THE UNIFIED EXAMINATION (2020)

INDEPENDENT CHINESE SECONDARY SCHOOLS

MALAYSIA
Senior Middle Level

PHYSICS
(SE12)

PAPER I MULTIPLE-CHOICE QUESTIONS

Date : 2 December 2020

Time: 2.00 p.m. - 3.10 p.m.
(l hour 10 minutes)

INSTRUCTIONS TO CANDIDATES

1. This subject comprises two papers:

Paper 1: Multiple-choice Questions (30%),
Paper 2: Subjective Questions (70%).
2. Complete Paper 1 within the 1 hour 10 minutes allocated. After a 15-minute interval, proceed to
complete Paper 2 within the 2 hours allocated.
3. Attempt all the 30 questions in Paper 1. For each question, select the correct or the best answer and
blacken the square corresponding to the same letter of the alphabet that you have selected. Mark the
answers with a 5B pencil on answer sheet 'O' (computer card).
4. The prescribed electronic calculators may be used.
5. The following constants may be used in your calculations, unless stated otherwise:
* Speed of light c=3.0 × 10 ms
8 −1

* Acceleration due to gravity g=9.8 m s−2

* Density of water ρw =1000 kg m
−3

* Electron charge e=−1.6 × 10−19 C

* Electron mass
−31
m e =9.1 ×10 kg
* Permittivity of free space ε 0=8.85 ×10
−12
Fm
−1

* Relative permittivity of air ε r=1

* Planck's constant h=6.63 ×10−34 J s
* Rydberg constant
7 −1
R H =1.097 ×10 m
* Avogadro's number N A =6.02× 10 mol
23 −1

* Boltzmann constant k =1.38 ×10 J K

−23 −1

* Permeability of free space μ0 =4 π ×10−7 T m A−1

* Molar gas constant R=8.31 J mol K
−1 −1

6. Paper 1 consists of eight printed pages.

Do Not Turn Over This Page Until You Are Told To Do so

(Candidate's Name: _________________________ Index Number:

___________________________

Dong Zong Copyright Reserved

U2K20-1 A ball rolls on a floor with uniform deceleration. If s is the displacement and t is the time taken by the
ball, which of the following produces a straight-line graph?
s s
A A. against t B. s against t C. s against t 2 D. 2 against t
t t
Ans: given distance s, time t and uniform deceleration a < 0,
1
s=ut+ a t 2
2
s 1
=u+ at
t 2
As compare to the equation of a straight line,
s
y=mx+c y ❑ , x ❑ t
→ t →

U2K20-2 Fig.U2K20-2 shows four persons P, Q, R and S performing an

acrobatic show. P is standing on the shoulder of S and stretches out
his hands to support Q and R while S is pushing Q and R. A person
behind Q is holding Q so that the whole system is in static
equilibrium. If all their masses are m and that they have equal arm
lengths, which of the following statements is true? (Gravitational
acceleration is g)
D A. The normal reaction on S from the ground is 4 mg.
B. The reaction on P from S is mg.
C. R experiences a leftward resultant force.
D. The resultant force on Q is zero.
Ans: The whole system is in static equilibrium; the resultant force is zero.

U2k20-3 A coin drops freely at a certain height from the floor of a lift. The coin reaches the floor in a time span
of t1 if the lift is stationary and in a time span of t2 if the lift is moving at uniform velocity. Which of the
following about the relation between t1 and t2 is true?
D A. t1 < t2 if the lift is going up
B. t1 > t2 if the lift is going up
C. t1 < t2 no matter the lift is going up or down
D. t1 = t2 no matter the lift is going up or down
Ans: The coin drops freely, u = 0,
If the lift kept stationary, the relative acceleration is a=g−0=g
1
2
1
2 √
s=ut+ a t 2= g t 2t 1= 2 h
g
If the lift is going up with uniform velocity, the relative acceleration is a=g+0=g
1 2 1 2
s=ut+ a t = g t t 2= 2 h
2 2 √ g
If the lift is going down with uniform velocity, the relative acceleration is a=g+0=g
1
2
1
2 √
s=ut+ a t 2= g t 2t '2= 2 h
g
The relative velocity of the initial velocity of the coin and the lift is zero (for inertia frame)

