 3d
V3 = kq [
1 1
2kq
− ]=−
3d d
3d
Q11.
Three capacitors are connected in parallel. Each has plate area A and plate
spacing d. What must be the spacing of a single capacitor of plate area A if its
capacitance equals that of the parallel combination?
A)
B)
C)
D)
E)
d/3
d/2
3d/2
d/4
d/6
Ans:
Ci =
ε0 A
d
Cp = 3Ci =
3ε0 A
d
Cx = Cp
ε0 A
3ε0 A
d
=
⇒x=
x
d
3
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 6
Q12.
A steady current passes through a wire. The amount of charge that passes through the
wire in 2.0 s is 1.8 C. How many electrons pass through the wire in 5.0 s?
A)
B)
C)
D)
E)
Ans:
𝑖=
2.8 × 1019
1.1 × 1019
2.5 × 1019
1.9 × 1019
2.3 × 1019
q
= constant
t
2 s → 1.8 C
5s → q
⇒q=
5 × 1.8
= 4.5 C
3
Charge quantization:
q = Ne
⇒N=
q
4.5
=
= 2.8 × 10−19
e
1.6 × 10−19
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Q13.
Page: 7
In the circuit shown in FIGURE 2: E2 = 44 V and E3 = 20 V. Calculate the emf E1.
Figure 2
I1
I2
A)
B)
C)
D)
E)
28 V
44 V
12 V
72 V
20 V
Ans:
Take the right loop:
ε3 − 5I2 + ε2 − 6I = 0
ε3 + ε2 − 6I = 5I2
20 + 44 − 24 = 5I2 ⇒ I2 = 8A
⇒ I1 = 4A
Take the big loop:
ε3 + 4I1 − ε1 + 4I1 − 6I = 0
ε3 + 8I1 − 6I = 0
20 + 32 − 24 = ε1
ε1 = 28 V
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 8
Q14.
Calculate the potential difference (Vb – Va) between points a and b in the circuit
shown in FIGURE 3.
Figure 3
A) – 4.00 V
B) + 3.00 V
I
C) – 6.00 V
D) + 1.50 V
I
E) – 5.25 V
Ans:
Take the left loop:
I
12 − 2I − 4I = 0
⇒I=
12
= 2A
6
Now proceed from a ⟶ b:
Va + 4 − 4I = Vb
⇒ Vb − Va = 4 − 4I = 4 = −4 V
Q15.
A capacitor of capacitance C = 4.0 mF is discharged through a resistor of resistance R
= 4.0 kΩ. How long will it take the capacitor to lose half of its initial stored energy?
A)
B)
C)
D)
E)
5.5 s
9.2 s
1.4 s
3.2 s
6.3 s
Ans:
Discharging Process:
U=
⇒
2t
q2
1
=
∙ q20 ∙ e−2t/τ = U0 e− τ
2C
2C
2t
U0
2t
τ. ln2
= U0 e− τ ⇒ 2 = e2t/τ ⇒ ln2 =
⇒ 𝑡=
2
τ
2
⇒ t=
RC ∙ ln2
4.0 × 4.0 × ln2
=
= 5.5 s
2
2
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 9
Q16.
Several identical lamps are connected in parallel. What happens if one lamp burns
out?
A)
B)
C)
D)
E)
The power dissipated in each of the remaining lamps will not change.
The power dissipated in each of the remaining lamps will increase.
The power dissipated in each of the remaining lamps will decrease.
The current passing through each of the remaining lamps will decrease.
The current passing through each of the remaining lamps will increase.
Ans:
A
Q17.
When two unknown resistors are connected in series with a battery, the battery
delivers 225 W of electric power, and carries a total current of 5 A. For the same total
current, 50 W is delivered when the resistors are connected in parallel. Determine the
resistance of each resistor.
A)
B)
C)
D)
E)
3Ω, 6Ω
7Ω, 7Ω
9Ω, 2Ω
3Ω, 8Ω
6Ω, 4Ω
Ans:
Series: R s = R1 + R 2
P = I2 R s ⇒ R s =
P
225
=
= 9Ω
2
I
25
The only possible combination is choice A.
Q18.
A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 × 107
m/s, and experiences an acceleration of 2.00 × 1013 m/s2 in the positive x direction
when its velocity is in the positive z direction. Determine the magnetic field.
