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MARKING
SCHEME
KEDAH
PHYSICS TRIAL
STPM 2010
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ANSWER SCHEME – PHYSICS TRIAL 2010 PAPER 1
1 D 11 A 21 A 31 A 41 B
2 D 12 D 22 A 32 D 42 D
3 A 13 D 23 D 33 C 43 D
4 B 14 D 24 A 34 B 44 A
5 B 15 C 25 D 35 C 45 D
6 B 16 B 26 C 36 C 46 B7 B 17 A 27 D 37 C 47 A
8 D 18 C 28 B 38 B 48 D
9 B 19 C 29 A 39 D 49 B
10 B 20 A 30 C 40 D 50 B
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MARKING SCHEME
KEDAH
PHYSICS TRIALSTPM 2010
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2
PAPER 2
No Solution Marks
PART A
1 (a)
(b)
0.2100
75
kW 5.1
Fv P
11.38.950
1500
ms F
P v
1
1
1
1,1
2
2
324
GT
r M
23
s
E
E
s
E
s
T
T
r
r
M
M
2
7
63
8
11
102.3
104.2
108.3
105.1
E
s
M
M
5105.3 E
s
M
M
1
1
1
1
3 (a)
3 (b)
g
l T 2
8.9
12 T
sT 01.2
ol l h 10cos
)10cos(2 02 l l g v
)10cos(2 ol l g v
)10cos1)(8.9(2 ov
155.0 msv
1
1
1
1
1
1
4 (a)
4 (b)
4 (c)
Young’s modulus =
l e A
F
= A
l
e
F
=
=
Strain energy = Fe2
1
= )100.2(1002
1 3
= 0.10 J
Assumption: Extension is within the limit of proportionality.
1
1
11
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3
5 (a)
(b)
CR
t
oeQQ
(1 – 0.63) Qo = Qo 6108.1
75.0
C e
6108.1
75.0
C = ln 0.37
C = 4.19 x 10-7 F
U =C
Q2
2
1 ………………………… (1)
U’ =C
Q 2)37.0(
2
1 ……………………(2)
(2) (1)U
U ' = 0.372
U’ = 0.137U Therefore, % of energy remains = 13.7%
1
1
1
1
1
6 (a)
(b)
(c)
Gain =
k
k
1
10+ 1 = 11
V in =k k
k
1.08.1
8.1
x 2
= 29.1
8.1 = 1.89 V
V o = 1.89 x 11 = 20.8 VSaturation occur, voltmeter reading = 9 V
V o = 112508.1
8.1
k k
k
= 0.76 V
1
1
11
1
1
7 (a)
(b)
(c)
min
hc = eV
V =e
hc
min
=
V 1910
834
106.11054.1
100.31063.6
= 1.81 x 104 V
For smallest angle of diffraction, n = 12d sin 160 = 1(1.54x10-10)
= 2.79 x 10-10 m
1
1
1
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4
Each corner of a cube contains8
1 of an ion. Thus, only one ion occupies a
volume of d 3.
Therefore, no. of ions in 1 m3 of NaCl =3
1
d =
3101079.2
1
= 4.6 x 1028
1
1
8 (a)
8 (b)
(i) random: At any instant, the chance of a radioactive atom decaying is the
same for all radioactive atoms./ Cannot be predicted
(ii) spontaneous: The process cannot be controlled by physical conditions or
external factors
(i) Number of proton decay in 5 year = 510
103
2
1
34
33
= 1 proton
(ii) mc =
h
m =c
h
m/me = 1.0
1
1
1
1
1
PART B
9 (a) Newton’s second law of motion :
Net or resultant force acting on an object is directly proportional to the rate of
change of momentum.
During collision suppose F1 and F2 are acting on two objects m1 and m2.
