Q1.
A transverse sinusoidal wave travelling on a string is given by: y (x,t) = 0.150 sin
(15.7 x – 50.3 t) (SI units). At a certain instant, point A is at the origin, and point B is
the closest point to A along the x axis where the motion is 60.0o out of phase with the
motion at A. What is the distance between points A and B?
A)
B)
C)
D)
E)
Ans:
m
m
m
m
m
360° ⟶ λ � d = 60 × λ = λ = 1 �2π� = π
360
6
6 k
3k
60° ⟶ d
=
Q2.
0.0667
0.133
0.267
6.28
3.14
π
= 0.0667 m
(3)(15.7)
Two identical sound sources emit sound waves of wavelength λ and are separated by a
distance d. What is the lowest non-zero value of d for which constructive
interference occurs everywhere along the line that passes through the two sources?
Consider only points which do not lie between the two sources.
A)
B)
C)
D)
E)
λ
λ/4
2λ
λ/2
4λ
Ans:
Consider point P:
∆ L = L1 − L2 = d
Constructive interference: ∆ L = mλ
⇒ d = mλ = 0, λ, 2λ, …
Lowest (non − zero)value of d = λ
𝐒𝐒𝟏𝟏
d
𝐒𝐒𝟐𝟐
𝐋𝐋𝟏𝟏
𝐋𝐋𝟐𝟐
𝐏𝐏
Phys102
Term: 112
Q3.
Final
Saturday, May 19, 2012
Code: 20
Page: 1
A 200-g ice cube at 0.0 oC is dropped into 350 g of water at 20 oC. What is the
temperature of the mixture when it reaches thermal equilibrium?
A)
B)
C)
D)
E)
0.0
–13
+24
–24
+13
o
C
C
o
C
o
C
o
C
o
Ans:
Heat to melt ice : Qi = mi LF = 0.2 × 333 = 66.6 k J
Heat gained from water : Qw = mw . cw . ∆T = 0.35 × 4190 × 20 = 29.3 k J
Since Qw < Qi : ice will not melt completely
Q4.
⇒ Tf = 0 ℃
An ideal monatomic gas undergoes an isobaric process at a pressure of 80 kPa from
an initial volume of 20 L to a final volume of 50 L. What is the change in the internal
energy of the gas?
A)
B)
C)
D)
E)
3.6 kJ
2.4 kJ
4.0 kJ
7.0 kJ
zero
Ans:
W = p ∆V = nR∆T
5
5
5
Q = nCp ∆T = (n) � R� (∆T) = nR∆T = p∆V
2
2
2
3
∆Eint = Q − W = p ∆V = 1.5 × 8.0 × 104 × 30 × 10−3 = 3.6 k J
2
Q5.
Four moles of an ideal monatomic gas undergo a free expansion to twice the initial
volume. What is the change in the entropy of the gas in the process?
A)
B)
C)
D)
E)
Ans:
∆S = �
23 J/K
29 J/K
35 J/K
17 J/K
0
dQ
1
Q
W
nR T 𝑙𝑙n(Vf /Vi )
= � dQ = =
=
T
T
T
T
T
⇒ ∆S = n. R. 𝑙𝑙n(Vf /Vi ) = 4 × 8.31 × 𝑙𝑙nZ = 23 J /K
Phys102
Term: 112
Q6.
Final
Saturday, May 19, 2012
An electron and a proton are fixed. The direction of the electric field halfway between
the electron and the proton:
A)
B)
C)
D)
E)
Q7.
Q8.
+
�E⃗P �E⃗e
+62 nC
–62 nC
+31 nC
–31 nC
zero
�⃗out is due to the charge on the outer surface
E
(−992)(0.75)2
kq 0
𝐸𝐸0 R2
𝐸𝐸0 = 2 ⇒ q 0 =
=
= −62 nC
R
k
9.00 × 109
q net = q in + q 0 ⇒ q in = q net − q 0 = 0 − (−62) = + 62 nc
In a certain region of space, a uniform electric field is in the positive x direction. A particle with
negative charge is moved from x = 20 cm to x = 60 cm. Which of the following statements is
CORRECT?
A)
B)
C)
D)
E)
Ans:
is toward the electron

