SP015 Past Years PSPM Chapter 6 -14 Ver 2020

PSPM SP015



PSPM 2006/2007 SF017/2 No. 6

43. What is the length of a bottle that has a natural fundamental frequency of 440 Hz?
-1
(Speed of sound = 343 m s ) [3 m]



PSPM 2009/2010 SF017/2 No. 5

44.


34 cm






FIGURE 10.6
FIGURE 10.6 shows a narrow cylindrical tube filled with water and forming a 34 cm
air column. When blown at the open end, a loud sound is heard.

(a) Explain the production of the loud sound. [2 m]
-1
(b) Given the velocity of sound in air is 340 m s , calculate the frequency of the loud
sound. [2 m]



PSPM 2016/2017 SF016/2 No. 7(a)(iii)

45. The first overtone standing wave is formed in a 30 cm closed pipe. Sketch the wave and
-1
calculate its frequency. The speed of sound is 340 m s . [3 m]



Doppler Effect

PSPM 2016/2017 SF016/2 No. 7(b)(i)

46. What is Doppler effect for sound wave? [1 m]



PSPM JUN 1999/2000 SF035/2 No. 10(b)(i)

47. When a police patrol car sounds its siren and approaches an observer at rest, the
frequency of sound heard by the observer is higher than its original frequency. Explain
this statement. [2 m]




51

PHYSICS PSPM SEM 1 1999 - 2017


PSPM JUN 2000/2002 SF035/2 No. 10(c)(ii)

48. f

660


550

x
Observer’s position

FIGURE 10.7

A patrol car emitting siren of frequency 600 Hz is approaching, passing, and moving
away from an observer at rest. The changes of frequency heard by the observer are
shown in FIGURE 10.7. What is the frequency heard by the observer when the patrol
car passes him? Determine the speed of the patrol car.
-1
(Speed of sound in air = 330 m s ) [8 m]



PSPM 2003/2004 SF017/2 No. 5(b)

49.


S 1 S 2

P
FIGURE 10.8

FIGURE 10.8 shows two identical sources S 1 and S 2 which are moving towards an
observer P. The two sound sources S 1 and S 2 produce notes of frequency 500 Hz. If the
-1
-1
speed of S 1 and S 2 are 20 m s and 23 m s respectively, calculate the beat frequency
-1
heard by P. (Speed of sound = 340 m s ) [4 m]



PSPM 2004/2005 SF017/2 No. 12(d)

50. The siren of a police car emits sound at frequency f. If the car is moving with speed v s
toward a stationary observer, the frequency heard by the observer is given by
f
f '
  1 v s  
 v 

where v is the speed of sound in air. Show that the frequency of the sound heard by the
observer increases.
[1 m]



52

PSPM SP015



PSPM 2007/2008 SF017/2 No. 13(a)

51. A siren of an ambulance emits a sound of frequency 800 Hz. The ambulance moves at
-1
25 m s away from a stationary observer and towards the wall of a building. Calculate
-1
the beat frequency heard by the observer. Speed of sound is 330 m s . [5 m]



PSPM 2011/2012 SF016/2 No. 7(b)(ii)

-1
52. What frequency will be heard by a moving observer travelling at 25 m s towards a
stationary siren that emits a sound of frequency 280 Hz?
-1
(Velocity of sound in air = 343 m s ) [2 m]



PSPM 2015/2016 SF016/2 No. 7(b)

53. A man stands by the roadside when an ambulance passes by him with constant velocity
-1
18 m s . The ambulance emits siren with frequency 256 Hz. The speed of sound is
-1
340 m s .

(a) Calculate the apparent frequency of the siren heard by the man before the
ambulance passes him. [2 m]

(b) Sketch a graph of the apparent frequency against distance travelled. [1 m]

(c) Sketch a graph of siren intensity against distance. [1 m]



PSPM 2016/2017 SF016/2 No. 7(b)(ii)

-1
54. A car is travelling at 25 m s emits a sound of frequency 1100 Hz approaches a
stationary observer. Calculate the apparent frequency of the sound heard by the
-1
observer. The speed of sound is 340 m s . [2 m]



PSPM 2017/2018 SF016/2 No. 7(c)

-1
55. A submarine that travels in the water at a speed of 8 m s and emits a sonar wave of
frequency 1400 Hz is approaching another stationary submarine. Given the speed of
-1
sound in water is 1530 m s , calculate the apparent frequency as detected by an
observer in the stationary submarine. [2 m]








53

PHYSICS PSPM SEM 1 1999 - 2017


PSPM CHAPTER 11: DEFORMATION OF SOLIDS
___________________________________________________________________________


Stress and Strain


PSPM 2011/2012 SF016/2 No. 7(c)(i)

1. Define stress and strain. [2 m]



PSPM 2014/2015 SF016/2 No. 7(c)(i)

2. Explain plastic deformation of an elastic material. [2 m]



PSPM 2015/2016 SF016/2 No. 7(c)

3. Why is Hooke’s law not applicable in the case of plastic deformation? [2 m]



PSPM 2007/2008 SF017/2 No. 10(c)

4. State ONE physical property of a material that is used as car bumper. Explain your
answer. [4 m]



PSPM 2012/2013 SF016/2 No. 7(d)(ii)

5. Sketch a labeled stress-strain graph of a ductile material. [2 m]





Young’s Modulus



PSPM JUN 2000/2002 SF025/2 No. 10(a)(ii)

6. Define Young’s modulus and state its validity condition. [2 m]






54

PSPM SP015



PSPM JUN 2000/2002 SF025/2 No. 10(b)

7.
load F (N)
A
300


200

100

2 4 6 8 10 12 elongation (mm)
0
FIGURE 11.1


FIGURE 11.1 shows an elongation of a metal wire with diameter 1.5 mm and original
length 1.0 m exerted by a load F until it breaks at A. Base on the graph,

(a) determine the stress on this wire when it breaks. [3 m]

(b) calculate the Young’s modulus of the wire. [4 m]




PSPM 2003/2004 SF017/2 No. 6

8. tension (N)

80

60
40

20

0 elongation (mm)
1 2 3 4 5 6 7 8 9

FIGURE 11.2

FIGURE 11.2 shows the elongation of a wire as tension is increased. Calculate the
work done to elongate the wire to 9 mm. [3 m]








55

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2004/2005 SF017/2 No. 6

9.

0.2 m

10.0 kg


0.2 m


5.0 kg

FIGURE 11.3

FIGURE 11.3 shows a 10.0 kg weight suspended by a 0.2 m vertical copper wire with
2
cross-sectional area 5 mm . A 5.0 kg weight is suspended by a similar wire from the
bottom of the 10.0 kg weight. Calculate

(a) the strain of each wire. [2 m]

(b) the extension of each wire. [2 m]

11
-2
(Young’s Modulus of copper is 1.10  10 N m )




PSPM 2006/2007 SF017/2 No. 13(c)(ii)

10. A wire of length L and cross-sectional area A elongates by e when a tensile force F acts
on it. If k and Y are the force constant and the Young’s modulus of the wire,
respectively, obtain an expression for k in terms of Y. [3 m]




PSPM 2008/2009 SF017/2 No. 6

2
11. A 4.0 m iron rod with cross-sectional area 0.5 cm extends by 1.0 mm when a 225 kg
mass is suspended from one of its ends. Calculate the

(a) Young’s modulus of the rod. [2 m]

(b) elastic energy stored in the rod. [2 m]







56

PSPM SP015



PSPM 2010/2011 SF017/2 No. 6

12. A spring stretches by 4 mm when a 1.5 kg mass is suspended at its end. Calculate the
spring constant. [3 m]



PSPM 2010/2011 SF017/2 No. 13(b)

13.

