Q1.
A transverse sinusoidal wave, travelling in the positive x direction along a string,
has an amplitude of 20 cm. The transverse position of an element of the string at t =
0, x = 0 is y = + 3.0 cm, and the element has negative velocity. What is the phase constant of this
wave?
A)
B)
C)
D)
E)
Ans:
0.15 rad
6.1 rad
1.4 rad
0.29 rad
0.14 rad
y = ym . cos (kx − ωt + Φ)
v = −ωym . sin (kx − ωt + Φ)
y0
+3.0
y0 = ym . cosΦ ⇒ cosΦ =
=
⇒ Φ = 81.4° = 1.4 rad
20
ym
v0 = −ωym . sinΦ < 0
Q2.
A transverse sinusoidal wave on a string is described by the wave function y (x,t) = 0.15
sin (0.80x – 50t), where x and y are in meters and t is in seconds. The mass per unit length of the string
is 12 g/m. What is the average power transmitted by the wave?
A)
B)
C)
D)
E)
Ans:
v=
Pavg
=
21 W
42 W
10 W
14 W
32 W
ω
50
=
= 62.5m/s
k
0.80
1
= µvω2 ym 2
2
1
× 0.012 × 62.5 × 2500 × 0.0225
2
= 21 W
Phys102
Term: 112
First Major
Wednesday, March 07, 2012
Code: 20
Page: 1
Q3.
Two identical sinusoidal transverse waves travel along a stretched string. Each wave
has an amplitude y m and wavelength λ. If the phase difference between the two waves
is 0.20 λ, what is the amplitude of the resultant?
A)
B)
C)
D)
E)
Ans:
1.6 y m
0.80 y m
0.45 y m
3.5 y m
4.8 y m
0.2λ × 2π
λ → 2π
�=
= 0.4 π rad
0.2λ → ϕ
λ
amp = 2ym . cos
ϕ
= 2ym . cos(0.2𝜋) = 1.6 ym
2
Q4.
A standing wave, having three nodes, is set up on a string fixed at both ends. If the
frequency of the wave is doubled, how many antinodes will there be?
A)
B)
C)
D)
E)
4
3
5
6
7
Ans:
Three nodes ⇒ n = 2
If frequency is doubled ⇒ n = 4 ⇒ 4 antinodes
Phys102
Term: 112
First Major
Wednesday, March 07, 2012
Code: 20
Page: 2
Q5.
A sound wave enters a tube at the source end, as shown in figure 1. At point P, the
sound wave splits into two waves that recombine at point Q. The radius of the
semicircle is varied until the first minimum is observed at the detector when r = 50.0
cm. What is the wavelength of sound?
Fig#1
P
A)
B)
C)
D)
E)
Q
114 cm
228 cm
456 cm
152 cm
304 cm
Ans:
First minimum: ∆L =
λ
2
ΔL = πr − 2r = (π − 2)r
⇒ (π − 2)r =
λ
2
⇒ λ = 2(π − 2)r = 2(π − 2)(50.0) = 114 cm
Q6.
A source emits sound with equal intensity in all directions. If the displacement amplitude is
doubled, the sound level increases by
A)
B)
C)
D)
E)
Ans:
6 dB
3 dB
2 dB
4 dB
5 dB
1
ρvω2 sm 2
2
∴ If sm is doubled ⇒ 𝐼 is multiplied by 4
𝐼2
4𝐼1
Δβ = 10. log � � = 10. log � � = 6 dB
𝐼1
𝐼1
I=
Phys102
Term: 112
First Major
Wednesday, March 07, 2012
Code: 20
Page: 3
Q7.
A tuning fork with a frequency of 512 Hz is placed near the top of the tube shown in
figure 2. The water level is lowered, by opening the valve, so that the length L slowly
increases from an initial value of 20.0 cm. Determine the next value of L that
corresponds to a resonance. Take the speed of sound in air to be 343 m/s.
Fig# 2
A)
B)
C)
D)
E)
Ans:
fn =
nv
4L
⇒ Ln =
50.2 cm
16.7 cm
83.7 cm
117 cm
33.5 cm
nv
343
= (n) �
� = 0.167 n
4f
4 × 512
n = 1 ⇒ L = 0.167 m
n = 3 ⇒ L = 0.502 m
Q8.
A block, with a speaker attached to it, is connected to a spring and oscillates on a
frictionless table between points A and B, as shown in figure 3. The speaker emits
sound that is received by a person. The closest point of the block to the person is A,
the farthest point is B, and O is the equilibrium point of the spring (neither stretched
nor compressed). What position of the speaker corresponds to the highest frequency
observed by the person?
