TRIAL2012PAPER1(Teachers)Edited
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JAWATANKUASA PANEL TRIAL EXAM PHYSICS STPM NEGERI SEMBILAN DARUL KHUSUS
P where P is the power dissipated when a current I I flows. The e.m.f. Ec induced in a coil by changing magnetic flux is equal to the rate of change of flux, 1.The e.m.f. Eb of a battery is given by Eb =
EC =
A
dφ . Which of the following is a unit for magnetic flux? dt
ms-1 A
ANSWER :
B
ms-2A-1
C
kgm2s-2A
kgm2s-2A-1
D
dφ= Edt= (kgm2s-2 A-1)(s) = kg m2 s-1 A-1
D
; E = P/I = (kgm2s-2)(A-1)
2. The graph shows how the force acting on a body varies with time
Force( N) 6 2 2
0
Time / S
6
Assuming that the body is moving in a straight line, by how much does its momentum change? A 40 Kgms-1
B 36 kgms-1
C 20 kgms-1
D 16 kgns-1
ANSWER : C : Change in momentum = area under graph
= (2)(2) + ½ (2+6)(4) = 20 kgms-1
3. A lorry and a car experience a head-on collision. Among the following quantities which possess the same magnitude for both vehicles? A only impulse C Forces and impulse
B only forces D Impulse and speed variation
ANSWER: C
4. A raindrop of mass, m is falling vertically through the air with steady speed v. The raindrop experiences a retarding force, kv due to the air, where k is a constant. The acceleration of free fall is g. Which expression gives the kinetic energy of the raindrop?
2
A
mg k
B
mg 2 2k 2
C
m3 g 2 k2
D
m3 g 2 2k 2
ANSWER : D mg =kv v= mg/k K.E. = ½ mv2 = 1/2m (mg/k)2 = m3g2/2k2 5. A body moves in a circle with increasing angular velocity. At time t, the angle θ swept out by the body and its angular velocities ω are as follows: t (s) 2 4 6 8
θ (rad) 14 44 90 152
ω (rad s-1) 11 19 27 35
The angular acceleration of the body A gradually increases and is 4.5 rads-2 when t = 6s B is constant at 4 rad s-2 C is constant at 8 rad s-2 D increases at a constant rate and is 15 rad s -2 when t = 6 s
ANSWER : B 6. A ball rolls down an inclined plane and its velocity at the bottom of the plane is V r. When the same ball slides down the inclined plane without rolling, its velocity at the bottom of the plane is Vs. Which of the following is correct?
A Vr > Vs, because the rolling ball has both linear and angular acceleration B Vr > Vs because the gain in kinetic energy causes the rolling ball to move faster C Vr < Vs, because work is done against friction when the ball rolls. D Vr < Vs, because a rolling ball has translational and rotational kinetic energy. (ANSWER:D)
3
7.
L=100.0 cm
O x
40.0 N
60.0 N
A student carries a uniform rod of 30.0 N and length L= 100.0 cm. Two loads 40.0 N and 60.0 N is hung at both ends as shown in the above figure. The value of x so that the system is in equilibrium is A 15.8 m
B 20.6 cm
C 40.0 cm
D 42.3 cm
ANSWER : D 0.5 m
x Refer point S : Use Static equilibrium for rigid body: (60)(x) = (30)(0.5-x) +(40)(1-x) 60x = 15-30x+40-40x S 130x = 55 60N X =55/130 =0.423m = 42.3 cm
W=30 N
8.
40 N
Earth
.
A
B R
4R
3R
P
The above figure shows a stationary object P is placed at point A away from the surface of the Earth. M and R are the mass of the Earth and radius of the earth respectively. If the object falls freely from point A to point B and the resistance of the air is neglected. What is the speed of the object as it reaches the point B? A
GM 6R
B
2GM 3R
C
4
3GM 4R
D
GM 4R
F
ANSWER : D Use principle oh conservation of energy; K.E gain = The change of gravitational potential energy
1 GMm GMm mV 2 = − − (− ) 2 8R 4R 1 GMm 1 1 1 GMm mV 2 = ( − ) ⇒ mV 2 = 2 R 4 8 2 8R
r
0
⇒V =
GM 4R
9. A particle is moving such that the force F acts on it changes with the distance r from a fixed point as shown.
Which graph shows the relationship between the potential energy EK of the particle and the distance r?