A


5 cm
F 150 cm



B



FIGURE 11.4

A wire AB, of length 150 cm and diameter 1 mm is fixed at both ends. A force F pulls
the wire at the midpoint and causes a displacement of 5 cm as shown in FIGURE 11.4.
If the Young modulus of steel is 2 GPa, calculate the

(a) magnitude of F. [7 m]

(b) energy stored in the wire. [2 m]



PSPM 2011/2012 SF016/2 No. 7(d)

14. A 20.0 kg mass is hung from a 2.0 m long vertical wire. If the wire is elongated by
3.0 mm, calculate the strain energy stored in the wire. [2 m]



PSPM 2012/2013 SF016/2 No. 7(d)(iii)

2
15. A wire fixed to a ceiling has a length of 1.5 m and cross-sectional area 0.2 cm . The
wire stretches by 0.15 cm when a 50 kg load is attached to its free end. Calculate the
Young’s modulus of the wire. [2 m]






57

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2013/2014 SF016/2 No. 7(d)

16. F F



material 1 material 2



FIGURE 11.5

FIGURE 11.5 shows an identical force F acting on two identical rods but made of
different materials. What concept will be used to determine which rod will bend more?
Explain your answer. [2 m]



PSPM 2014/2015 SF016/2 No. 7(c)(ii)

11
-2
17. A wire of diameter 0.5 mm has Young’s modulus 2  10 N m . Calculate the strain if
it is extended by 150 N load. [3 m]



PSPM 2015/2016 SF016/2 No. 7(c)

18. A 50 cm wire has Young’s modulus 175 GPa and diameter 0.25 mm. Calculate the
force needed to elongate the wire 1.5 mm. [3 m]



PSPM 2016/2017 SF016/2 No. 7(c)

5
19. A solid cylinder 10 m high and 10 cm in diameter is compressed by a 1  10 kg load.
Calculate the strain energy stored in the cylinder. The Young’s modulus of cylinder is
11
1.9  10 Pa. [4 m]



PSPM 2017/2018 SF016/2 No. 7(d)

20. A 20 cm cylindrical brass rod with diameter 6 cm is held vertically on its one circular
flat end. A load of 5 kg is placed on its upper end. Given the Young’s modulus of brass
-2
10
is 9.1  10 N m , calculate the strain energy of the rod. [4 m]








58

PSPM SP015



PSPM CHAPTER 12: HEAT CONDUCTION AND
THERMAL EXPANSION
___________________________________________________________________________


Conduction

PSPM 2017/2018 SF016/2 No. 8(a)(i)

1. Define heat. [1 m]




PSPM JUN 1999/2000 SF015/2 No. 6(a)

2. Define thermal conductivity. [1 m]




PSPM JAN 1999/2000 SF015/2 No. 13(c)(i)

3. State the factors that influence the thermal conductivity of a conductor. [1 m]




PSPM JAN 1999/2000 SF015/2 No. 13(c)(ii)

4. On the same axes, sketch the temperature versus length graph for an insulated
conductor and non-insulating conductor. [3 m]




PSPM 2017/2018 SF016/2 No. 8(a)(ii)

5.

27 C 100 C


FIGURE 12.1

FIGURE 12.1 shows a rod with both ends at different temperatures. The right half of
the rod is insulated while the left half is not insulated. Sketch the graph of temperature
against distance of the rod. [2 m]





59

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2002/2003 SF017/2 No. 7

6.

X Y
Q Cold
Hot P

FIGURE 12.2

A metal bar XY contains a part of other metal PQ as shown in FIGURE 12.2. Sketch
temperature versus length graph for the bar if it is

(a) insulated. [2 m]

(b) not insulated. [2 m]



PSPM JUN 1999/2000 SF015/2 No. 6(b) Edited

3
7. A water tank of measurement 2  2  1 m is filled up at temperature 96 C. The tank is
1 cm thick. Calculate the rate of heat loss through conductivity on each wall of the tank
if the temperature outside the tank is 22 C. Given thermal conductivity of the tank is
-1
-2
-1
3.78  10 W m K . [4 m]



PSPM JUN 2000/2002 SF015/2 No. 13(b)

2
8. A cylindrical container with a cross-sectional area of 2.0 m and its bottom made from
2.0 cm thick iron contains water at temperature 100 C. This cylinder is then placed on
a heater where 1 kg of water is evaporated every 5 minutes.

6
-1
(a) If latent heat of vaporization L = 2.26  10 J kg , calculate the heat provided to
the water in 1 second. [3 m]

-1
-1
-1
(b) If thermal conductivity of iron is given as 50.2 J s m K , what is the
temperature at the bottom surface of the cylinder that touches the heater? [3 m]



PSPM 2004/2005 SF017/2 No. 7

2
9. A slab of thermal insulator has cross-sectional area 200 cm and thickness 5.0 cm. Its
-1
-1
thermal conductivity is 0.2 W m C . If the temperature difference between opposite
surfaces is 75 C, what is the total heat flow through the slab in 1 hour? [3 m]






60

PSPM SP015



PSPM 2005/2006 SF017/2 No. 7

10.


80 C M N 20 C




FIGURE 12.3

Two metallic rods M and N with similar length and cross-sectional area are joined
together and insulated. At steady state, the temperatures at their ends are shown in
FIGURE 12.3. The thermal conductivity of M is twice that of N. Calculate the
temperature at their adjoint end. [3 m]




PSPM 2007/2008 SF017/2 No. 7

11. How much heat flows through a sheet of copper 1.2 m × 0.8 m of thickness 3.0 mm for
1 hour if the temperatures of the surfaces are 30 C and 28 C respectively? (Given
-1
-1
thermal conductivity of copper = 401 W m K .) [4 m]




PSPM 2011/2012 SF016/2 No. 8(b)

12. An insulated spherical glass bulb of radius 2.5 cm, thickness 0.4 mm whose outer
surface temperature is 30 C dissipates 55 W of heat.

-1
-1
-1
(a) Given the thermal conductivity of glass is 0.84 J s m C . Calculate the
temperature of the inner surface of the bulb. [3 m]

(b) Sketch a labelled temperature against distance graph of the heat conduction
through the bulb. [2 m]




PSPM 2013/2014 SF016/2 No. 8(b)(i)

2
13. Calculate the heat transferred in 24 hours through a 2.4 m metal sheet of thickness
1 cm when the temperature difference between the surfaces is 0.5 C. Given the thermal
-1
-1
conductivity coefficient of the metal is 16 W m K . [2 m]




61

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2015/2016 SF016/2 No. 8(a)(ii)

14. copper iron

160 C 40 C


8 cm 5 cm

FIGURE 12.4

A composite rod is made by joining a copper rod of diameter 4 cm with an iron rod of
similar diameter. The rod is insulated and its ends are kept at two different temperatures
as shown in FIGURE 12.4. The coefficient of thermal conductivity of copper and iron
-1
-1
-1
-1
are 385 W m K and 80 W m K respectively. Determine the temperature at the
joint. [4 m]



Expansion

PSPM 2011/2012 SF016/2 No. 8(a)(i)

15. Define the coefficient of linear thermal expansion. [1 m]


PSPM 2014/2015 SF016/2 No. 8(a)

16. Define the coefficient of area thermal expansion,  . [1 m]


PSPM 2015/2016 SF016/2 No. 8(a)

17. Define the coefficient of volume thermal expansion. [1 m]


PSPM JAN 1999/2000 SF015/2 No. 6

3
18. Volume of mercury in a thermometer bulb at a temperature of 10 C is 0.11 cm . If the
2
cross-sectional area of the capillary tube is 0.01 mm , how high will the mercury rise at
room temperature of 27 C?
-4
-1
(Coefficient of volume thermal expansion of mercury = 1.8  10 K ) [4 m]


PSPM JUN 1999/2000 SF015/2 No. 13(a)(i)

3
19. A 1000 cm glass thermos is filled with mercury at temperature 0 C. When it is heated
3
o
at temperature 100 C, 15 cm of mercury spilt out. Determine the coefficient of volume
expansion of mercury by assuming that the glass thermos does not expand. [2 m]


62

PSPM SP015



PSPM JUN 2000/2002 SF015/2 No. 13(c)

20. When metal rod A of 100 cm long is heated from 0 C to 100 C, it experiences
elongation of 0.080 cm. While metal B of the same length will experience elongation of
0.035 cm at the same temperature range.