A)
B)
C)
D)
E)
– 20.9 ĵ
+ 20.9 ĵ
+ 20.9 k̂
– 20.9 k̂
+ 20.9 î
(mT)
(mT)
(mT)
(mT)
(mT)
⃗
F
⃗
v
⃗B
Ans:
⃗ B = ma⃗ = 1.67 × 10−27 × 2.00 × 1013 î = 3.34 × 10−14 î (N)
F
⃗FB = qv
⃗ × ⃗B
⃗B must be in the (– y) direciton
∴ The only possibily is 𝐀
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 10
Q19.
A proton moves in a circle of radius 5.10 cm in a 0.566 T magnetic field. What value
of electric field could make its path a straight line?
A)
B)
C)
D)
E)
1.57×106 V/m
2.41×106 V/m
6.18×106 V/m
9.21×106 V/m
5.88×106 V/m
Ans:
circle: qvB =
mv 2
qBR
→v=
R
m
qB2 R
line: FB = Fe → qvB = qE → E = vB =
m
=
1.6 × 10−19 × (0.566)2 × 0.051
1.67 × 10−27
= 1.57 × 106 V/m
Q20.
A coil with a magnetic dipole moment of 1.45 A.m2 is oriented initially with its
magnetic dipole moment antiparallel to a uniform 0.835 T magnetic field. What is the
change in the potential energy of the coil when it is rotated 180o so that finally its
magnetic dipole moment is parallel to the field?
A)
B)
C)
D)
E)
– 2.42 J
– 6.12 J
– 1.10 J
– 5.30 J
– 7.08 J
Ans:
U = −μ
⃗ ∙ ⃗B
Ui = μB
} ∆U = Uf − Ui = −2μB = (−2)(1.45)(0.835) = − 2.42 J
Uf = − μB
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 11
Q21.
A wire carrying a 15-A current is directed along the positive x axis, and is
perpendicular to a magnetic field. A magnetic force per unit length of 0.12 N/m acts
on the wire in the negative y direction. Determine the magnetic field in the region
through which the wire passes.
A) + 8.0 k̂ (mT)
B) + 3.0 k̂ (mT)
C) + 4.0 ĵ (mT)
D) – 4.0 ĵ (mT)
E) + 3.0 î (mT)
Ans:
⃗L
⃗FB = iL
⃗ × ⃗B
⃗ must be in the (+Z)direction
⇒ B
⃗B
F
FB = iLB
B=
1 FB
1
∙
=
× 0.12 = 8.0 × 10−3 T
i L
15
Q22.
A uniform magnetic field exerts a torque on each of the current-carrying single loops
of wire shown in FIGURE 4. The loops lie in the xy-plane, each carrying the same
value of current in the directions shown. The magnetic field points in the positive xdirection. Rank the coils by the magnitude of the torque exerted on them by the field,
from largest to smallest.
A)
B)
C)
D)
E)
A,C,B
A,B,C
B,A,C
C,A,B
B,C,A
Figure 4
Ans:
τ = μB = 2AB = (iB)A
A: 1 × 2 = 2
π
B: = 0.78
4
A, C, B
1
C: × 1 × 3 = 1.5
}
2
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 12
Q23.
Two long wires are oriented so that they are perpendicular to each other, as shown in
FIGURE 5. What is the magnitude of the magnetic field at a point midway between
them if the top one carries a current of 20.0 A, and the bottom one carries a current of
5.00 A?
Figure 5

IT = 20.0 A
10.0 cm ⃗
B2
⃗1
B
B=?
10.0 cm

IB = 5.00 A
Bottom wire
A)
B)
C)
D)
E)
4.12 × 10-5 T
5.22 × 10-3 T
3.02 × 10-5 T
7.47 × 10-3 T
6.25 × 10-6 T
Ans:
⃗B1 → right
⃗ 2 → out of the page
B
μ0 IT
4π × 10−7 × 20
B1 =
=
= 4.00 × 10−5 T
2πd1
2π × 0.1
B2 =
μ0 IB
4π × 10−7 × 5
=
= 1.00 × 10−5 T
2πd2
2π × 0.1
⇒ Bnet = √B12 +B22 = 4.12 × 10−5 T
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 13
Q24.
FIGURE 6 shows four different sets of wires that cross each other without touching.
The magnitude of the current is the same in all four cases, and the directions of
current flow are as indicated. For which configuration will the magnetic field at the
center of the square formed by the wires be equal to zero?
Figure 6
A)
B)
C)
D)
E)
3
1
2
4
The field is equal to zero in all four cases
Ans:
A
Q25.