EK
EK
0
0
r
r
A
B EK
EK
r
0
0
C
D
Suggested Answer : B The particle is moving in simple harmonic motion. K +U = ETotal ,
5
r
Since F is a straight line with negative gradient, EP is a quadratic curve with a minimum turning point, So that EK is a quadratic curve with a maximum turning point base on the conservation of energy. 10. Three object X, Y and Z are in simple harmonic motion as shown. X
Mass between two springs
Z YY Vibrating mass
Simple pendulum
In which system(s) will the period increase if the mass increases? A B C D
X only Y only X and Y only Y and Z only
Answer: C
: For the spring systems, T = 2π
m k
Pendulum ,T = 2π
⇒
l g
11. A pendulum is driven by a sinusoidal driving force of frequency f. Which graph best shows how the amplitude a of the motion of the pendulum varies with f?
A
B
C D Suggested Answer : A Graph A shows the phenomenon of resonance i.e. the amplitude a is high when driving force frequency equals the natural frequency of the pendulum.
6
12. Which of the following represents a transverse wave of amplitude 4 m, wavelength 1 m and frequency 2 Hz propagating in the positive x direction? A B C D
y = 4 sin (4πt+2πx) y = 4 sin ( 4πt-2πx) y = 4 sin (2πt+ πx) y = 4 sin (2πt-πx)
Answer : B
:
k=
2π
λ
=
2π = 2π ω = 2πf = 2π (2) y = A sin(ωt − kx) 1 , ,
13. A displacement-time graph is shown for a particular wave.
A second wave of similar type has twice the intensity and half the frequency. When drawn on the same axes, what would the second wave look like?
Suggested answer : B
7
Intensity α (amplitude)2 2 × intensity α (
amplitude)2 New amplitude =
A = 1.4 A
The frequency is halved means its period is doubled, i.e. the wave takes a longer time to completenone cycle. 14. The threshold of hearing for the human ear is normally at the intensity of 10 -12 Wm-2. The corresponding intensity level for sound intensity is 0 dB. If the intensity level is 100 dB, what is the sound intensity? A 10-11 Wm-2
B 10-7 Wm-2
C 10-10 Wm-2
D 10-2 Wm-2
Answers: D The intensity level = 10 log
I I0
⇒
100= 10 log
I ⇒⇒ 10 −12
I = 10-2 Wm-2
15. The length ℓ of an air column is slowly increased from zero while a note of constant frequency is produced by a loudspeaker placed above it.
l
When ℓ reaches 17 cm, the first loud sound is heard. The wavelength of the sound wave is A
8.5 cm
B
17 cm
C
34 cm
D
68 cm
Suggested answer: D When ℓ = 17 cm, there is an antinode at the loudspeaker and a node at the plunger. λ/4 = ℓ, λ = 68 cm 16. The diagram shows the variation of potential energy U between two atoms with their distance of separation, r for a solid expanding due to an increasing temperature.
8
When the temperature of the solid is raised from the temperature of absolute zero, the potential energy of the atom increases from Uo to U1. The separation of the atoms compared with the separation of the atoms at equilibrium will increase by A
r2 – r 1
B
Ans:D
r3 – r 1
C
r4 – r 1
D
r3 – r 2
Increasement= r3 – r2
17. Two wires have the same diameter and length. One is made of copper, the other brass. The wires are connected together end to end. When the free ends are pulled in opposite direction, the two wires must have the same A Stress
B Strain
C Young’s modulus
D Ultimate strength
Ans: A. Both have same cross sectional area and experience same force. 18. Nitrogen gas at a pressure of 1.01 speed of the molecules is A 256 m s-1
× 10 5 Pa has a density of 2.5 kg m −3 . The root mean square
B 318 m s-1
C 348 m s-1
D 384 m s-1
Answer: C Applying P = 1.01
1 ρ 3
× 10 5 = 1 × 2.5 < c 2 > 3
< c2 >
=
121200
= 348 m s −1 19. An ideal gas undergoes adiabatic changes. Which of the following graphs show the relationship between the temperature T and its volume, V?