(a) Calculate the coefficient of linear expansion of metal A and metal B. [2 m]


(b) 100 cm



Metal A Metal B
FIGURE 12.5

If metal rod C is a combination of both metals A and B as shown in
FIGURE 12.5 experiences elongation of 0.070 cm when heated from 0 C to
100 C, determine the original length of both metals. [4 m]




PSPM 2001/2002 SF015/2 No. 7

21. A man uses a measuring tape made from iron. If the length of the tape is 500.00 cm at
room temperature 25 C, calculate the extension of the length of the tape at temperature
35 C.
-1
-5
(Coefficient of linear expansion of iron,  iron = 1.2  10 K ) [3 m]



PSPM 2006/2007 SF017/2 No. 7(a)

22. The surface of a space shuttle is hot due to air resistance during the re-entry into the
Earth atmosphere. The temperature of the shuttle during this re-entry is 215 C.
Calculate the fractional change in its length if the take-off temperature of the shuttle is
-1
-6
23 C and its coefficient of linear expansion is 24  10 K . [2 m]



PSPM 2008/2009 SF017/2 No. 7

-5
-1
23. The coefficient of linear expansion of aluminium is 2.6 × 10 K .

(a) What is meant by this value? [1 m]

(b) An aluminium cube with side 0.2 m is heated from 20 C to 80 C. Calculate the
increase in volume of the cube. [3 m]




63

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2009/2010 SF017/2 No. 7(b)

24.

1.5 m





gap = 5.0 mm (the scale is exaggerated)

FIGURE 12.6

A 1.5 m steel rod is attached to a fixed wall as shown in FIGURE 12.6. At 30 C, a gap
between the rod and the opposite wall is 5.0 mm. By neglecting the expansion of the
walls, at what temperature will the gap close? The coefficient of linear expansion for
-1
-6
steel is 12 × 10 C . [3 m]




PSPM 2011/2012 SF016/2 No. 8(a)(ii)

25. A pair of eyeglass frame is made of epoxy plastic. At 20 C, the frame has circular lens
holes of radius 2.20 cm. If the coefficient of linear expansion of the epoxy is
-1
-4
3.0  10 C , to what temperature must the frame be heated so that a lens of radius
2.21 cm fits the frame? [3 m]




PSPM 2012/2013 SF016/2 No. 8(b)

26. The length of a metal bar is 300 cm at –30 C. Given the coefficient of linear expansion
-6
-1
of the metal bar is 11  10 C , calculate the length of the bar at 80 C. [3 m]

















64

PSPM SP015



PSPM 2014/2015 SF016/2 No. 8(b)

27.
brass plate


aluminium
rod



0.02 mm

FIGURE 12.7

FIGURE 12.7 shows an aluminium rod with radius 5 cm and having a clearance of
0.02 mm completely around it within a hole in a brass plate at 20 C. The coefficient of
-1
-5
-1
-5
linear thermal expansion of brass and aluminium are 1.9  10 C and 2.4  10 C
respectively.

(a) Calculate the lowest temperature if both metals are heated until the clearance is
zero. [3 m]

(b) Would such a tightly fit be possible if the plate is aluminium and the rod is brass
and both metals are heated? Explain your answer. [2 m]





PSPM 2016/2017 SF016/2 No. 8(a)

3
28. A 200 cm glass cylinder is filled to the brim with mercury at 27 C. Calculate the
amount of overflow when the temperature of the system increases to 100 C. The
-1
-6
-1
-5
coefficient of linear expansion of the glass and mercury are 4  10 K and 6  10 K
respectively.
[4 m]
















65

PHYSICS PSPM SEM 1 1999 - 2017


PSPM CHAPTER 13: GAS LAWS AND KINETIC THEORY
___________________________________________________________________________

Ideal Gas Equation

PSPM 2003/2004 SF017/2 No. 8(a)

1. What is ideal gas? [1 m]



PSPM 2012/2013 SF016/2 No. 8(c)(i)

2. Write the ideal gas equation. [1 m]



PSPM 2009/2010 SF017/2 No. 14(a)

3. The equation of ideal gas is pV = nRT = NkT. State the meaning of n, N, and k. [3 m]


PSPM JUN 1999/2000 SF015/2 No. 14(b)

-3
4. Oxygen gas with mass of 75.02  10 kg is filled into a cylinder at pressure
2 atmosphere and temperature 0 C. Determine the initial volume and final pressure of
the gas if the gas is compressed to half of its initial volume and its temperature is
-26
500 C. Mass of an oxygen molecule = 5.32  10 kg. [5 m]



PSPM JAN 2000/2001 SF015/2 No. 7

23
5. A carbon dioxide gas of mass 22.0 g contains 3.01  10 molecules at pressure 1.0 atm
and temperature 0 C. Calculate the number of molecules of oxygen gas which has of
1
4
volume of carbon dioxide gas at pressure 1.5 atm and temperature 68 C. [4 m]



PSPM JAN 2000/2001 SF015/2 No. 14(a)

6. A tank is filled with 2.00 litres of air at pressure 2.0 atm and temperature 293 K. The
tank is then sealed with glue to prevent the air leakage. After that the tank is heated
until its pressure becomes 4.0 atm.

(a) Calculate the final temperature by assuming the volume is constant. [3 m]

(b) If the temperature is maintained as 293 K, and the gas expand until the pressure
becomes 1.0 atm, calculate the final volume. [3 m]



66

PSPM SP015



PSPM 2004/2005 SF017/2 No. 8

7. Nitrogen (N 2) gas in cylinder at STP has a mass of 112 g. When the gas is compressed
at pressure 10 atm, the temperature of the gas increases to 700 C. Calculate the initial
and final volume of the gas. (Mass of 1 mole of N 2 gas is 28 g.) [4 m]



PSPM 2012/2013 SF016/2 No. 8(c)(ii)

3
8. The pressure of a 50 cm ideal gas at 25 C is 75 Pa. Determine the number of
molecules of the gas. [3 m]



PSPM 2013/2014 SF016/2 No. 8(b)(ii)

9. The mass of an empty 50 litres gas cylinder is 4.8 kg. The cylinder is filled with
nitrogen gas up to pressure of 60 atm. Given the room temperature 29 C and the
molecular mass of nitrogen 28 g, calculate the new mass of the cylinder. [3 m]



Kinetic Theory of Gases


PSPM 2006/2007 SF017/2 No. 14(a)

10. State four assumptions of kinetic theory of gases. [3 m]



PSPM JUN 1999/2000 SF015/2 No. 7(a)

11. State the assumptions required about molecules to obtain the relation between pressure
2
and density of ideal gas, p = <v >. [2 m]
1
3



PSPM 2003/2004 SF017/2 No. 8(b)

12. According to the theory of kinetics gas, the pressure of an ideal gas, p is given as
1
2
p =  <v > where  is the gas density.
3

2
(a) State what is meant by the symbol <v >. [1 m]
2
(b) Express the rms velocity of the molecule in terms of <v >. [1 m]
(c) If the gas is composed of polyatomic molecule, how would the expression in (b)
changes? [1 m]



67

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2007/2008 SF017/2 No. 8(b)

13. By how many percent will the rms speed of an ideal gas molecules change if the gas
temperature increases by 50%? [3 m]


PSPM 2010/2011 SF017/2 No. 8(b)

14. The root mean square (rms) speed of gas molecules is to be reduced by 1%. If the gas
temperature is 27 C, at what temperature should it be cooled down? [3 m]


PSPM 2015/2016 SF016/2 No. 8(b)(ii)

-4
15. Calculate the rms speed of helium atoms at temperature 1  10 K. Molar mass of
-1
helium is 4 g mol . [2 m]


PSPM 2017/2018 SF016/2 No. 8(b)(i)

-3
3
16. A 5  10 m tank contains nitrogen at temperature 27 C and pressure 1.2 atm. The gas
pressure increases to 2.5 atm when the tank is heated. Given the molar mass of nitrogen
-1
is 28 g mol , calculate the change in the rms speed of the nitrogen molecules. [4 m]