is toward the proton
cannot be determined because the field is zero

depends on the distance between the two particles

is perpendicular to the line joining the two particles

-
A conducting spherical shell has an outer radius of 0.75 m and a net charge of zero. A
point charge q is placed at the center of the shell. The electric field just outside its
surface is 992 N/C pointing radially toward the center of the sphere. What is the
charge on the inner surface of the shell?
A)
B)
C)
D)
E)
Ans:
Code: 20
Page: 2
The potential energy of the charge increases.
The potential energy of the charge decreases.

The potential energy of the charge remains the same.

The particle moves to a region of higher electric potential.

The particle moves to a point of the same electric potential.
�⃗ = E𝑖𝑖̂ , ∆ r⃗ = 0.4 𝑖𝑖̂
E
�⃗. ∆ r⃗ = − 0.4 E ⇒ ∆ 𝑉𝑉 < 0
∆ V = −E
∆ U = q. ∆ V ⇒ ∆ U > 0
(−) (−)
Phys102
Term: 112
Q9.
Final
Saturday, May 19, 2012
Two electrons are initially far way from each other. They are projected toward each
other, with each having a speed of 1.0 × 103 m/s. At what separation between the
electrons will they momentarily stop?
A)
B)
C)
D)
E)
Ans:
Code: 20
Page: 3
0.25 mm
1.3 cm
0.32 mm
1.8 cm
0.65 cm
K 𝑖𝑖 + U𝑖𝑖 = K𝑓𝑓 + U𝑓𝑓
U𝑖𝑖 = 0 (far away); K𝑓𝑓 = 0 (momentarily stop)
𝑘𝑘𝑒𝑒 2
1
∴ U𝑓𝑓 = K 𝑖𝑖 ⇒
= 2 × 𝑚𝑚𝑣𝑣𝑖𝑖 2
𝑟𝑟
2
2
9
9 × 10 × (1.6 × 10−19 )2
ke
=
= 2.5 × 10−4 m = 0.25 mm
⇒r=
m𝑣𝑣𝑖𝑖 2
9.11 × 10−31 × 1.0 × 106
Q10. For the combination of capacitors shown in FIGURE 1, what is the potential
difference across the 4.0-µF capacitor?
Fig# 1
A)
B)
C)
D)
E)
Ans:
C34 =
3.0 V
6.0 V
7.5 V
9.0 V
4.5 V
C1
C3
C3 × C4
2×4
4
=
= μ𝐹𝐹
C3 + C4
2+4
3
Q34 = C34 . V =
Q34 = Q3 = Q4
⇒ V4 =
4
× 9 = 12 μC
3
Q4
12
=
= 3.0 V
C4
4
C4
C2
Phys102
Term: 112
Q11.
Final
Saturday, May 19, 2012
Code: 20
Page: 4
A 5.00-mV voltage is applied to a copper rod, initially at 20.0 oC, resulting in a
current Io. When the rod is heated to a temperature Tx, the current is reduced to Io/3. If
the temperature coefficient of resistivity of copper is α = 4.10 × 10-3 K-1, what is the
temperature (Tx) of the rod? Assume that the dimensions of the rod do not change.
A)
B)
C)
D)
E)
Ans:
I=
508 °C
488 °C
60.0 °C
6.70 °C
20.0 °C
V
V
V
; I0 =
; Ix =
R
R0
Rx
I0
V Rx
Rx
I0
R 0 (1 + α∆T)
=
.
=
⇒
=
Ix
R0 V
R0
I0 /3
R0
⇒ 1 + α∆T = 3 ⇒ α∆T = 2 ⇒ ∆T =
∴ T𝐹𝐹 = T0 + ∆T = 508 ℃
2
α
= 488 ℃
Q12. A copper (Cu) rod and an aluminum (Al) rod of the same length and different cross
sectional areas are connected in series as shown in FIGURE 2. The resistivities and
cross sectional areas are: ρCu =1.69 × 10-8 Ω.m, ρAl = 2.75 × 10-8 Ω.m , AAl = 0.400
cm2 and ACu =0.200 cm2. The ratio of the magnitude of the electric field along the