A wire with a weight per unit length of 0.080 N/m is suspended directly above a
second parallel wire. The top wire carries a current of 30 A, and the bottom wire
carries a current of 60 A. Find the separation between the wires so that the top wire
will be held in place by magnetic repulsion.
A)
B)
C)
D)
E)
4.5 mm
2.1 mm
5.3 mm
7.5 mm
1.7 mm
Ans:
FB = mg
μo i1 i2 L
μo i1 i2
4π × 10−7 × 30 × 60
= mg ⇒ d =
=
= 4.5 × 10−3 m
mg
2πd
2π
×
0.080
2π ( L )
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Q26.
The value of the line integral
Page: 14
 
B
 .d s around the closed path in FIGURE 8
is 1.4 × 10-5 T.m. What are the direction and magnitude of I3?
Figure 8
A)
B)
C)
D)
E)
19 A, into the page
19 A, out of the page
3.1 A, out of the page
3.1 A, into the page
8.0 A, into the page
Ans:
⃗ ∙ ds = μ0 Inet
∮B
∮ ⃗B ∙ ds = μ0 (I2 + I3 )
∮ ⃗B ∙ ds
− I2
μ0
⇒ I3 =
1.4 × 10−5
=
+ 8.0 = +19 A , into the page
4π × 10−7
Q27.
Two coplanar and concentric circular loops of wire carry currents of I1 = 6.0 A and I2
= 3.0 A, as shown in FIGURE 7. If r1 = 12 cm. What are the value of r2 and the
direction of I2 such that the net magnetic field at the center of the two loops is zero?
Figure 7
A)
B)
C)
D)
E)
r2 = 6.0 cm, Counterclockwise
r2 = 6.0 cm, Clockwise
r2 = 3.0 cm, Counterclockwise
r2 = 3.0 cm, Clockwise
r2 = 4.0 cm, Counterclockwise
Ans:
I2 must be counter clockwise
B1 = B2 :
μ0 ii (2π) μ0 i2 (2π)
=
4πr1
4πr2
⇒ r2 =
i2
3
∙ r1 = × 12 = 6.0 cm, counterclockwise
i1
6
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 15
Q28.
A 10-turn square coil, with a resistance of 0.60 Ω and side length 5.0 cm, is placed in
an external magnetic field, as shown in FIGURE 9. The magnetic field decreases
uniformly from 0.050 T to zero in 10 ms. What is the current induced in the loop?
A)
B)
C)
D)
E)
Figure 9
0.21 A, clockwise
0.21 A, counterclockwise
0.42 A, clockwise
0.42 A, counterclockwise
Zero
Ans:
Lenz law:
Induced current must produce a magnetic field
increasing into the page.
→ Clockwise
εind =
dΦB
d
dB
ΔB
(NBA) = NA
=
= NA
dt
dt
dt
Δt
⇒ Iind =
NA ΔB
10 × 25 × 10−4 0.050
=
×
= 0.208 A ≈ 0.21 A
R Δt
0.60
0.010
Q29.
FIGURE 10 shows a 50-turn coil, with a diameter of 1.0-cm, inside a solenoid with a
diameter of 2.0 cm. The solenoid is 8.0 cm long, has 1200 turns, and carries the
current shown on the right in the figure. What is the magnetic flux through the coil at
time t = 0?
Figure 10
A)
B)
C)
D)
E)
3.7 × 10-5 Wb
4.2 × 10-5 Wb
5.3 × 10-5 Wb
8.1 × 10-5 Wb
1.2 × 10-5 Wb
Ans:
ΦB = Nc ∙ Bs ∙ AC = Nc ∙ AC ∙ BS = (Nc ∙ πrc2 ) ∙ (μ0 ∙ is ∙
=
NS
)
Ls
50 × π × 25 × 10−6 × 4π × 10−7 × 0.5 × 1200
= 3.7 × 10−5 Wb
0.080
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0
Phys102
Fi
nal-131
Monday, December 30, 2013
Coordinator: xyz
Page: 16
Q30.
A force F is applied to pull a conducting rod out of a magnetic field at constant speed
v and power input P, as shown in FIGURE 11. The rod moves on conducting rails.
The force is increased so that the constant speed of the rod is doubled to 2v. What are
the new force and power input?
Figure 11
A)
B)
C)
D)
E)
2F and 4P
2F and 2P
4F and 2P
4F and 4P
2F and P
Ans:
ε
BLv
B2 L2
F = iLB = ∙ LB =
∙ LB = (
)v
R
R
R
P=
ε2 B2 L2 v 2
=
R
R
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-30-s-0-e-1-fg-1-fo-0