Molecular Kinetic Energy and Internal Energy

PSPM 2007/2008 SF017/2 No. 8(a)

17. Write an equation for the average translational kinetic energy of molecules of an ideal
gas in terms of the absolute temperature of the gas. [1 m]


PSPM 2016/2017 SF016/2 No. 8(b)(i), (ii)

18. (a) Define the degree of freedom of gas molecules. [1 m]

(b) State the principle of equipartition of energy. [1 m]


PSPM 2006/2007 SF017/2 No. 14(b)

19. Distinguish heat and internal energy in the context of kinetic theory of gas. [3 m]


PSPM JUN 1999/2000 SF015/2 No. 7(b)

20. Explain the concept of internal energy of an ideal gas and state the types of degree of
freedom. [2 m]



68

PSPM SP015



PSPM JUN 2000/2002 SF015/2 No. 7

21. A closed container contains 0.3 mol of nitrogen gas. What is the internal energy, U of
-1
the system if the root min square velocity, v rms of the nitrogen molecules is 180 m s ?
-1
(Molar mass of nitrogen = 28 g mol ) [3 m]



PSPM 2002/2003 SF017/2 No. 8

22. Nitrogen gas N 2 of 14 g is filled into a container at STP. Calculate the internal energy U
of the gas. [3 m]



PSPM 2009/2010 SF017/2 No. 14(c)

-1
23. The molar mass of oxygen is 32 g mol . At 300 K, calculate the

(a) root mean square speed of the oxygen molecules. [3 m]

(b) internal energy of 3 mol oxygen. [3 m]



PSPM 2014/2015 SF016/2 No. 8(c)

-27
24. A balloon is filled with helium at 25 C. The mass of a helium atom is 6.65  10 kg.
Calculate the

(a) root mean square speed of the helium atom. [1 m]

(b) kinetic energy of 0.5 mole helium atom. [2 m]



PSPM 2016/2017 SF016/2 No. 8(b)(iii)

3
25. Two moles of a polyatomic ideal gas has volume of 0.05 m at pressure 250 kPa.
Calculate the internal energy and temperature of the gas. [4 m]



PSPM 2017/2018 SF016/2 No. 8(b)(ii)

26. At the same temperature, which gas a greater energy per mole: a diatomic gas or
monoatomic gas? Explain your answer. [3 m]






69

PHYSICS PSPM SEM 1 1999 - 2017


PSPM CHAPTER 14: THERMODYNAMICS
___________________________________________________________________________


First Law of Thermodynamics

PSPM 2012/2013 SF016/2 No. 8(d)(i)

1. State the first law of thermodynamics. [1 m]



PSPM 2011/2012 SF016/2 No. 8(c)(ii)

2. If a system loses 1200 J of heat when 800 J of work done on it, what is the change in its
internal energy? [2 m]




Thermodynamics Processes


PSPM 2010/2011 SF017/2 No. 14(a)

3. What is meant by the isothermal process? [1 m]



PSPM 2013/2014 SF016/2 No. 8(a) Edited

4. Define the following thermodynamics processes:

(a) Isobaric. [1 m]

(b) Adiabatic [1 m]



PSPM 2001/2002 SF015/2 No. 14(a)(i)

5. Explain the meaning of adiabatic compression process. [1 m]



PSPM 2015/2016 SF016/2 No. 8(c)(ii)

6. Explain why a gas becomes colder when it expands adiabatically. [4 m]





70

PSPM SP015



PSPM JUN 2000/2002 SF015/2 No. 8

7.
Pressure
P Q

T

R
S
Volume

FIGURE 14.1

On a p versus V graph as shown in FIGURE 14.1, name the thermodynamic process
from

(a) P to Q [1 m]

(b) Q to R [1 m]

(c) Q to T [1 m]

(d) Q to S [1 m]




PSPM 2003/2004 SF017/2 No. 14(b)

8. Sketch a graph of pressure p versus volume V of 1 mole of ideal gas. Label and show
clearly the four thermodynamic processes. [5 m]





Thermodynamics Work

PSPM 2003/2004 SF017/2 No. 14(a)(ii)

9. Write an expression representing the work done by an ideal gas at variable pressure.
[1 m]


PSPM 2013/2014 SF016/2 No. 8(c)

10. Derive the equation for the work done in an isothermal process. [3 m]





71

PHYSICS PSPM SEM 1 1999 - 2017


PSPM JAN 1999/2000 SF015/2 No. 14(b)

-2
11. p (N m )
K L
5



1
J M
3
V (m )

FIGURE 14.2

FIGURE 14.2 shows the graph of pressure p versus volume V of an ideal gas system.
The change in internal energy for the gas from K to M is +400 J, while the work done
along the path of KLM is +100 J.

(a) How much thermal energy must be added to the system during the K to L to M
process? [2 m]

(b) Determine the work done by the system for the process from M to J. [3 m]

(c) Calculate the thermal energy with the environment during M to K process. [2 m]






PSPM JUN 1999/2000 SF015/2 No. 14(c)

12. pressure

2p o A



p o C
B
V o volume
0.5V o
FIGURE 14.3

A 0.5 mole of ideal gas change from A to C as shown in FIGURE 14.3.
-2
-2
5
3
If V o = 3.0  10 m and p o = 1.5  10 N m , determine

(a) the total change of internal energy in the gas. [5 m]

(b) the total heat supplied to the gas. [3 m]




72

PSPM SP015



PSPM JAN 2000/2001 SF015/2 No. 14(b)

13. p (atm)
C
p f


2 A
B
2.0 10.0 V (l)

FIGURE 14.4

FIGURE 14.4 shows an ideal gas which is compressed from a volume of 10.0 l to 2.0 l
at constant pressure of 2.0 atm. Heat is then supplied to the system at constant volume
while pressure and temperature is let to change until the system finally reaches the
initial temperature at C. Calculate
(a) the total work done in the above process. [3 m]
(b) the final value of pressure, p f. [3 m]



PSPM 2001/2002 SF015/2 No. 8

5
3
3
14. An ideal gas expands from volume 50 cm to 60 cm at constant pressure 1.01  10 Pa.
If the gas temperature before expansion is 25 C, determine
(a) the temperature after expansion. [2 m]
(b) the total work done by the gas to expand. [2 m]



PSPM 2001/2002 SF015/2 No. 14(b)

15. One mole of a monoatomic ideal gas is at initial temperature of 650 K, while initial
pressure and initial volume are p o and V o respectively. At the beginning stage, this ideal
gas goes through an isothermal expansion process which causes its volume to become
2V o. Then this gas goes through an isovolumetric process and returns to its initial
pressure. Finally this gas goes through an isobaric process causing it to return to its
initial temperature, pressure, and volume.

(a) Sketch a graph of pressure versus volume for the whole process. [4 m]

(b) By using the first law of thermodynamics, prove that the total heat for the whole
process is
Q = (nRT ln 2 – p oV o)
where n is the number of mole, R is the molar gas constant while T is the absolute
temperature. Calculate the total of heat for the whole process. [8 m]

(c) State whether heat is absorbed or released. [1 m]




73

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2002/2003 SF017/2 No. 14(c)(i), (ii), (iv)

3
16. Two moles of helium gas is at the temperature 27 C and filled a volume of 0.02 m
initially. The helium gas expands at constant pressure until its volume becomes two
times greater. Then it expands adiabatically until its temperature returns to its initial
value which is 27 C.

(a) Sketch a graph of p versus V for both the processes. [2 m]

(b) Calculate the temperature of helium gas at the end of the expansion process at
constant pressure. [2 m]
(c) Calculate the total work done by the helium gas in the expansion process at
constant pressure. [3 m]



PSPM 2003/2004 SF017/2 No. 14(c)(ii)

17. A monoatomic ideal gas at pressure p and volume V is compressed isothermally until its
5
3
-2
new pressure is 3p. If p and V for the gas are 1.2  10 Pa and 1  10 m respectively,
calculate the work done on the gas during isothermal compression. [3 m]



PSPM 2005/2006 SF017/2 No. 14(a), (b), (d)

18. Two identical containers X and Y has movable piston to allow gas expansion. Both
containers contain nitrogen at 30 C. Gas in container X expands adiabatically while
gas in container Y expands isothermally to twice their original volumes. Molecular
-1
mass of nitrogen is 28 g mol . Assume nitrogen behaves like an ideal gas.