U2k20-4 An object is projected upward at an initial velocity u and at an angle θ with the horizon. If the kinetic
energy at the highest point is half of its initial kinetic energy, what is the value of θ ?
B A. 30 B. 45 C. 60 D.
90
Ans: For a projectile motion, the speed of the body at the highest point is v=v x =ucosθ ,
Given that the kinetic energy at the highest point is half of its initial kinetic energy

-2-
Dong Zong Copyright Reserved
 1 1 2 1 1 2 2 1 2 2 1 1
Ek , maxheight = E k ,initial m v highest = × mu ( ucosθ ) = u cos θ= cosθ=± θ=45 °
2 2 2 2 2 2 √2
U2k20-5 Two satellites P and Q are moving in the same circular orbit around the Earth. If the mass of P is twice
that of Q, what is the ratio of the tangential velocity of P to Q?
1
B A. B. 1 C. √ 2 D. 2
√2
Ans: Same circular orbit, r 1=r 2=r
For the satellite is in circular motion,
F G=F C
2
m1 m E m1 v
G 2 =
r r

√
v= G E
m
r
The tangential velocity of the satellite is inversely proportional to the radius of the circular orbit and
independent of the mass of the satellite. Then the ratio of the tangential velocity of the two satellites is
1: 1.

U2k20-6 An object falls in air with uniform velocity and penetrates a layer of sand with uniform retardation until
it stops. If R represents the resultant force acting on the object and s represents the displacement, which of
the graphs below best illustrates the motion of the object?
A A. R B. R C. R R

s s s s
0 0 0 0

For the object falls in air with uniform velocity, a=0 ;∑ F=ma=0 .
For the object penetrates a layer of sand with uniform retardation until is
stops, a=constant < 0
∑ F=ma=constant <0
From the graphs given, the best answer is A.

U2k20-7 As shown in Fig. U2k20-7, the two blocks A and B with different
masses are connected by a light inextensible string through a
frictionless pulley. The inclined surface is smooth. Initially, A and B are
at rest on the same level. When the string between the blocks is cut, A
falls vertically while B slides down the inclined surface. From the time
the string is cut to the time they touch the ground, which of the
following statements is correct?
AD A. Their change in speed are the same
B. Their change in mechanical energy are different
C. Their change in gravitational potential energy are equal
D. Their average power of the gravity are the same
Ans:
Object A Object B
Initial velocity 0 0
Distance h h
moved sinθ
Acceleration g a=gsinθ
Velocity as v =u +2 asv=√ 0+2 gh¿ √ 2 gh
√
2 2
h
reaches the v 2=u2 +2 asv= 0+2 gsinθ × ¿ √ 2 gh
sinθ
ground
Time taken 1 1
s=ut+ a t 2 s=ut+ a t 2
2 2

Dong Zong Copyright Reserved

 √ 2h h 1 2
t A= = gsinθ t
g sinθ 2

ΔE Δ E=Δ E k + Δ E p
t B=
Δ E=Δ E k + Δ E p
√ 2h
gsin 2 θ

( ) ( )
(Energy is 1 1
conserved) ¿ mA × 2 gh−0 + ( 0−mA gh )¿ 0 ¿ mB × 2 gh−0 + ( 0−mB gh )
2 2
¿0
ΔEp Δ E p=0−m A gh¿−m A gh Δ E p=0−mB gh¿−mB gh
m A ≠ mB