Fig#3
A)
B)
C)
D)
E)
Point O travelling to the right. → moving TOWARD with maximum speed 
Point O travelling to the left.  → moving AWAY with maximum speed
Point A.  → at rest
Point B.  → at rest
The frequency is the same everywhere.  → see above
Phys102
Term: 112
Q9.
First Major
Wednesday, March 07, 2012
Code: 20
Page: 4
A block of ice, with a mass of 6.30 kg at 0.00 oC, is added to a certain amount of
water. The water is cooled from 100 to 0.00 oC. What is the mass of water used?
Assume that all the ice melted.
A)
B)
C)
D)
E)
5.01 kg
2.04 kg
3.42 kg
1.50 kg
6.05 kg
Ans:
melting ice: Q1 = mi Lf = 6.30 × 3.33 × 105 = 2.10 × 106 J
cooling water: Q2 = mw . cw . ∆T = mw . (4190)(0 − 100)
= −0.419 × 106 mw
Q1 + Q2 = 0: Q1 = −Q2
2.10 × 106 = 0.419 × 106 mw
Q10.
⇒ mw = 5.01 Kg
Consider a copper slab of thickness 19 cm and cross sectional area 5.0 m2. Heat is
conducted from the left to the right of the slab at a rate of 1.2 MW. If the temperature
on the left of the slab is 102 oC, what is the temperature on the right of the slab? (The
thermal conductivity of copper is 400 W/m∙K)
A)
B)
C)
D)
E)
Ans:
Pcond =
∴ ∆T =
= 114℃
o
–12 C
+12 oC
–20 oC
+20 oC
+42 oC
k. A. ∆T
L
Pcond . L
1.2 × 106 × 0.19
=
k. A
400 × 5.0
TL − TR = 114 ⇒ TR = TL − 114 = 102 − 114 = −12℃
102 ℃
Q
Phys102
Term: 112
Q11.
First Major
Wednesday, March 07, 2012
Code: 20
Page: 5
Consider benzene liquid filling completely a steel container of volume 100 cm3 at
20.0 oC. Calculate the coefficient of volume expansion of benzene when 0.372 cm3 of
benzene spills over from the tank as its temperature is raised to 50.0 oC. (Neglect the
thermal expansion of the container)
A)
B)
C)
D)
E)
Ans:
1.24 × 10-4 /Co
3.68 × 10-4 /Co
5.32 × 10-4 /Co
2.34 × 10-5 /Co
4.33 × 10-4 /Co
∆V = β. V0 . ∆T
∴β=
Q12.
∆V 1
0.372
1
.
=
×
= 1.24 × 10−4 (℃)−1
V0 ∆T
100
30.0
Which of the following statements are WRONG?
1. Two objects are in thermal equilibrium if they have the same temperature.
2. In an isothermal expansion, the work done by the gas is equal to the change in
its internal energy.  ∆T = 0 ⇒ ∆Eint = 0 but W ≠ 0
3. In an adiabatic process, heat lost by the gas is equal to the work done on the
gas.  adiabatic: Q = 0
4. In an isochoric process, heat exchanged is equal to the change in internal
energy of the gas. W = 0
A)
B)
C)
D)
E)
Q13.
2 and 3
1 and 2
1 and 3
1 and 4
2 and 4
Phys102
Term: 112
First Major
Wednesday, March 07, 2012
Code: 20
Page: 6
Consider an isothermal compression of an ideal gas at a temperature of 0.00 oC. The
initial pressure of the gas is 1.0 atm and the final volume is 1/5 the initial volume.
Find the number of moles of the gas if it loses 365 J of heat during the process.
A)
B)
C)
D)
E)
0.10 moles
0.63 moles
2.0 moles
3.4 moles
1.5 moles
Ans:
Isothermal: T = constant ⇒ ∆Eint = 0 ⇒ Q = W
Q = nRT𝑙n(Vf /Vi )
n=
Q14.
Q
−365
=
= 0.10
8.31 × 273.15 × 𝑙n(1/5)
RT. 𝑙n(Vf /Vi )
For a temperature increase of ΔT, a certain amount of a monatomic ideal gas requires
50 J of heat when heated at constant volume, and 83 J of heat when heated at constant
pressure. How much work is done by the gas in the second situation?
A)
B)
C)
D)
E)
Ans:
33 J
130 J
67 J
90 J
50 J
Qp = nCp ∆T
� Q − QV = n�Cp − CV �∆T = nR∆T
QV = nCV ∆T p
Second Process: isobaric ⇒ W = p∆V = nR∆T
∴ W = Qp − QV = 83 − 50 = 33 J
Q15.