(a) State the difference in the speed of the piston in containers X and Y during
expansion. [2 m]

(b) Using the same axes, sketch and label the p-V graphs of the expansion
experienced by the gas in containers X and Y. Assume that the initial state of the
process is p 1, V 1, and T 1. [3 m]

(c) If container Y contains 0.03 mol of nitrogen gas, calculate the work done during
the expansion. [2 m]



PSPM 2007/2008 SF017/2 No. 14(b)

3
19. A 0.5 mole of ideal gas is filled in a 0.5 m container at 30 C. The gas is isothermally
3
compressed to a volume of 0.3 m . Then the gas expands isobarically to its original
volume.
(a) Sketch a p-V diagram of these two thermodynamic processes. [3 m]
(b) Calculate the total work done in the processes. [10 m]



74

PSPM SP015



PSPM 2008/2009 SF017/2 No. 8

20. A gas filled metal container shrinks when it is cooled.
(a) What happen to the internal energy of the gas? [1 m]

(b) Is it the surrounding or the gas that does the work? Explain your answer. [2 m]



PSPM 2008/2009 SF017/2 No. 14(d)(ii), (iii)

21. One mole of N 2 gas at 0 C is heated to 90 C at constant pressure. The change in
internal energy is 1863 J. Calculate
(a) the work done by the gas. [2 m]

(b) the heat added to the gas. [2 m]



PSPM 2009/2010 SF017/2 No. 8

22. A gas expands isothermally from volume V 1 to volume V 2 and performs mechanical
work 1200 J.

(a) Sketch the p-V graph of this process. [2 m]

(b) Calculate the heat gained by the gas. [2 m]



PSPM 2010/2011 SF017/2 No. 14(b)

23. One cubic meter air initially at 27 C and atmospheric pressure is compressed
isothermally to half of its original volume. Then the air is allowed to expand
isobarically back to its original volume.

(a) Using the same axes, sketch and label a p-V diagram of these two thermodynamic
processes. [3 m]
(b) Calculate the pressure of the air after the isothermal compression. [2 m]

(c) Calculate the final temperature of the air. [2 m]

(d) Calculate the total work done for the whole process. [7 m]



PSPM 2011/2012 SF016/2 No. 8(c)(iii)

3
3
24. Three moles of an ideal gas expands from 50 cm to 120 cm at a constant temperature
of 30 C. Calculate the heat transferred to the surroundings by this process. [3 m]




75

PHYSICS PSPM SEM 1 1999 - 2017


PSPM 2012/2013 SF016/2 No. 8(d)(ii), (iii)

25. (a) In an isovolumetric process, a system is supplied with 250 J of heat. What is the
change in its internal energy? [2 m]

3
-4
(b) What is the work done by 1.56  10 moles of ideal gas that expands from 65 cm
3
to 135 cm at a constant temperature of 40 C? [3 m]




PSPM 2013/2014 SF016/2 No. 8(d)

26. p (N m )
-2

8  10 B C
4




4
2  10 A

-3 -3 3
2  10 6  10 V (m )

FIGURE 14.5

FIGURE 14.5 shows a p-V graph for a series of thermodynamic processes, ABC. In
process AB and BC, 160 J and 600 J are added to the system respectively. Calculate the
change of the internal energy during the process ABC. [5 m]




PSPM 2014/2015 SF016/2 No. 8(d)

5
27. A 15 liter gas cylinder contains helium gas with pressure 1.01  10 Pa at 25 C. When
heated, the gas undergoes an isochoric process.

(a) Calculate the mass of helium gas. The atomic molar mass of helium gas is 4 g.
[2 m]

(b) If 500 J of heat is added, calculate the change in the internal energy of the gas and
sketch the graph of pressure versus volume for the isochoric process. [2 m]

5
(c) If the gas cylinder can withstand a pressure up to 4.55  10 Pa, calculate the
maximum quantity of gas at 45 C. [2 m]





76

PSPM SP015



PSPM 2016/2017 SF016/2 No. 8(c)

28. p (kPa)

1000 A



500 B



0 0.02 0.04 V (m )
3

FIGURE 14.6

FIGURE 14.6 shows a p-V graph of an isothermal process at 600 K for 2 moles of ideal
4
gas in a cylinder with frictionless piston. If the internal energy at point A is 2.5  10 J,
does the heat being absorbed or released during the process from A to B? Justify your
answer. [5 m]





PSPM 2017/2018 SF016/2 No. 8(c)

29.
p (N m )
-2


4 B C
7  10



2  10
4
A
3
V (m )
-3
-3
3  10 6  10

FIGURE 14.7

FIGURE 14.7 shows a series of thermodynamic processes ABC. During the process
AB and process BC, 120 J of heat and 500 J of heat are added respectively. Calculate
the change in the internal energy in the process ABC. [5 m]






77

PHYSICS PSPM SEM 1 1999 - 2017


FINAL ANSWER FOR CHAPTER 6: CIRCULAR MOTION

1. Centripetal acceleration is the acceleration of an object moving in circular path whose direction is
toward the centre of the circular path and whose magnitude is equal to the square of the speed divided by
the radius.
2. Yes, it does because there is a change in velocity.
3. Acceleration is the rate of change of velocity. In circular motion, velocity is always perpendicular to
radius of the circular path. So, the change in velocity is always pointing towards the centre of the circular
path.
-2
-2
5. a = 3.38  10 m s
6. Centripetal force is the force needed to keep an object in circular motion.
Centripetal force is the net force required to maintain a body in a circular motion.
7. (a) Since the direction of the velocity is constantly changing, the body accelerates. According to
Newton’s second law, a force F = ma, acts on the body causing it to accelerate.
mv 2
(b) F  acting towards the centre of the circle.
r
8. (a) A satellite orbiting earth.  Gravitational force.
(b) A ball attached to a string that swirls horizontally.  Tension
9. (a) Frictional force. (b) Gravitational force.
2
-1
2
10. (a) At A, mg  R  mr  ; At B, sin R   mr  (b)  max = 2.21 rad s
11. F c = 31.25 N, T max = 55.78 N, T min = 6.73 N
-1
12. (b)  max = 63.17, v max = 5.3 m s
1 g tan 
13. f 
2 r
mv 2


14. (b) F  mg cos  N  (c) Normal force (N = 0)
net
r
15. (a) T = 31.84 N (b) T = 28.16 N
-1
16. (a) v max = 19.4 m s
-1
-1
17. (a) v = 6.26 m s (b)  = 3.13 rad s (c) T = 0.29 N
-1
(d) v mid = 5.27 m s (e) Lower velocity
5
18. (a) F net = 3.15  10 N
(b) The force that enables the car to successfully negotiate the curve is friction between tyre and road.
-1
19. (b) v = 1.19 m s (c) T = 1.32 s
-1
-1
20. (b) v = 1.63 m s (c)  = 4.53 rad s
(d) If the angle remains unchanged but a longer string is used, the angular velocity will decrease
because  is inversely proportional to length of the string  .
-2
21. (a) a c = 17.2 m s (b) N = 665.1 N
-2
-1
22. (a)  = 1.46 rad s (b) a = 1.54 m s (c) F = 92.4 N
(d) Travelling in a straight line, tangent to the circle the instant they let go.
23. (b) By leaning towards the centre, the tyres tend to skid away from the centre. Frictional force prevents
the skid. The force f is the centripetal force for the cyclist to be in circular motion.
v 2
(c)  
rg
24. (a)  = 53.7 (b) T = 4.1 N
2
(c) If the speed is increased,  or r will increase because v  r tan 
25. (a) T a = 112.1 N (b) T b = 60.2 N