aluminum rod to that along the copper rod (EAl / ECu) is
Fig#2
A)
B)
C)
D)
E)
0.814
3.25
1.23
1.00
2.00
Ans:
E = ρJ =
ρI
A
EA𝑙𝑙
ρA𝑙𝑙 . I Acu
ρA𝑙𝑙 . I Acu
=
.
=
.
Ecu
AA𝑙𝑙 ρcu . I
ρcu AA𝑙𝑙
=
2.75 0.2
×
= 0.814
1.69 0.4
Phys102
Term: 112
Q13.
Final
Saturday, May 19, 2012
Code: 20
Page: 5
A 6.00-V ideal battery is used to power a device whose resistance is 200 Ω. If the
battery can move a charge of 240 C, how long will it last?
A)
B)
C)
D)
E)
Ans:
I=
I=
2.22 hours
0.556 hours
35.6 hours
8.89 hours
11.8 hours
6
V
=
= 0.03 A
200
R
Q
Q
240
⇒ t = =
= 8000 s = 2.22 h
t
I
0.03
Q14. A 0.20-A current flows through a metallic rod of length 1.0 m when connected to a
1.0-V battery. The rod is then cut into four identical pieces each having a length of
0.25 m, which are then connected in parallel to the same 1.0-V battery. The new
current delivered by the battery is
A)
B)
C)
D)
E)
Ans:
R0 =
V
1.0
=
= 5.0 Ω
I0
0.20
R0
= 1.25 Ω ⟵ for each piece
4
R i 1.25
=
=
= 0.3125 Ω ⟵ equivalent resistance
4
4
Ri =
R eq
3.2 A
0.40 A
0.80 A
0.050 A
0.72 A
⇒ If =
V
1.0
=
= 3.2 𝐴𝐴
R eq
0.3125
Phys102
Term: 112
Final
Saturday, May 19, 2012
Code: 20
Page: 6
Q15. In the circuit of FIGURE 3, the current I1 = 3.0 A. What is the value of current I3?
Fig#3
A)
B)
C)
D)
E)
A
5.0 A
1.0 A
13 A
7.0 A
6.0 A
Loop
Ans:
Consider the loop shown:
+12 − (6 × 3) + 3I2 = 0
⇒ I2 =
−12 + 18
= +2.0 A
3
∗ Consider the junction A:
I3 = I1 + I2
= 3 + 2 = 5.0 A
Q16. A capacitor of capacitance C is connected to a 12-V battery, as shown in FIGURE 4.
First, switch S2 was open, and switch S1 was closed until the capacitor is fully
charged. Then, S1 is open and S2 is closed. If the voltage across the capacitor decays
and reaches 6.0 V after 0.10 s, the capacitance C is equal to
Fig# 4
A) 24 µF
B) 11 µF
C) 14 µF
D) 140µF
E) 47 µF
Ans:
Consider the discharging process:
VC = V0 e−t/τ ⇒ et/τ =
V0
V
t
V0
t
= 𝑙𝑙n � � ⇒ τ =
V
τ
V
𝑙𝑙n � V0 �
t
⇒ RC =
V
𝑙𝑙n � V0 �
t
1
0.10
1
⇒C=
=
×
= 2.4 × 10−5 F = 24 μF
3
12
R 𝑙𝑙n �V0 �
6 × 10
𝑙𝑙n � 6 �
V
∴
Phys102
Term: 112
Final
Saturday, May 19, 2012
Code: 20
Page: 7
Q17. FIGURE 5 shows a portion of an electric circuit that includes a battery connected in series to a resistor
of resistance R. Considering the variation of the electric potential along the circuit shown in the figure,
which one of the following statements is CORRECT?
Fig# 5
A) The current in R flows to the left and the positive
terminal of the battery is at point
1.
B) The current in R flows to the left and the positive
terminal of the battery is at point
2.
C) The current in R flows to the right and the positive
terminal of the battery is at point
1.
D) The current in R flows to the right and the positive
terminal of the battery is at point
2.
E) The current is zero.



