Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 1
Q1.
The distance between two successive minima of a transverse wave is 2.76 m. Six
crests of the wave pass a given point along the direction of travel every 14.0 s. Find
the wave speed?
A) 0.986 m/s
B) 0.892 m/s
C) 0.267 m/s
D) 1.012 m/s
E) 0.768 m/s
Ans:
f 
5 .0 0
 0 .3 5 7 H z  v   f  ( 2 .7 6 m ) ( 0 .3 5 7 H z )  0 .9 8 5 7 m /s
1 4 .0 s
Q2.
The speed of waves on a thin wire is 150 m/s. The density of the material that the
wire is made of is 5000 kg/m3. The wire has a 0.500 mm diameter circular crosssection. What is the tension in the wire?
A)
B)
C)
D)
E)
22.1 N
76.2 N
88.4 N
0.147 N
63.7 N
Ans:
ρ  m a s s /V o lu m e  m /(A L )=  /A
v 

  A


    v   A 1 5 0 
2
2
 0 .5  1 0
 5000   
2

3



2
1 5 0 
2
 2 2 .0 9 N
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 2
Q3.
Two identical waves, with amplitude A, travel simultaneously through the same
medium and in the same direction with phase difference  . If the amplitude of the
resulting superposition is A/2, what is the possible value of  ?
151o
45o
37o
55o
91o
A)
B)
C)
D)
E)
Ans:
y m  2 y m c o s


2
A
 2 A cos
2

   2 cos
1
2
1
 2 .6 3 6 r a d  1 5 1
o
4
Q4.
Figure 1 shows two loudspeakers, (1) and (2), above each other. They are driven by
the same source at frequency of 450 Hz. An observer is sitting at point O, at the same
distance from each speaker. What minimum upward vertical distance speaker (1)
should be moved to in order to create destructive interference at point O? [Note: speed
of the sound in air is 343 m/s]
Figure 1
A)
B)
C)
D)
E)
1.43
2.80
1.22
1.01
2.15
m
m
m
m
m
Ans:
 
v

3 4 3 m /s
f
 0 .7 6 2 m
450 H z
r1  r2 
1 .5 0  8 .0 0
2
 8 .1 4 m
2
T o c r e a te d e s tr u c tiv e in te r f e r e n c e a t p o in t O ,
r1  r2 
r1 

2
 0 .3 8 1 m  r1  r2  0 .3 8 1  8 .5 2 m
(1 .5 0  y )  8 .0 0
2
2
 8 .5 2  y 
8 .5 9  1 .5 0  1 .4 3 m
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 3
Q5.
At a distance of 2.0 m from a point source of sound, the sound level is 80 dB. What
will be the sound level at a distance of 4.0 m from this source? (Assume that the point
source radiates uniformly in all directions)
A)
B)
C)
D)
E)
74 dB
83 dB
62 dB
16 dB
12 dB
Ans:
 I 
  
 I0 
  1 0 lo g 
 I2 
 r1 
 r1 
  1  1 0 lo g 
  1 0 lo g  2   2 0 lo g 

2
 I1 
 r2 
 r2 
2
 r1 
 2m 
  8 0  2 0 lo g 
  74 dB
 4m 
 r2 
 2   1  2 0 lo g 
Q6.
A fixed alarm is emitting sound waves of frequency 520 Hz. You are on a motorcycle,
traveling directly away from the alarm. How fast you must be traveling if you detect a
frequency of 490 Hz? [Note: Speed of the sound in air = 343 m/s]
A)
B)
C)
D)
E)
19.8 m/s
22.2 m/s
27.1 m/s
11.9 m/s
16.3 m/s
Ans:
f f (
v  vd
v
)  v d  v (1 
f
f
)  3 4 3 (1 
490
520
)  1 9 .8 m /s
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 4
Q7.
A 0.90 m long pipe is open at one end but closed at the other end. If it currently
resonates with a harmonic of wavelength 0.72 m, what is the wavelength of the next
higher harmonic in this pipe?
A)
B)
C)
D)
E)
0.51 m
0.33 m
0.21 m
0.74 m
0.88 m
Ans:
 
4L
, n  1, 3 , 5 , 7 , .......
n
0 .7 2  0 .9 0
 n  4
n  5   

4L

4  0 .9
n
4  0 .9
T h e n e x t n is n  7   
 0 .7 2 m
5
 0 .5 1 m
7
Q8.
A point source, of sound waves, radiates uniformly in all directions. At a distance of
20 m from the source the sound level is 51 dB. What is the total power output of the
source?
A)
B)
C)
D)
E)
6.3×10-4 Watt
9.6×10-4 Watt
1.1×10-4 Watt
9.8×10-5 Watt
4.2×10-5 Watt
Ans:
P  4 r I  4 r  I 0  1 0
2
2