Power of P=Fvcosθ ¿ m A gvcos 0 °¿ m A g √2 gh P=Fvcosθ¿ m B gvcos(90 °−θ)

gravity *** [m A g=mB gsinθ ] ¿ m B gsinθ √ 2 gh

U2k20-8 Two balls P and Q are moving on a smooth horizontal surface in the same direction with momentum 10
kg m s–1 and 15 kg m s–1 respectively. Ball P is moving behind ball Q, before it makes direct collision with
Q. What are the possible momentum of the balls after the collision?
C Momentum of P Momentum of Q
A 12 kg m s –1
13 kg m s–1
B 10 kg m s–1 15 kg m s–1
C 8 kg m s–1 17 kg m s–1
D - 10 kg m s–1 35 kg m s–1
Ans: According to principle of conservation of linear momentum,
m1 v 1+ m2 v 2=m1 u1 +m2 u2m1 v 1−m1 u1 =−( m2 v 2−m2 u2)

在后方运动物体与前方物体发生碰撞后，如果它继续前进的话，它的速率一定会变小，则动量也一
定会变小；如果要以相同大小的速度反弹，则前方物体的质量必须远大于后方物体的质量 mP ≫ mQ，
此情况下，P 物的动量为 – 10 kg m s–1, 但前方物体碰撞后的速度保持原本的速度，这是不可能的。
(−10 )2 352 50 1
又 碰 撞 后 的 总 能 量 Tf E = + = +612.5 > 碰 撞 前 的 总 能 量
2m p 2 m Q mp mQ
( 10 )2 152 50 1
ETi = + = +112.5 , 这对于碰撞来说是不可能获得更多的动能的。
2m p 2 m Q m p mQ

,
U2k20-9 The displacement of a free-falling object is divided into three equal portions. If the initial velocity is
zero and the time taken in the second portion is 2 seconds, find the time taken in the third portion.
C A. 0.65 s B. 1.04 s C. 1.53 s D. 2.00 s
Ans: Given: u = 0, a = g,
1 2 1 2
s=ut+ a t = g ( t 1−0 )
2 2
t 1=
1 2 1
√2s
g
2
2 s=ut + a t = g ( t 2−0 )
2 2
t 2=
1 2 1
√4s
g
2
3 s=ut + a t = g ( t 3−0 )
2 2
t 3=
√6s
g

-4-
Dong Zong Copyright Reserved
 t 3 −t 2√ √
=
6s
g
−
4s
g √ 6−2
=

√ √ 2 s 2−√ 2
t 2−t 1 4s
−
g g
t 3−t 2= √ 6−2
× 2=1.53( s)
2−√ 2

U2k20-10 By using a constant volume gas thermometer, the pressure of a gas at temperatures 0 C and 100 C
are measured to be 1.2105 Pa and 1.8105 Pa respectively. Find the temperature when the pressure of that
gas is 1.65105 Pa.
D A. 19.0 C B. 25.0 C C. 27.5 C D. 75.0 C
pθ − p 0 5
1.65× 10 −1.2 ×10
5
Ans: θ= × 100 ° C= × 100° C=75.0 ° C
p100 − p0 5
1.8× 10 −1.2 ×10
5

U2k20-11 Fig.U2k20-11 shows the variation of temperature  with /C

time t of a heating object. If the object is initially in solid state
and it constantly absorbs 2000 J per minute during the 80
experiment, which of the following statements is true?
D A. The specific latent heat of fusion of the object is 6000 J kg–1 60
B. After 10 minutes of heating, the object starts to vaporise.
40
C. After 4 minutes of heating, the object completely changes
into liquid state. 20
D. The specific heat capacity at liquid state is lower than that at
solid state of the object. t/min
Ans: 2 4 6 8 10
A: When the solid is melting,
Q=Pt L f =Pt
m
Pt 2000 × (7−4 )
Lf = = =6000 m−1 ( J kg−1)
m m
B: Cannot be determined.
C: After 4 minutes of heating, the object starts to melt.
D: For solid state,
Pt 2000 × ( 4−0 )
Q=Pt m c s ΔT =Pt c s = = =200 m−1 ( J kg−1)
m ΔT m× 40
For liquid state,
Pt 2000 × ( 10−7 ) −1 −1
c l= = =150 m (J kg )
m ΔT m× 40
⟹ c l <c s