Phys102
Term: 112
First Major
Wednesday, March 07, 2012
Code: 20
Page: 7
Figure 4 shows a cycle undergone by 1.0 mole of an ideal diatomic gas. The
temperatures are T 1 = 400 K, T 2 = 700 K, and T 3 = 555 K. Calculate the net work
done in one cycle.
Fig# 4
A) 1.7 kJ by the gas
B) 1.7 kJ on the gas
C) 3.8 kJ on the gas
D) 3.8 kJ by the gas
E) 0.52 kJ by the gas
Ans:
Wnet = Q31 + Q12 {Q23 = 0}
= ncp (T1 − T3 ) + ncV (T2 − T1 )
= �1 ×
7R
5R
(T1 − T3 )� + �1 ×
(T − T1 )�
2
2 2
R
R
(7T1 − 7T3 + 5T2 − 5T1 ) = (2T1 + 5T2 − 7T3 )
2
2
8.31
(800 + 3500 − 3885) = +1724 J
2
=
= +1.7k J
by the gas
Q16.
The temperature of an ideal gas is reduced from 100 oC to 20 oC. The rms speed of its
molecules decreases by:
A) 11%
B) 22%
C) 47%
D) 55%
E) 71%
Ans:
3RT
3R
vrm𝑠 = �
= � . √T
M
M
% difference =
√293 − √373
√373
v2 − v1
�T2 − �T1
× 100 =
× 100
v1
�T1
× 100 = −11.4% → 11%
Phys102
Term: 112
First Major
Wednesday, March 07, 2012
Code: 20
Page: 8
Q17.
Two moles of a monatomic ideal gas are taken through the cycle shown in figure 5.
The change in entropy in going from b to c is:
Fig# 5
A)
B)
C)
D)
E)
+ 17.3 J/K
– 17.3 J/K
– 2.17 J/K
– 34.6 J/K
+ 34.6 J/K
Ans:
b → c ∶ isochoric
dQ
ncV dT
Tc
=�
= ncV . 𝑙n � �
T
Tb
T
pc
= n. cV . 𝑙n � �
pb
2p0
J
= 2 × 1.5 × 8.31 × 𝑙n �
� = +17.3
p0
K
∆s = �
Q18.
In four experiments, 3 moles of an ideal gas undergo reversible isothermal
expansions, starting from the same volume but at different temperatures. The
corresponding p-V plots are shown in figure 6. Rank the four situations a, b, c, and d
according to the change in the entropy
of
the gas, greatest first.
Fig#6
A)
B)
C)
D)
E)
Ans:
(a and c) tie, then (b and d) tie.
a, then b, then c, then d.
d, then c, then b, then a.
(b and d) tie, then (a and c) tie.
(a and d) tie, then (b and c) tie
𝑉𝑓
Isothermal: ∆s = nR. 𝑙n � �
𝑉𝑖
⇒ a and c: same ∆S
b and d: same ∆S
AND: ΔSa,c > ΔSb,d
Phys102
Term: 112
Q19.
First Major
Wednesday, March 07, 2012
Code: 20
Page: 9
A Carnot engine whose cold reservoir is at 15 oC has an efficiency of 34%. Then, the
temperature of the hot reservoir is fixed while that of the cold reservoir is decreased.
What should the temperature of the cold reservoir be in order to make the efficiency
of this engine equal to 36%?
A)
B)
C)
D)
E)
Ans:
6.3 o C
280 oC
160 oC
7.2 oC
12 oC
TL
TL
→
= 1 − 𝜀 ⇒ TL = TH (1 − 𝜀)
TH
TH
1 − 𝜀2 1 − 0.36
= TH (1 − 𝜀2 ) TL2
�
=
=
= 0.97
= TH (1 − 𝜀1 ) TL1
1 − 𝜀1 1 − 0.34
𝜀 =1−
TL2
TL1
⇒ TL = (0.97)(15 + 273) = 279 K = 6.3℃
Q20.
A Carnot air conditioner takes heat from a room at 21 °C and transfers it to the
outdoors, which is at 35 °C. For each two joules of electric energy required to operate
the air conditioner, how many joules are removed from the room in the form of heat?
A)
B)
C)
D)
E)
Ans:
42 J
21 J
1.7 J
3.3 J
2.0 J
TL
21 + 273
=
= 21
TH − TL
35 − 21
QL
K=
→ QL = K. W = 21 × 2 = 42 J
W
K=