Q18. A positive charge q is moving with velocity v in a uniform external magnetic field B .
�⃗ experiences maximum magnetic force if
The charge �F⃗B = q
�⃗ v × B
A)
B)
C)
D)
E)
The direction of
The direction of
The direction of
The direction of
The direction of


.
v is perpendicular to the direction of B


v is the same as the direction of B 
.


.
v is opposite to the direction of B


o
v makes an angle of 45 with the direction of B
.


o
.
v makes an angle of 135 with the direction of B

Q19. A proton travels through both a uniform magnetic field B and a uniform electric field


E . The magnetic field is given by B = 2.5ˆi (mT). At one instant, the velocity of the

proton is v = 2.0 × 103 ˆj (m/s) and the net force acting on it is zero. Find the electric

field E in units of V/m. Ignore the gravitational force on the proton.
A) +5.0 k̂
B) –5.0 k̂
C) +5.0 ĵ
D) –5.0 ĵ
E) –5.0 k̂ +5.0 ĵ
Ans:
�⃗e + F
�⃗B
�⃗net = F
F
�⃗e + F
�⃗B = −F
�⃗e
⇒0= F
�⃗ ⇒ �E⃗ = −�v
qv
�⃗ × �B⃗ = −qE
�⃗ × �B⃗�
= −[(2 ĵ) × (2.5 𝑖𝑖̂)] × 10−3 × 103
= +5.0 𝑘𝑘� (V/m)
Phys102
Term: 112
Final
Saturday, May 19, 2012
Code: 20
Page: 8
Q20. An electron of speed 2.0 × 107 m/s circles in a plane perpendicular to a uniform
magnetic field. The radius of the orbit is 25 cm. The magnitude of the magnetic field
is:
A)
B)
C)
D)
E)
4.6 × 10-4 T
1.6 × 10-6 T
6.3 × 10-6 T
2.0 × 10-5 T
3.2 × 10-4 T
Ans:
magnetic force = centripetal force
qvB =
=
Q21.
mv 2
mv
⇒ B=
R
qR
9.11 × 10−31 × 2 × 107
= 4.6 × 10−4 T
1.6 × 10−19 × 0.25
A 100-turn coil, in the shape of a right triangle, has two equal sides. A current of 3.0
A passes through the coil. A uniform external magnetic field of magnitude 1.0 T is
perpendicular to the hypotenuse of the coil, as shown in FIGURE 6. Find the
magnitude of the total magnetic force on the coil.
Fig # 6
A)
B)
C)
D)
E)
zero
30 N
0.30 N
0.60 N
60 N
Ans:
Net magnetic force on a closed current loop
in a uniform magnetic field is zero.
Phys102
Term: 112
Q22.
Final
Saturday, May 19, 2012
Code: 20
Page: 9
The coil in FIGURE 7 has its plane parallel to the xz plane, and carries current i =
1.0 A in the direction indicated. The coil has 8.0 turns and a cross sectional area of
4.0 × 10-3 m2, and lies in an external uniform magnetic field that is given by