FINAL ANSWER FOR CHAPTER 7: GRAVITATION

1. Newton’s law of gravitation states that the gravitational force of attraction between two masses (m 1 and
m 2 ) is directly proportional to the product of the masses and inversely proportional to the square or the
distance (r) between them.
23
2. F = 3.53  10 N
3. (b) x = 6.74 m
4. Gravitational field strength is the gravitational force per unit mass.
78

PSPM SP015



-2
5. a g = 8.38 m s
-2
6. a g = 3.73 m s
-2
7. a g = 9.83 m s
5
8. h = 3.37 × 10 m
9. (b) M E = 80.59 M m
6
10. W = 3.79  10 N
-2
11. a g = 5.65 m s
-1
12. v = 7894 m s , T = 5094 s
GM r 3
13. v  , T  2
r GM
8
14. r Jupiter = 1.71  10 m
-1
15. v = 3104 m s
-1
3
16. v = 7.55  10 m s
17. (a) Satellite moves at the same rotation as Earth. (b) Its direction is always changing.
-2
(c) a = 0.22 m s
-3
-1
-4
-2
18. (a) g = 1.85  10 m s (b)  = 3.93  10 rad s
-1
-2
19. (a) v = 7687 m s (b) T = 5497 s (c) a = 8.79 m s
20. T = 5181 s
7
21. r = 4.2  10 m
7
22. (b) r = 4.22  10 m
23
23. M = 2.2  10 kg


FINAL ANSWER FOR CHAPTER 8: ROTATION OF RIGID BODY

1. Torque is the tendency of a force to produce rotation @ a measure of turning effect of a force on a body.
2. F 1 = 42.92 N, F 2 = 55.18 N
3. (b) x = 0.42L
4. (b) N A = N B = 1.5 N
5. (a) T = 1667 N (b) R = 1590 N,  = 24.8
6. T = 5231 N
7. (b) x = 0.96 m
9. (b) F = 328.13 N
4
10. (b) T = 2.04  10 N (c) T decrease. 

11. (a) Two (2) conditions for equilibrium of rigid body: F  0 and   0
(c) N P = 60 N, R Q = 247.4 N (d)  = 76
12. Angular acceleration is the rate of change of angular velocity.
13. Instantaneous angular acceleration is the instantaneous rate of change of angular velocity.
-2
14. (a)  = –1000 rad s (b)  = 3581 rotations (c) l = 1800 m
-1
-2
(d) a t = –80 m s (e) t = 2 s from  = 2000 rad s
-2
15. (a)  = 2 rad s (b) t = 5 s from  o = 10 rad s -1
-1
16. (a)  = 10.8 rad s (b)  = 1.55 rotations
-1
-1
17. (a) f = 20 Hz (b)  = 40 rad s (c) v = 22 m s
-2
18. (a)  = 12 rad (b) a = 0.18 m s
-1
19. (a)  = –3.33 rad s (b)  = 3.25 revolutions
-2
20. (a)  = –0.698 rad s (b) N = 112.5 rev
21.  = 46.5 rotations
-1
-1
22. (a)  = 125.66 rad s (b) v = 3.77 m s (c) s = 20.1 m
23. Moment of inertia of a rigid body about an axis is the sum of product of mass and the square of radius of
the rotation.
n
2
I   m i r define all symbols
i
 i 1
24. Three (3) factors which affect the moment of inertia of a rigid body is massof the body, shape and axis of
rotation. @ perpendicular (shortest) distance of the body to the axis of rotation.
25. s    s = r, v    v = r , a    a = r, F     = rF
2
26. I = 152 kg m
28. K r : K s = 10:7
79

PHYSICS PSPM SEM 1 1999 - 2017


2 -1
29. (a) L = 5.63 kg m s (b)  = 0.80 N m
-1
30. v = 1.9 m s
31. h max = 0.29 m
-1
2
2
32. (a) I = 0.25 kg m (b) L = 0.785 kg m s (c) K = 1.23 J
33. (b) a = 3.50 m s -2
-2
34. (a)  = –47.12 rad s (b) t = 4 s (c) K r = –888.3 J
-1
-1
35. (a)  = 6 rad s (b) v = 1.2 m s
-1
36. (a) (i) U = 98.1 J (ii) v = 8.1 m s (iii) t = 2.47 s
1
(b) Solid cylinder   = I = constant    Since I solid < I hollow   solid >  hollow
I
-2
37. (a) a = 7.36 m s (b) T = 7.36 N
2 -1
(c) (i) K r = 23.12 J (ii) L = 0.68 kg m s
-1
38. (a)  = 2.86 N m (b)  = 6.01 rad s (c)  = 2 rev
39. Solid sphere will arrive first at Q since v solid > v hollow .
-1
40. (a)  = 0.175 rad s (b)  = 3.12 rev (c) K i = 8.58 J
-2
41. a = 0.75 m s
-2
2
42. (a) I = 3.78  10 kg m (b) W = 3109 J
43. Solid sphere will reach the bottom of the slope first. When I , v 
-1
-2
44. (a)  = 1.46 rad s (b) a = 1.54 m s (c) F = 92.4 N
(d) Travelling in a straight line, tangent to the circle the instant they let go.
45. t = 6.7 s
-2
46. (a)  = 6.3 rad s (b) t = 11.1 s (c) P = 2500 W
-2
47. (a)  = 1.2 N m (b)  = 0.71 rad s (c) a t = 1.07 m s -2
48. Angular momentum is the vector product between radius of rotation with the linear momentum. Angular


momentum, L  r  p or L = I or L = mrv
49. L = constant in a closed system
50. The boiled egg will rotate longer because the yolk in the boiled egg is harden thus its moment of inertia is
constant.   L = I = constant and I is constant thus  is also constant.
The egg yolk in the raw egg is soft and mobile thus its moment of inertia will change when r changes.
  L = I = constant and I is large thus  will decrease.
2 -1
2
51. (a) I system = 82.3 kg m (b) L = 450 kg m s (c) W = 1230 J
(d) The platform will rotate faster.
52. r f = 36 cm
-1
53.  = 2.5 rad s


FINAL ANSWER FOR CHAPTER 9: SIMPLE HARMONIC MOTION

1. (a) Simple harmonic motion is a periodic motion of a body without loss of energy with its acceleration
directed towards the equilibrium point and proportional to the displacement from the equilibrium
point.
(b) No, because its acceleration is not proportional to its displacement.
2. The equation shows that acceleration of the motion is directly proportional to the displacement but
always in opposite direction.
3. Simple harmonic motion is a periodic motion without energy loss due to existence of restoring force and
absence of dissipative force.
4. Acceleration is directly proportional to the displacement of the object from the equilibrium position but
always in opposite direction.
Acceleration always point towards the equilibrium position.
5. A point is in circular motion. Projection of the moving point on any diameter is a simple harmonic
motion.
6. Force is proportional to the magnitude of displacement. Force is opposite direction to displacement.
7. The direction of acceleration in SHM is always opposite to displacement.
-1
-1
8. (a)  = 4.19 rad s (b) v max = 0.84 m s
-1
-3
9. E = 1.58  10 J, k = 0.12 N m
-1
-1
10. (a) v = 2.83 m s (b) v = 2.11 m s (c) E = 2 J
-1
11. (a) T = 1.57 s (b) (i) v = –1.75 m s
(b) (ii) The body moves to the left towards the equilibrium point.
80

PSPM SP015



-1
12. (a) T = 8.89 s (b) v max = 21.2 m s
-2
13. (a) a = 30.12 m s (b) F = 27.11 N
14. y =  4.3 cm
-1
-2
15. (a) f = 0.2 Hz (b) v = –0.48 cm s (c) a = –0.50 cm s
-2
(d) E = 3.95  10 J
16. (a) x = 12.5 sin 3.49t where x is in cm and t is in seconds.
-1
-2
(b) a = 91.4 cm s , v = 34.9 cm s (c) x = 8.84 cm
-1
-2
17. (a) f = 1 Hz (b) v = –25.1 cm s (c) a = 0 cm s
18. (a) x = 2 sin 3t where x is in cm and t is in s.
-2
-1
(b) v max = 18.8 cm s (c) a = 169 cm s
19. h max = 0.13 m
20. (a) A = 2 cm, T = 4 s (b) y  2 sin (  2 ) t where y is in cm and t is in second
-1
22. (a) A = 0.2 m (b)  = 10 rad s
24. A = 15 cm, f = 1.25 Hz
25. (a) A = 0.13 m (b) t = 0.07 s
-1
26. (a) A = 5 cm (b) f = 0.25 Hz (c)  = 1.57 rad s
(d) y  5 sin   t where y is in cm and t is in seconds