U2k20-12 The air tank of an air compressor contains 6.0 L of air at pressure
1.0 atm. If 9.0 L of air at pressure of 1.0 atm is added to the tank and the
total air in the tank is compressed isothermally to its initial volume of 6.0
L, what is the final pressure of the air in the tank? (Assume the air is an
ideal gas)
A A. 2.5 atm B. 2.0 atm C. 1.5 atm D. 1.0 atm
Ans: According to principle of conservation of mass,
p V p V p V p ×6.0 1.0 ×6.0 9.0× 1.0
nT =n 1+ n2 T T = 1 1 + 2 2 T = + pT =2.5(atm)
R TT R T 1 R T 2 RT RT RT

U2k20-13 An object is placed on the principal axis at point P, 40 cm in front of a spherical mirror. The size of
1
the real image formed is 3 times the size of the object. To obtain an image of the size of the object, how
3
far the object should be moved from point P?
C A. 40 cm B. 60 cm C. 80 cm D. 100 cm
Ans: Case 1: m1 = +3, u1 = 40 cm,

Dong Zong Copyright Reserved

 1
Case 2: m2 =
3
From case 1, we known that the 3 times magnified and real image, it is a concave mirror. And in case 2, the
1
size of the object image should be a real image.
3
1 1 11 1 1 4 1
+ = + = = f =30( cm)
u1 v 1 f u1 3 u1 f 3× 40 f
For case 2,
1 1 1
1 1 1 + =
+ = u2 1 30 u❑2=120 ( cm ) Δ u=120−40=80(cm)
u2 v 2 f u
3 2

U2k20-14 As shown in Fig.U2k20-14, light ray is refracted at the boundary MN

between vacuum and a medium, which of the following statements is
correct?
D A. The deviation angle of the light is 60 .
B. The refractive index of the medium is 0.577.
C. Light travels from vacuum into the medium. Fig.U2k20-14
D. The speed of light in the medium is 1.732× 108 m s−1.
Ans:
A: angle of deviation, Δ θ=90 °−30 ° −30° =30°
B: The refractive index of a medium with respects to vacuum must be larger than 1.
n1 sin i 1=n2 sini 2n1 sin 30 °=1× sin 60 °n1 =√ 3
C: The light travels from the medium into vacuum.
D: The speed of light in the medium,
n1 v 1=n2 v 2√ 3 v 1=1 ×(3× 108 )v1 =√ 3 × 108 (m s−1)

U2k20-15 Fig.U2k20-15 shows the position of an object and its

image formed by a convex lens, which of the following
substitutions into the lens formula is correct?
1 1 1 1 1 1
B A. = + B. = +
f 3 cm 6 cm f 9 cm 18 cm
1 1 1 1 1 1
C. = + D. = + Fig.U2k20-15
f 9 cm 12 cm f 12 cm 18 cm
Ans: The position of an object in front of a convex lens and its real image is in a relation of the nearer the object
the farther is its image.
If the object distance is 9 cm, then its image distance is 18 cm.
If the object distance is 12 cm, then its image distance is 12 cm.

U2k20-16 A white light source is used by a student in a Young's double slits experiment. If each slit is covered
by a blue glass and yellow glass respectively, which of the following interference patterns will be observed?
D A. yellow and blue color interference fringes B. yellow color interference fringes
C. blue color interference fringes D. no interference fringes
Ans: The condition for obvious interference is coherent sources, two sources with constant phase difference and
same in frequency and amplitude. The two sources, one is covered by a blue glass and the other is yellow,
two light emerged from the glasses are difference in frequency. Then, they are not a coherent source, there is
no clear interference pattern will be observed.