B = −2⋅. 0ˆi (mT). Find the torque (in units of µN.m) on the coil due to the magnetic

field B .
Fig# 7
A)
B)
C)
D)
E)
Ans:
–64 k̂
+64 k̂
+12 î
–12 î
– 64 ĵ
�⃗ = 8 × 1.0 × 4.0 × 10−3 (− ĵ) = −32 × 10−3 ĵ (A. m2 )
�μ⃗ = NiA
�⃗ × �B⃗ = (−32 ĵ × −2 𝑖𝑖̂) × 10−3 × 10−3 = −64 × 10−6 k� (N. m)
�τ⃗ = μ
= −64 k� (μ N. m)
Q23.
Ans:
FIGURE 8 shows cross sections of two long straight wires. The left hand wire
carries current i1 = 5.0 A, directed out of the page. In order to produce a zero net
magnetic field at point P, what should be the current i2?
�⃗2
Fig# 8
𝐵𝐵
A) 10 A, into the page
B) 10 A, out of the page
C) 2.5 A, into the page
D) 2.5 A, out of the page
E) 5.0 A, into the page
�B⃗𝟏𝟏 is down ∴ �B⃗𝟐𝟐 must be up
⇒ 𝑖𝑖2 must be into the page
For cancellation: B1 = B2
μ0 𝑖𝑖1
μ0 𝑖𝑖2
=
⇒ 𝑖𝑖2 = 2𝑖𝑖1 = 10 A
(2𝜋𝜋)(2𝑑𝑑)
2πd
�⃗1
𝐵𝐵
Phys102
Term: 112
Final
Saturday, May 19, 2012
Code: 20
Page: 10
Q24. Three long parallel wires are arranged as shown in FIGURE 9, where d = 50 cm. The
current into the page is i1 = 3.0 A. The currents out of the page are i2 = 0.25 A and i3
= 4.0 A. What is the magnitude of the net force per unit length acting on the wire
carrying current i2 due to the currents in the other wires?
Fig# 9
A)
B)
C)
D)
E)
7.0 × 10-7 N/m
1.0 × 10-7 N/m
1.8 × 10-7 N/m
1.0 × 10-6 N/m
2.4 × 10-7 N/m
F1
F3
Ans:
The two forces are in the same direction.
∴ Fnet = F1 + F3 =
μ0 𝑖𝑖1 𝑖𝑖2 L μ0 𝑖𝑖2 𝑖𝑖3 L
+
2πd
2πd
μ0 𝑖𝑖2
4π × 10−7 × 0.25
(𝑖𝑖
)
(3 + 4)
∴ Force per unit lenght ∶ f =
+ 𝑖𝑖3 =
2π × 0.5
2πd 1
= 7.0 × 10−7 n/m
Q25.
A long straight wire carrying a 3.0-A current enters a room through a window that is
�⃗ . ����⃗
𝑑𝑑𝑑𝑑 around the
2.0 m high and 1.5 m wide. The absolute value of the path integral ∮ 𝐵𝐵
window frame is
A)
B)
C)
D)
E)
3.8 × 10-6
2.5 × 10-7
3.0 × 10-7
2.0 × 10-7
1.6 × 10-5
T.m
T.m
T.m
T.m
T.m
Ans:
����⃗ = 𝜇𝜇0 . 𝑧𝑧𝑒𝑒𝑒𝑒𝑒𝑒 = 4𝜋𝜋 × 10−7 × 3.0 = 3.8 × 10−6 T. m
�⃗ . 𝑑𝑑𝑑𝑑
� 𝐵𝐵
Phys102
Term: 112
Final
Saturday, May 19, 2012
Code: 20
Page: 11
Q26. A current is set up in a wire loop that is formed as shown in FIGURE 10, where R1 =
2.0 cm and R2 = 4.0 cm. The loop carries a current of 5.0 A, as shown in the figure.
What is the magnetic field at the center of the loop (C)?
Fig# 10
A) 3.9 × 10-5 T out of the page
B) 3.9 × 10-5 T into the page
C) 1.2 × 10-4 T out of the page
D) 1.2 × 10-4 T into the page
E) 7.9 × 10-5 T into of the page
Ans:
B1 =
B2 =
𝜇𝜇0 𝑖𝑖𝜙𝜙
⟶ out of the page
4πR1
𝜇𝜇0 𝑖𝑖𝜙𝜙
⟶ into the page
4πR 2
B1 > B2 because R 2 > R1
∴ Bc = B1 − B2 =
=
𝜇𝜇0 𝑖𝑖𝜙𝜙 1