2
2
2
27. (a) a = – x or a = – A sin ( t   )
-1
(c) v = 8.65 m s (d) E = 2.03 J (f) T stays the same.
-1
29. (a) A = 8 cm (b)  = 7.85 rad s (c) E = 0.04 J
-1
30. (a) v (t) = –10 sin 2t where v is in m s and t is in s.
-2
(c) a max = 20 m s (d) E = 10 J
31. Elongation percentage for the cable of the pendulum is 26.6%.
32. m = 800 kg
33. (a) Oscillation period will decrease. (b) Oscillation period remains the same.
34. m = 0.7 kg
-2
35. g = 1.6 m s
36. T = 0.4 s
37. m = 0.54 kg
-1
38. (a)  = 1.05 rad s (b) Additional mass that should be added to the spring is 38.89 kg.
-1
39. (a) T = 0.93 s (b) v max = 0.68 m s (c) E = 0.04 J
40. (a) T = 0.549 s
(b) If the spring with the same load is allowed to oscillate horizontally on a frictionless surface, the
period will be the same because same load and same spring is used.
-1
41. (a) E = 0.028 J (b) v = 0.97 m s (c) T = 0.01 s


FINAL ANSWER FOR CHAPTER 10: MECHANICAL AND SOUND WAVES

-1
-1
1. T = 0.2 s, f = 5 Hz,  = 10 rad s , k = 5.24 rad m
-1
-1
2. (b) A = 15 m,  = 6.67 m,  = 3 rad s , f = 1.5 Hz, T = 0.67 s, v = 10 m s
3. Direction of wave propagation is to the right.
-1
4. (a)  = 0.94 m (b) v = 46.97 m s
(c) Direction of wave propagation is to the left. (d)  = 3 rad

5. (a) v  24 cos (8 t  0.4x ) cm s -1
y
-1
(b) a   192  sin (8 t  0.4x ) cm s -2 (c) v = 0.63 m s

y
 t x 
6. y  Asin 2    

 T  
7. (a) y = A sin  t (b) y 1  A sin (t   2 )

-1
-1
-1
8.  = 44 m,  = 50 rad s , k = 22 rad m or k = 0.14 rad m
9. Two important properties of progressive wave.
- Each of particle oscillate with the same amplitude.
- Progressive wave is able to transfer energy from one to another point.
11. (a)  = 0.5 m (b) y = 0.2 sin (3770t – 12.57x) where y and x are in meter and t is in second.
81

PHYSICS PSPM SEM 1 1999 - 2017


12. (a) Speed of propagation of wave is the speed of wave front motion @ speed of wave profile when
propagating.
(b) Vibrational speed of a particle is the speed of the particle when vibrating at equilibrium @ speed of
wave energy transferring.
13. (a) Wave propagation velocity is the velocity of wave profile or velocity of energy propagation.
(b) f = 143.2 Hz
 2
-1
-1
-1
14. (a)  = rad s (b) k = rad cm (c) v = 1.25 cm s
2 5
 2
-1
(d) v y = 2.51 cm s (e) y = 2 sin ( t – x) cm
2 5
-1
-3
15. (a) y = 15 × 10 sin (25t – 5x) (b) v y = –1.17 m s
-1
16. v = 261.8 m s
17. v = 0.02 m s -1
18. (a) Direction of wave is to the left @ to negative x-axis. (b)  = 6.28 m
-1
-1
19. (a) v = 747.6 cm s (b) v max = 3768 m s
20. y = 0.12 sin (t – 2.5x) or y = 0.12 sin (3.14t – 7.9x) where y and x in m and t in s.
-1
-1
21. (a) k = 50 rad m or k = 157.1 rad m (b) f = 600 Hz
(c) y = 0.02 sin 50 (24t –x) where y and x in m and t in s.
22. Stationary wave is produced by the superposition of two identical progressive waves but moving
towards each other resulting in a steady waveform or series of uniform loops.
Formation of stationary wave is the superposition of two waves having the same speed, frequency and
amplitude travelling in the opposite direction.
-1
23. v = 10 m s
24. Distance between two successive nodes = 1.5 m
25. (a) Standing wave or stationary wave (b) y = 12 cos 3x sin 2t
26. Sound intensity is power per unit area.
@ Sound intensity at a point is the rate of sound wave energy per unit area.
2
27. The intensity is directly proportional to the square of amplitude  I  A .
1
The intensity is inversely proportional to the square of distance  I  .
r 2
28. (a) Sound intensity is inversely proportional to the square of distance from the point source.
(b) If the coil speed is increased, wave propagation speed will remain the same because v =  means
f
wave does not depend on the speed of coil.
29. (c) f o = 81.65 Hz (d) amplitude = 0.03 m (e) y = 0.77 cm
-1
30. v = 34.2 m s ,  = 0.68 m
-1
31. v = 160 m s
-1
32. (a)  = 0.04 kg m (b) T = 0.4 s
-3
-1
-3
-1
33. (a)  1 = 2.55  10 kg m (b)  2 = 2.66  10 kg m (c) f 2 = 342.8 Hz
-1
34. v = 9.04 m s
-1
-3
35. (a)  = 2.64  10 kg m (b)   . 0 625 m (c) f = 1320 Hz
36. d new = 0.5d original
37.  = 0.75 m
-4
38. (a) y = 2  10 sin (3142t – kx) where y and x are in m and t is in s. (b) T = 157.5 N
39. (b) f o = 107.5 Hz, f 7 = 752.5 Hz
40. (a)  = 0.28 m (b)  = 0.56 m
41. (a) f 1 = 425 Hz, f 3 = 1275 Hz (b) f 2 = 1700 Hz
-3
42. (b) a’ = 7.64  10 m
)
(c) y  02.0 cos ( 5  x sin 410 t  where y and x are in meter and t is in second

4
43.  = 0.195 m
44. (a) Air disturbance causes the air column to vibrate at its natural frequency.
@ The air blown resonates with the natural frequency.
(b) f = 250 Hz
45. f = 850 Hz
46. Doppler effect for sound wave is the apparent change in the frequency of sound as a result of relative
motion between the sound and the source.
-1
48. f a = 600 Hz, v s = 30 m s
49. f beat = 5 Hz
82

PSPM SP015



51. f b = 121.9 Hz
52. Frequency heard by a moving observer is 300.41 Hz.
53. (a) f o = 270.3 Hz
54. f a = 1187 Hz
55. f a = 1407 Hz


FINAL ANSWER FOR CHAPTER 11: DEFORMATION OF SOLIDS

F ΔL

1. stress,   , strain, 
A L o
2. Plastic deformation: Material undergoes permanent extension when stressed or forced beyond its elastic
proportionality limit. When the force is gradually reduces to zero, its stretched length will always be
longer then its original length.
3. Hooke’s law is not applicable in the case of plastic deformation because plastic deformation occurs
beyond the proportionality limit.
1
4. Strong @ elastic @ brittle. F   Contact time is increased, impulsive force will decrease.
t
6. Young’s modulus is the ratio of stress to strain if elasticity limit is not exceeded.
10
-2
-2
8
7. (a)  break = 1.98  10 N m (b) Y = 7.07  10 N m
8. W = 0.43 J
-5