U2k20-17 A uniform stretched wire of length 0.8 m has a fundamental frequency of 256 Hz. If the wire is
shortened to 0.2 m under the same tension, what will be the new fundamental frequency of this wire?
C A. 64 Hz B. 256 Hz C. 1024 Hz D. 2048 Hz
Ans: For fundamental mode in a stretched wire,
λ
l=
2
λ=2 l

-6-
Dong Zong Copyright Reserved
 v 1 T
f 0= =
λ 2l μ√
√
1 T2
f 02 2 l 2 μ2 l 1
= =

√
f 01 1 T 1 l2
2 l 1 μ1
l1 0.8
f 02= f 01= × 256=1024 (Hz)
l2 0.2

t x
U2k20-18 A progressive wave equation can be represented by y= Acos 2 π ( + ). What is the phase of the
T λ
particles at position x ?
πt 2 πt πx 2 πx
A. B. C. D.
T T λ λ
t x
Ans: The phase of the particle at position x is 2 π ( + ).
T λ

t x
A progressive wave equation can be represented by y= Acos 2 π ( + ). What is the phase difference
T λ
between the particles at position x and x = 0?
πt 2 πt πx 2 πx
A. B. C. D.
T T λ λ
2 πx
Ans: The phase difference between the particle at x and origin is .
λ

U2k20-19 As shown in Fig.U2k20-19, a block connected to the wall

with a light spring performs simple harmonic motion of amplitude
A
A on a smooth horizontal surface. If the amplitude is reduced to
2
, which of the following quantities would be halved?
A A. The maximum velocity of the block.
B. The maximum kinetic energy of the block. Fig.U2k20-19
C. The period of oscillation of the block.
D. The maximum elastic potential energy stored in the spring.
Ans:
' ' 1 1
A: v max= A ω= Aω= v max
2 2
2
1 1 A 1 1
B, D: E'k , max =E'p ,max = k A ' 2= k ( ) = × k A 2
2 2 2 4 2
C: T =2 π
√ m
k
, it is independent of amplitude. It remains unchanged.

U2k20-20 When an electron makes a transition from energy level E3 to energy level E2, a green line is produced
in the visible spectrum. Which of the following transitions is most likely to produce a blue line?
C A. E2 to E1 B. E3 to E1 C. E4 to E2 D. E4 to E3
Ans: If it is a transition of an electron in a hydrogen atom. Only the transition of E6, E5, E4, E3 to E2 radiated
visible light spectrum. But the visible light spectrum for the transition from E3 to E2 for a hydrogen is red
light.

U2k20-21 Which of the following statements are true when the intensity of the light source in a photoelectric
effect experiment is increased?
I. The number of photoelectrons emitted from the metal surface per second increases.
II. The average kinetic energy of the photoelectrons emitted from the surface of the metal is greater.
III. The total energy of the photons emitted by the source per second increases.
IV. The speed of photoelectrons emitted from the metal surface is greater.
Dong Zong Copyright Reserved
B A. I, II B. I, III C. II, III D. II, IV
Ans: The intensity of the light source in a photoelectric effect experiment is increased, only the rate of number of
photoelectrons emitted from the metal’s surface increased. While the maximum initial kinetic energy of the
photoelectron does not alter by the intensity of the incident light.
E
E
The total energy of the photons emitted by the source per second, t . depends on the intensity
Intensity= t
A
of the light source.

U2k20-22 A laser pen with a power P emits monochromatic light of frequency f and the number of photons
emitted per second is N. Find the Planck's constant h, in terms of P and N.
C A. h=
N
B. h=
√P C. h=
P √f
D. h= 2
√ fP fN fN N P
Ans: The power of a laser pen,
P
E Nhf h=
P= = N
t t f
t

U2k20-23 When a beam of X-ray of wavelength 3.1 ×10−10 m is directed at a glancing angle of 26.5 to a salt
crystal, a second order maximum diffraction is observed. Find the crystal lattice constant d of the crystal.
C A. 3.5 ×10−10 m B. 5.2 ×10−10 m C. 6.9 ×10−10 m D. 8.6 ×10−10 m
Ans: Second order maximum diffraction is observed, n = 2,
−10
nλ 2× 3.1×10
2 dsinθ=nλ d= ¿ ¿ 6.95 ×10−10( m)
2 sinθ 2 sin 26.5 °