ln T
ln T
ln V
ln V A
ln T
B
ln T
ln V 9
ln V C
D
Answer : A
TVγ-1 = constant ln T + ln V γ −1 = constant ln T = c – (γ-1) ln V 20 . When n mole of an ideal gas is heated at constant pressure, the internal energy of the gas increases by ∆U when the temperature increases by ∆T. The molar heat capacity of the gas at constant pressure cp,m is A
p∆V∆U n∆T
B
∆U∆T np∆V
C
∆U + p∆V n∆T
D
p∆V − ∆U n∆T
Answer : C
First law of thermodynamics : ∆Q = ∆U + W
⇒
nCpm ∆T = nCvm ∆T + nR∆T
21. A combined rod consists of a rod Y of length 2ι and thermal conductivity, λ1 in series with a rod Z of length ι and thermal conductivity λ2 of the same cross-sectional area. The combined rod is perfectly lagged. The ends of the combined rod are maintained at 100 oC and 0 oC. If the temperature of the junction is 60oC, what is the ratio of
A 1/3
B
2/3
C
λ1 ? λ2
3/2
D
3
Answer : D dQ (100 − 60) (60 − 0) dQ dθ = kA = λ1 A = λ2 A = dt dx dt 2l l 22. Figure shows two metal rods each of length 0.40 m that are placed between two blocks with a temperature difference of 40K so that the heat can flow from the hotter block to the colder block. The two metal rods are insulated and both have same thermal conductivity of 35 Wm -1K-1. If the crosssectional areas of the two metal rods are 1.0 x 10-4 m2 and 5.0x 10-4 m2, calculate the total rate of heat transfer between the two blocks.
A 0.5 W B 1.0 W C 2.1 W D 3.0 W
10
Answer : C dQ 40 40 −4 = (35)(1.0 x10−4 ) Total + (35)(5.0 x10 ) = 2.1W dt 0.40 0.40
23.
2cm R
-1 µC
3cm S
T
+2 µC
+3µC
Three point charges -1µC, +2µC and +3µC are fixed along a straight line as shown in the diagram. Calculate the resultant force acts on charge S by the charge R and T? A 11 N towards T
C 15 N towards R
B 15 N towards T
D 105 N towards R
ANSWER : D Direction of both electric force; FSR & FST which act on S are toward left 2x10-6
24.
+
150 -
A pair of metal plates is placed vertically and parallel to each other. They are separated by a distance of 8.0 cm. One plate is charged positively while the other is charged negatively as shown in the above figure. A charged ball of mass 50 g and charge q is suspended in the middle of the space between the plates. If the p.d. across the plates is 210.0 V and makes the string which suspends the ball becomes inclined to the vertical by an angle of θ= 15o. The magnitude of charge q is
ANSWER C A. 100.0: μC B. 80.5 μC T cos 150 = mg (1) T Sin 15o = qV/d (2)
C. 50.1 μC
150
D. 40.6 μC
T F = qE=q(V/d)
Eq(2)÷Eq(1) Tan 150 =qv/mgd mg 11 0 q= mgdtan 15 /V = (50 x 10-3)(9.81)(0.08)tan 150 / 210 =5.01x 10-5= 50.1 μC
25 A capacitor shown in the circuit below is not charged initially. Which of the following graphs best shows the variation of the potential difference V across the resistor R with time t after switch S is closed?
Answer : D 26 Current I flows through wire P with the charge carriers moving with drift velocity v. What is the velocity of the charge carriers if the current I flows through another wire of the same material but with its radius doubled that of the radius of the first wire? 1 v 4
A
B
1 v 2
C
v
D
2v
Answer: A For the first wire: I = πr 2 nve
(
)
For the second wire: I = 4πr 2 nv ' e
(
)
Since I is constant, hence I = (πr 2 )nve = ( 4πr 2 )nv ' e v' =
1 v 4
27. The electric conductivity of a metal will decrease if its temperature increases because A B C D
the speed of free electrons decreases. the number of free electrons decreases. the metal ions vibrates more vigorously. the average velocity of metal ions decreases.
Answer: C
12
When the temperature increases, metal ions will vibrate with greater amplitude and hence will block the flow of free electrons. 28. The figure below shows an arrangement of three resistors with resistance of R, R and 10.00 Ω respectively. Given that resultant resistance between X and Y is 5.00 Ω .
Z R
R
X
10.00
Y
What is the resultant resistance between X and Z? A 2.25 Ω
B 3.75 Ω
C 3.90 Ω
D 4.25 Ω
Answer: B
R XY = 5.00 −1
1 1 + 10 . 00 2 R R = 5.00Ω
= 5.00
−1
1 1 R XZ = + 5 . 00 15 .00
= 3.75Ω
29 A milliammeter with negligible internal resistance and a full scale deflection of 1 mA is modified for use as a voltmeter with a full scale defection of 10 V. The correct modification for the required circuit is A B C D
a resistor of 102 is connected in parallel to the milliammeter. a resistor of 103 is connected in parallel to the milliammeter. a resistor of 104 is connected in series to the milliammeter. a resistor of 105 is connected in series to the milliammeter.