-5
-5
-4
9. (a)  1 = 2.68  10 ,  2 = 8.9  10 (b)  L = 5.36  10 m, L = 1.8  10 m
1
2
YA
10. k 
L
11
-2
11. (a) Y = 1.77 × 10 N m (b) U = 1.10 J
-1
3
12. k = 3.68  10 N m
13. (a) F = 0.464 N (b) U = 0.0058 J
14. U = 0.294 J
10
15. Y = 2.45  10 Pa
16. The concept of Young’s modulus will be used to determine which rod will bend more.
1
Y   Y lower e higher  bend more
e
Material with smaller value of Young’s modulus will bend more.
-3
17.  = 3.8  10
18. F = 25.8 N
19. U = 3224.5 J
-7
20. U = 9.35  10 J


FINAL ANSWER FOR CHAPTER 12: HEAT CONDUCTION AND THERMAL
EXPANSION

1. Heat is energy transfer due to temperature difference.
Q  1   x 
2. k     
t  A   T 
3. Factors that influence the thermal conductivity of a conductor is the type of conductor.
7. P loss = 559.4 W
-1
8. (a) Q = 7533 J s (b) T = 101.5 C
9. Q = 21.6 kJ
10. T = 60 C
8
11. Q = 9.24  10 J
12. (a) Temperature of the inner surface of the bulb is 33.3 C.
8
13. Q = 1.66  10 J
14. T j = 130.1 C
15. Coefficient of linear thermal expansion is the fractional increase in length of a solid per unit rise in
temperature.
83

PHYSICS PSPM SEM 1 1999 - 2017


16. Coefficient of area thermal expansion,  is the fractional change in area per unit change in temperature.
17. Coefficient of volume thermal expansion is the change in volume per unit of its original volume per
unit change in temperature.
18. h = 11.011 m
-1
-5
19.  Hg = 15  10 C
-1
-6
-5
-1
20. (a)  A = 8  10 C ,  B = 3.5  10 C (b) L oA = 77.78 cm, L oB = 22.22 cm
-4
21.  L = 6  10 m
 L
-3
22. = 4.61  10
L
-1
-5
-5
23. (a) 2.6  10 K means for temperature increase of 1 K, fractional increase in length is 2.6  10 .
-5
(b) V = 3.74  10 m 3
24. T = 308 C
25. The frame must be heated to 35.15 C.
26. L = 3.0036 m
27. (a) T = 100.3 C (b) No.  aluminium >  brass
3
28. V = 2.5 cm


FINAL ANSWER FOR CHAPTER 13: GAS LAWS AND KINETIC THEORY

1. Ideal gas is gas that obeys all the law of gas such as Boyle’s law, Charles’ law, and law of pressure.
2. Ideal gas equation: pV = nRT
3. n = number of moles of the gas, N = number of molecules in the gas,
k = Boltzmann constant
3
4. V i = 0.026 m , p f = 11.45 atm
22
5. N = 9.03  10 molecules
6. (a) T f = 586 K (b) V f = 4 litres
3
3
7. V 1 = 89.6 dm , V 2 = 31.9 dm
17
8. N = 9.09  10 molecules
9. m new = 8.19 kg

10. Four assumptions of kinetic theory of gases.
 Gas consists of identical spherical molecules.
 Molecules move in random motion.
 Intermolecular forces are negligible.
 Volumes of molecules are negligible.
 Collisions between molecules are negligible.
 Collision time can be neglected compared to time between collision

11. Assumptions about molecules to obtain the relation between pressure and density of ideal gas,
2
1
p =  < v > is assumptions of kinetic theory of gases.
3
 Gas molecules collide elastically with each other and with the walls of the container  kinetic
energy and momentum are conserved. Collision of gas molecules with the walls of the container
produces pressure on the walls.
 Gas molecules are at constant, free, and random motion to all directions with different velocity
 Forces between molecules are negligible except during collision.

2
12. (a) < v > represents mean square velocity of gas molecules.
(b) rms velocity =  v 2  (c) The expression in (b) remains the same.
13. v rms increases by 22%.
14. T 2 = 294 K
-1
15. v rms = 0.79 m s
-1
16. v = 229 m s
3
17. K tr = kT
2
18. (a) Degree of freedom of gas molecules is the number of ways in which an atom or molecule can
absorb or store energy.


84

PSPM SP015



18. (b) Principle of equipartition of energy states that energy is shared equally among the active degrees
1
of freedom, each active degrees of freedom has an average energy equal to kT .
2
Principle of equipartition of energy states that energy of a system is shared equally among its
available degree of freedom.
Principle of equipartition of energy states that an average energy of each degree of freedom of a
molecule is ½kT.
19. Heat is the energy that being transferred due to temperature difference.
Internal energy is the sum of all the energy of all the molecules in a gas.
20. Internal energy U of a gas equals the total amount of average kinetic energy, K and potential energy
which contains in any gasses.
The capacity of the internal energy depends on the state of the system. For ideal gases, potential energy
can be neglected because the force of exertion between molecules does not exist so the internal energy, U
equals the total of average kinetic energy of the molecule, K.
f f
U  NkT or U  nRT

2 2
 internal energy depends on temperature and degrees of freedom
Types of degree of freedom are translational, rotational, and vibration.
21. U = 227 J
22. U = 2837 J
-1
23. (a) v rms = 483 m s (b) U = 18.7 kJ
3
-1
24. (a) v rms = 1.36  10 m s (b) K = 1857 J
4
25. U = 3.75  10 J, T = 752 K
26. At the same temperature, diatomic gas a greater energy per mole because U  f.
f diatomic = 5 and f monoatomic = 3 so U diatomic > U monoatomic


FINAL ANSWER FOR CHAPTER 14: THERMODYNAMICS

1. First law of thermodynamics: Q = U + W
where Q = thermal energy, U = internal energy, W = work done by the gas
2. U = –400 J
3. Isothermal process is a thermodynamic process where the temperature is constant.
5. Adiabatic process is a process where heat is not absorbed or released.
6. Adiabatic compression process is a process where heat is not absorbed or released during compression.
7. (a) P to Q: Isobaric (b) Q to R: Isovolumetric
(c) Q to T: Isothermal (d) Q to S: Adiabatic
9. W   V f p dV
V i
11. (a) Q KLM = 500 J (b) W MJ = –20 J (c) Q MJK = –420 J
12. (a) U ABC = 0 J (b) Q ABC = 2250 J
13. (a) W ABC = –1621 J (b) p f = 10 atm
14. (a) T 2 = 357.6 K = 84.6 C (b) W = 1.01 J
15. (b) Q = –1657 J (c) Heat is released.
16. (b) T 2 = 600 K (c) W 1 = 4986 J
3
17. W = –1.32  10 J
18. (a) Piston X moves a lot faster than piston Y during expansion. (c) W = 52.4 J
19. (b) W = 196.3 J
20. (a) Internal energy of gas decreases.
(b) Surrounding does the work. Gas volume decreases hence negative work done.
21. (a) W = 747.9 J (b) Q = 2610.9 J
22. (b) Q = 1200 J
5
4
23. (b) p 2 = 2.03  10 Pa (c) T 3 = 600 K (d) W = 3.11  10 J
24. Q = 6.62 kJ
25. (a) U = 250 J (b) W = 0.297 J
26. U ABC = 440 J
27. (a) m = 2.44 g (b) U = 500 J (c) m = 10.32 g
28. Q is absorbed. Q = 6912 J
29. U = 410 J



85

PHYSICS PSPM SEM 1 1999 - 2017




PREFIX


Prefix Symbol Factor Prefix Symbol Factor

yotta- Y 10 24 deci- d 10 -1
-2
zeta- Z 10 21 centi- c 10
-3
18
exa- E 10 milli- m 10
15
-6
peta- P 10 micro-  10
12
tera- T 10 nano- n 10 -9
9
giga- G 10 pico- p 10 -12
6
mega M 10 femto- f 10 -15
3
kilo- k 10 atto- a 10 -18
2
hecto- h 10 zepto- z 10 -21
1
deka- da 10 yocto- y 10 -24




THE GREEK ALPHABET

  Alpha   Nu

  Beta   Xi

  Gamma   Omicron
  Delta   Pi

  Epsilon   Rho

  Zeta   Sigma
  Eta   Tau

  Theta   Upsilon

  Iota  ,  Phi
  Kappa X  Chi

  Lambda   Psi

  Mu   Omega







86