10
 4  2 0  1 0
2
12
 10
51
10
 6 .3  1 0
4
W a tt
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 5
Q9.
The coefficient of linear expansion of iron is 10–5 °C-1. The volume of an iron cube,
having an edge of 5.00 cm, will increase if it is heated from 10.0 °C to 60.0 °C by
A)
B)
C)
D)
E)
0.188 cm3
0.375 cm3
0.225 cm3
0.750 cm3
0.625 cm3
Ans:
α = 10−5 ; L = 5; ∆T = 60 − 10; v0 = L3 ;
∆v = 3αv0 ∆T = 0.188 cm3
Q10.
In an insulated container, 250 g of ice at 0 oC are added to 500 g of water at 18 oC.
What is the final temperature of the system?
A)
B)
C)
D)
E)
0 oC
5 oC
25 oC
17 oC
100 oC
Ans:
QL(lost by ice) = mL = 250 g ⨯ 333 J /g = 83250 J
QG(gained by water) = M c dT = 500 g ⨯ 4.190 (18 − 0) J /g = 37710 J
∵ QL > QG
, so we still have some ice not melted ⇒ T = 0 ℃
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 6
Q11.
A system of an ideal gas undergoes the cyclic process shown in the Figure 2.
Calculate the work done by the system along the path XY.
A)
B)
C)
D)
E)
+90 J
−60 J
+60 J
zero
−90 J
Ans:
W = are under the curve
1
= 10(4 − 1) + (50 − 10)(4 − 1)
2
1
= 10(3) + (40)(3) = 30 + 60 = 90
2
J
Q12.
A bar of gold is in thermal contact with a bar of silver having the same length and
area, see Figure 3. One end of the connected bars is maintained at 80.0 oC and the
opposite end is at 30.0 oC. When the energy transfer reaches steady state, what is
𝑊
the temperature at the junction? [The thermal conductivity for gold = 314
and
the thermal conductivity for silver = 427
A) 51.2 oC
B) 58.8 oC
C) 70.8 oC
D) 33.4 oC
E) 42.7 oC
Ans:
kg
∆Tg A
∆Ts A
= ks
L
L
k g (80 − Tj ) = k s (Tj − 30)
k g (80) = k s (30) = Tj (k s + k g )
Tj =
k g (80) + k s (30)
= 51.2℃
ks + kg
𝑊
𝑚∙℃
𝑚∙℃
]
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 7
Q13.
Two moles of an ideal gas are in a 6.0×10-3 m3 container at a pressure of 5.0×105 Pa.
Find the average translational kinetic energy of a single molecule.
A) 3.7×10-21 J
B) 7.5×10-21 J
C) 9.4×10-21 J
D) 0.22×10-21 J
E) 5.7×10-21 J
Ans:
K avg =
3
kT
2
PV = nRT
6 × 105 × 5 × 10−3
T=
2 × 8.31
K avg =
3
× 1.38 × 10−23 × T = 3.7 × 10−21 J
2
Q14.
Which one of the graphs in Figure 4 best represents the variation of pressure with
volume for an isothermal process of an ideal gas?
A) 3
B) 2
C) 1
D) 5
E) 4
Ans:
A
Q15.
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 8
One mole of an ideal monatomic gas is initially at 300 K and 1.0 atm. The gas is
compressed adiabatically to 2.0 atm. What is the final volume of the gas?
A) 0.016 m3
B) 0.037 m3
C) 0.056 m3
D) 0.025 m3
E) 0.012 m3
Ans:
PV γ = constant
Pi (
nRTi γ
γ
) = Pf Vf
Pi
Vf =
1
1 × 8.31 × 300
(
)
21/γ
1.01 × 105
Vf = 0.016m3
Q16.
Find the change in entropy of a 100 g of ice at 0 oC that is isobarically heated slowly
to reach 80 oC water. [The heat of fusion for ice LF = 80 cal/g and the specific heat of
water 𝑐𝑤 = 1.0 cal/g.K].
A) 55 cal/K
B) 12 cal/K
C) 62 cal/K
D) 35 cal/K
E) 85 cal/K
Ans:
∆S =
=
Q17.
mL
Tf
+ mcln
T
Ti
100 × 80
273 + 80
+ 100 × 1 ln
= 55 cal/k
273
273
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 9
A sample of an ideal monatomic gas undergoes the reversible process from A to B as
displayed in the T-S diagram shown in Figure 5. The process is:
A) an isothermal expansion
B) a free expansion.
C) an isothermal compression.
D) a change of phase.
E) a constant-volume process.
Ans:
A
Q18.
Five moles of an ideal monatomic gas are allowed to expand isobarically. The initial
volume is 20.0 cm3 and the final volume is 100 cm3. Find the change in entropy of
the gas.
A) 167 J/K
B) 100 J/K
C) 66.9 J/K
D) 234 J/K
E) 33.4 J/K
Ans:
∆S = nR ln
Vf
Vi
+ nCv ln
Tf
Ti
pV = nRT ⇒
∆S = nRln
Vf
Vf
+ nCv ln
Vi
Vi
= n(𝑅 + Cv )ln
Q19.
Vf Tf
=
Vi Ti
Vf
5
100
= 5 ( ) (8.31)ln
= 167
Vi
2
20
J /K
Phys102
Coordinator: Prof. I. Nasser
First Major-151
Sunday, October 25, 2015
Zero Version
Page: 10
A Carnot heat engine absorbs 70.0 kJ as heat and expels 55.0 kJ as heat in each cycle.
If the low-temperature reservoir is at 120 oC, find the temperature of the hightemperature reservoir.
A)
B)
C)
D)
E)
227 oC
500 oC
153 oC
773 oC
35.9 oC
Ans:
|𝑄𝐶 | 𝑇𝐶
=
|𝑄𝐻 | 𝑇𝐻
55
273 + 120
=
70
273 + 𝑡ℎ
𝑡ℎ =
70
(273 + 120) − 273 = 227 ℃
55
Q20.
Is it possible to transfer energy as heat from a low-temperature reservoir to a hightemperature reservoir? Choose the right answer with the right explanation.
A) Yes, this is what a refrigerator does, but work must be done on the
refrigerator to make this happen.
B) No, this violates the second law of thermodynamics, if no work is being
involved.
C) No, this violates the zero’s law of thermodynamics.
D) Yes, this is what a heat engine does, and it can happen without the engine
doing work.
E) Yes, this is what a refrigerator does, and it can happen without doing work
on the refrigerator.
Ans:
A.