Answer: C V AB = IR 10 = (1x10 −3 ) R R = 10 4 Ω
30. The magnetic flux density B at a distance d from a long wire carrying a current I is given by
B=
µo I where µo is the permeability of free space. If a cable carries an alternating current of 20 A, 2πd
what is the distance of the cable so that the maximum magnetic flux density is 1 µ T? A 3.7 m
B 4.7 m
C 5.7 m
Answer: C
13
D 6.6 m
Maximum current =
2 I = 2 x 20 = 28.3 A
µo I 2πd µ I ( 4πx10 −7 )( 28.3) d = o = = 5.7m 2πB 2π (1.0 x10 −6 ) B=
31. Three long wires P, Q and R are arranged parallel with one another in a vacuum as shown in figure below.
P Q R
1.0 A 2.0 AA 1.0 AA
10.0 cm 10.0 cm
Given that the force per unit length between 2 wires, each carrying current of 1.0 A and separated by 10.0 cm is 2.0 x 10-6 Nm-1. What is the resultant force per unit length experienced by wire R? A 2.0 x 10-6 Nm-1
B 3.0 x 10-6 Nm-1
C 4.0 x 10-6 Nm-1
D 5.0 x 10-6 Nm-1
Answer: B 2.0 x10 −6 = 1.0 x10 −6 Nm −1 2 = 2 x 2.0 x10 −6 = 4.0 x10 −6 Nm −1
Fattraction ( P −R ) = Frepulsion ( Q −R )
Hence resultant force per unit length = (4.0-1.0)x10 -6= 3.0 x 10-6 Nm-1 32. Inserting a pure iron core into a solenoid can increase the self-inductance. This is because the iron core will A B C D
increase the current flow. strengthen the flux linkage. reduce the resistance of the solenoid. increase the mutual inductance between the core and the solenoid.
Answer: B B = µo nI
Since the permeability µ for the pure iron core is greater than µo , hence B is strengthened.
E
33. A conductor is placed in a magnetic field with magnetic flux Φ that varies with time t according to the relationship Φ = kt 2 , where k is a constant. Which of the following graphs represents the variation of induced e.m.f. E of the conductor with time t? A
B
0 E
0
t
14
t
C
D
E
E
t
0
t
0
Answer: D
dΦ d ( kt 2 ) = = 2kt dt dt ∴ Eα t E=
34 A sinusoidal current from an electric source is given by I = I o sin ωt . Which of the following expressions for the current is true when the amplitude and frequency of the current are increased to twice its initial amplitude and frequency respectively? 1 A I o sin 2ωt B 2 I o sin ωt C 2 I o sin ωt D 2 2 I o sin 2ωt Answer: D I ' = 2 I o sin 2π ( 2 f ) t = 2 I o sin 2ωt 35. A sinusoidal supply of frequency 100 Hz and r.m.s. voltage 12V is connected to a 2.2 μF capacitor. What is the r.m.s. value of the current? A
5.5mA
Answer : D Reactance of the capacitor is
B
26mA
XC =
C
0.42mA
1 2πfC
D
I rms =
1 2π (100)(2.2 ×10 −6 ) = 723Ω =
Vrms Xc
=17 mA
36. The diagram shows an op-amp circuit. Determine the values of the resistance R.
15
17mA
A B C D
2 kΩ 5 kΩ 60 kΩ 80 kΩ
Answer : D 37 The diagram shows the relationship between the energy of electromagnetic radiation and the wavelength of the waves.
Which of the following has the lowest energy? A infra-red B microwaves
C
ultra-violet
D
X-rays
Suggested answer: B The e.m. radiation with largest wavelength has the smallest energy. Order of the radiation in decreasing wavelength or increasing frequency is Microwaves, infra-red, ultra-violet, X-rays. Since microwaves has the largest wavelength among the four, its thus has the lowest energy. 38 Two converging lenses P and Q of focal lengths 10 cm and 5 cm respectively are arranged at a separation of 40 cm between each other. An object is placed at a distance of 15 cm from lens P as shown in the figure below.
Which of the following is true of the final image and its distance from lens Q?