1
� − �
4π R1 R 2
4π × 10−7 × 5 × π 1 1
� − � × 102 = 3.9 × 105 T ⟶ out of the page
4π
2 4
Q27. Two long wires are placed in the xy plane, as shown in FIGURE 11. Each wire
carries a current of 1.5 A, directed out of the page. If the distance d = 3.0 m, what is
the net magnetic field due to these wires at the origin?
Fig# 11
A) (+0.10 î – 0.10 ĵ ) (µT)
B) (+0.10 î + 0.10 ĵ ) (µT)
1
C) (–0.10 î – 0.10 ĵ ) (µT)
D) (–0.10 î + 0.10 ĵ ) (µT)
E) zero
Ans:
�⃗1 and B
�⃗2 are shown
The direction of B
2
�⃗2
B
�⃗net has (+) x − component and (– ) y − component
⟹ B
(Both components are equal in value)
B𝑖𝑖 =
𝜇𝜇0 𝑖𝑖
4𝜋𝜋 × 10−7 × 1.5
=
= 1.0 × 10−7 T = 0.1 μT
2𝜋𝜋𝜋𝜋
2𝜋𝜋 × 3
�B⃗1
Phys102
Term: 112
Final
Saturday, May 19, 2012
Code: 20
Page: 12
Q28. A conducting rectangular loop of wire is placed midway between two long straight
parallel wires as shown in FIGURE 12. The wires carry currents i1 and i2, as
indicated. If i1 is increasing and i2 is constant, then the induced current in the loop
Fig# 12
A)
B)
C)
D)
E)
Ans:
Q29.
is counterclockwise
is zero
is clockwise
depends on the value of i1 − i2
depends on the value of i1 + i2
XXXXX
XXXXX
XXXXX
As i1 increases, Φ𝐵𝐵 increases into the page.
⇒ The induced current must produce a magnetic field that is out of the page.
⇒ Induced current must be counterclockwise.
A 10.0-m long copper wire, with a resistance of 5.00 Ω, is formed into a square loop
and placed with its plane perpendicular to an external magnetic field that is increasing
at the constant rate of 10.0 mT/s, at what rate is thermal energy generated in the loop?
A)
B)
C)
D)
E)
0.780 mW
3.20 mW
4.35 mW
2.50 mW
2.10 mW
l
Ans:
L
wire
l
loop
𝜖𝜖 =
dΦB
d
dB
(BA) = A.
=
dt
dt
dt
⟹ 4𝑙𝑙 = 𝐿𝐿 ⇒ 𝑙𝑙 =
𝐴𝐴 = 𝑙𝑙 2
(2.5)4
ϵ2
A2 dB 2 𝑙𝑙 4 dB 2
P =
=
� � = .� � =
× (10 × 10−3 )2
R
R dt
R dt
5
= 7.8 × 10−4 W = 0.780 mW
𝐿𝐿
= 2.5 m
4
Phys102
Term: 112
Q30.
Final
Saturday, May 19, 2012
Code: 20
Page: 13
In FIGURE 13, a metal rod, of resistance 15 Ω, is forced to move with constant
velocity along two parallel metal rails, connected with a metal strip at one end. The
rod moves in a uniform magnetic field of magnitude 0.50 T that points directly out of
the page. If the rails are separated by L = 25 cm, and the speed of the rod is 0.55 m/s,
what is the current in the rod?
Fig# 13
A)
B)
C)
D)
E)
4.6 mA, up
4.6 mA, down
6.9 mA, up
6.9 mA, down
2.3 mA down
Ans:
The current must be up (clockwise) to reduce the flux.
𝜖𝜖 = B. L. 𝑣𝑣
𝜖𝜖
B. L. 𝑣𝑣
=
𝑅𝑅
𝑅𝑅
0.5 × 0.25 × 0.55
=
15
= 4.6 mA
𝑖𝑖 =