U2k20-24 The energy required to roast a chicken drumstick is 3000 J. What is the shortest time needed to roast
the drumstick if an oven is connected to 110 V voltage supply and 0.40 A current is used?
B A. 44.0 s B. 68.2 s C. 132.8 s D. 170.5 s
Ans: The electric energy of a roast machine,
H 3000
H=W H=VIt t= ¿ ¿ 68.18 ( s )
VI 110× 0.40

U2k20-25 Fig.U2k20-25 shows a point charge + Q and a point charge of

+Q
placed at coordinates 0 and d on the x-axis respectively. Which
4
of the following figures represents the equipotential lines around the Fig.U2k20-25
two charges?
B A.

Ans: The electric flux is directly proportional to the quantity of the charge. The electric field lines of charge
Q at O are about 4 times that of the charge ¼ Q at x = d. The electric field lines is represented by D.
Both the electric charges carry positive charge, they repel each other.
So, the equipotential lines around the two charges best represents by B.

U2k20-26 In Fig.U2k20-26, the position X is the midpoint between two

charges, + q and – q, which are separated by a distance of d. What is +q x -q
the electric field strength E and electric potential V at X?
C electric field strength E electric potential V d
A 0 0 Fig.U2k20-26

-8-
Dong Zong Copyright Reserved
 B 2q q
, towards left 2 π ε0 d
π ε 0 d2
C 2q 0
2 , towards right
π ε0 d
D 0 q
2 π ε0 d
Ans: The electric field strength,
1 q 1 q 2q
ET =E1 + E2 = + = (¿ the ¿)
() ()
4 π ε0 d 2
4 π ε0 d 2
π ε0 d
2

2 2
The electric potential,
1 q 1 −q
V T =V 1 +V 2= + =0
4 π ε 0 d 4 π ε0 d

U2k20-27 Fig.U2k20-27 shows a circuit in which a 12 V d.c. supply is

connected to 5 resistors. The internal resistance of the d.c. supply is
negligible. Find the current flowing through the 4  resistor.
B A. 0 B. 1.5 A
C. 3.0 A D. 4.0 A
Ans: The potential difference across AB is 12 V. A
Fig.U2k20-27
And the potential difference the 4  resistor is

( )
−1
1 1
+
R AC 2+2 4
V AC = ×V AB = ×12=6 ( V )
R AC + RCB
( 1 1 −1
2+2 4
+ ) +2 C B
The current passing through the 4  resistor is,
V AC 6
I= = =1.5( A)
4 4

U2k20-28 For a RC circuit in series, a resistor of 180  and a capacitor of 2.0 F are connected in series to an
a.c. voltage supply. If the frequency of the a.c. supply is 250 Hz, which of the following statements is true?
A A. The phase difference between voltage and current is 1.06 rad
B. The impedance of the circuit is 498 
C. The voltage is leading current
D. Total electric power is zero
Ans: Given: R = 180 , C = 2.0 F,
1 1 VR
XC= = =318.31 ( Ω )
2 πfC 2 π × 250 ×2 ×10−6
−1 X C −1 318.31
θ=tan =tan =1.06 rad VT
R 180 VC
The current leads voltage by 1.06 rad
Z=√ R 2+ X 2C = √1802 +318.312=365.68 ( Ω )
P=IVcosθ ≠ 0

U2k20-29 The magnetic flux through a coil of 5 turns changes with time t according to the equation ϕ =t 2. What
is the e.m.f. induced in the coil at the instant of t = 3 s?
B A. 15 V B. 30 V C. 45 V D. 60 V
Ans: The magnitude of the induced emf is
d ϕm d 2
ε =−N =−5 t =−5 ×2 t |t =3 =−5 ×2 ×3=−30(V )
dt dt
Dong Zong Copyright Reserved
U2k20-30 A lightning bolt on the equator is caused by the sudden flow of electrons vertically downward into the
earth. In which direction would the lightning be deflected by the magnetic field of the Earth?
B A. East B. West C. South D. North

-10-
Dong Zong Copyright Reserved