16
A B C D
Characteristic of image Real Virtual Real virtual
Suggested answer: A For lens P:
Distance of image from lens Q 10.0 cm 10.0 cm 6.0 cm 6.0 cm For lens Q: New object distance
39 A beam of monochromatic light of wavelength 600 nm is incident normally on a diffraction grating that has 3.0 × 105 lines per meter. What is the total number of diffracted fringes produced by light transmitted through this grating? A 5 B 8 C 10 D 11 Suggested answer: D Using d sin θ = nλ ( n = 5.6 ≈ 5 (integer) Total bright fringes seen is 5 (one side) + 1 (central bright fringe) + 5(the other side) = 11 40 The surface of a glass convex lens can be coated with a thin layer of a fluoride salt to reduce the reflection of light. What is the thinnest thickness of the salt layer if the coated surface does not reflect almost normal incident light of wavelength 6.0 × 10-7 m? (Reflective index of glass = 1.5 and refractive index of normal fluoride salt = 1.4) A 1.00 × 10-7 m C 1.50 × 10-7 m B 1.07 × 10-7 m D 2.00 × 10-7 m Suggested answer: B
The two reflected light at point P and point Q experience destructive interference to reduce the reflected light. This occurs when the path difference =
, that is
The minimum thickness of fluoride salt, 41. When the wavelength of a radiation that falls on a metal surface decreases, the maximum kinetic energy of the photoelectrons emitted from the metal surface will A increase.
17
B decrease. C not change. D increase and later becomes constant. K max = h
Answer: A
c
λ
− W will increase when λ decreases.
42 While conducting a photoelectric-effect experiment with light of a certain frequency, a reverse potential difference of 1.25 V is required to reduce the current to zero. What is the maximum speed of the emitted photoelectrons? A 1.55 x 104 ms-1
B 6.63 x 105 ms-1
C 4.40 x 1011 ms-1
D 1.65 x 1015 ms-1
Answer:
K max = eVo = (1.60 x10 −19 )(1.25) = 2.00 x10 −19 J K max =
1 2 mv max 2
⇒⇒
v max =
2 K max = m
2( 2.00 x10 −19 ) = 6.63x105 ms −1 9.11x10 −31
43 A light radiation with wavelength λ is incident perpendicularly onto a plane mirror and is then reflected. The intensity of the light is equal to the rate of photons, n , incident on the mirror. The force acting on the mirror due to the light radiation is nh A nhλ B 2nhλ C D
λ
2nh
λ
Answer: D Momentum of a photon, p =
h
λ
h Change of momentum of the photon, ∆p = p − ( − p ) = 2 p = 2 λ 2nh Force = Rate of change of momentum =
λ
44 A characteristic line corresponding with wavelength 1.85 x10 -11 m in an x-ray spectrum for gold is produced by an electron transition between 2 stationary states. If the lower stationary state is -80.9 keV, what is the energy of the upper stationary state? A -6.4 keV C -20.0 keV B -13.7 keV D -26.4 keV Answer B -80.9 – E1 = -67.2 E1 = -13.7 keV 45 If an electron collides with a target in an X-ray tube with a maximum speed v, the minimum X-ray wavelength produced is
. If the maximum electron speed is increased to 2v, find the minimum X-
ray wavelength produced in term of A
. C
18
B
D
Answer C
46 A population inversion means that A there are more atoms in one gas than in another. B there are more atoms in some excited state than in a lower state. C there are more states populated than unpopulated. D the lower states are filled rather than the higher ones. Answer B 47 If the mass of the
atom is 62.9296 u,
calculate the binding energy of the
atom is 1.00783 u, and neutron is 1.00867 u,
nucleus.
A 629 MeV
C 592 MeV
B 635 MeV
D 553 MeV
Answer D Total mass of constituent particles of
atom = 29 mp + 29 me + (63-29)mn = 29 m H + 34 mn = 29(1.00783 u) + 34(1.00867 u) = 63.5219 u
Mass defect,∆m = 63.5219 u – 62.9296 u = 0.5923 u Binding energy, EH = (∆m)c2 = 553 MeV 48 A
nucleus decays into a nucleus W after emitting a α-particle and two β-particle
consecutively. Find the number of protons and neutrons in W. Number of protons Number of neutrons A
92
146
B
92
142
C
90
146
D
90
142
Answer B +
+2
49 Nuclear fusion can occur A for two light nuclei, spontaneously.
19
B for light nuclei at temperature
108 K.
C if a thermal neutron bombards a light nucleus D during the collision of two light nuclei, each of kinetic energy Answer
B
50 Which of the following is a hadron? A Electron
C Tau
B Muon
D Meson
Answer D
20
4 keV
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