According to Newton’s second law :
F1 =t
umvm 1111
F2 =t
umvm 2222
According to Newton’s third law :
F1 = -F2
1
1
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5
t
umvm 1111 = -t
umvm 2222
m1v1 – m1u1 = - m2v2 + m2u2
m1u1 + m2u2 = m1v1 + m2v2
1
1
9 (b) (i) F
A 2T
2T T
B
3.0g
3.0g
(ii) For A :
F - 3T - 3.0g = 3(7.0)
For B :
2T - 3.0g = 3(10.5)
Solving the 2 equations :
F = 141.9 N
T = 30.5 N
1
1
1
1
1
1
9 (c) (i) For motion from A to B
fd = ½ mvo2 - ½ mvB
2
mgd = ½ mvo2 - ½ mvB
2 [ f = mg ]
gd = ½ vo2 - ½ vB
2
0.3(9.81)(2.0) = ½ (4)2 - ½ vB2
vB = 2.06 m s-1
(ii) For motion from B to C
½ mvC2 - ½ mvB
2 = mgh
vC2 - vB
2 = 2gh
vC = 4.89 m s-1
1
1
1
1
1
10 (a) (i) Number of independent ways in which a molecule can possess energy.
(ii) Boltzmann constant
(iii) Number of molecules in 1 mol of gas = N A where N A is the Avogrado
number.
Total kinetic energy of 1 mol of ideal gas = bT f
2( N A)
= RT f
2
, R is the molar gas constant
Internal energy of a gas
= total kinetic energy of molecules + total potential energy of the
molecules
For 1 mol of an ideal gas,
1
1
1
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6
Internal energy = total kinetic energy of molecules, as potential energy = 0
Therefore for 1 mol of an ideal gas,
Internal energy, U = RT f
2
Therefore
U = 2
f
R T
According to the first law of thermodynamics, Q = U + W
When a gas absorbs energy at constant volume, Q = U , as W = 0
Therefore Q = U =2
f R T
Molar heat capacity at constant volume, C V =T
Q
=2
f R
1
1
10 (b)(i) Using
2
2
1
1
T
V
T
V to find the final temperature
2
'2
273
'
T
V V
T 2 = 546 K
Q = C p T
= ( 2
f R + R) T
=2
7(8.31) (546 – 273) ( substitution)
= 7940 J (with unit)
(ii) Q = W as U = 0 for isothermal process.
W = RT ln ('
'2
V
V )
= (8.31) (273) (ln 2) (substitution)
= 1570 JTherefore Q = 1570 J (with unit)
1
1
1
1
1
1
10 (c) (i) Let the temperature of the junction be
(0.05) (25 - ) = 0.13 ( - 0) (substitution)
= 6.9 oC (with unit)
(ii) rate of heat flow through rubber
= k A temperature gradient
= 0.13 100 10-4
3102
09.6
(substitution)
= 4.5 J s-1 (with unit)
1
1
1
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7
5.0 mm
B
+ + + + + + + + +
11 (a) Mutual induction is the phenomenon where an e.m.f. is induced in a conductor
when the current in a neighbouring conductor is changing.
Primary coil (solenoid): N2 turns, length l , radius r .
Secondary coil (narrow): N1 turns, radius r .
When current I flows in primary coil, magnetic flux density in the
solenoid is
B = µonI ( n = N2/l )
magnetic flux linkage in the secondary coil
Φ = N1BA
= N1(l
I N 20 )( 2r )
Mutual inductance, M = Φ/I
M = l
r N N 2
210
2
1
1
1
11 (b)(i) V = IR = I x ρ
A
l
5 = I x 5 x 10-2 x 10 x 10-3 x6105.05
1
v
5 = 200I
I = 0.025 A
= 25 mA
(ii) I = Aven
22196 104106.1105.05
025.0
Aen
I v
= 25/16
= 1.563
= 1.6 m s-1
(iii) F = Bqv
= 0.050 x 1.6 x 10-19 x 1.56
= 1.248 x 10-20
= 1.2 x 10-20 N
(iv) VH = Bvd
= 0.05 x 1.56 x 5 x 10-3
= 0.39 mV
OR VH =ntq
BI
=19322
106.1105.0104
025.0050.0
= 0.00039
= 0.039 mV
1
1
1
1
1
1
1
1
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8
0.50 mm
10 mm
I
- - - - - - - - - - - - --
correct faces ----------- (1)
correct signs ----------- (1)
12 (a) (i)
Applying Kirchhoff’s 2nd rule to upper loop
I1 5 + (I1 + I2) (2 + 1) = 12 + 6
Applying Kirchhoff’s 2nd rule to lower loop
I2 9 + (I1 + I2) (2 + 1) = 8 + 6
Ammeter reading = I1 = 2.0 A
(ii) I2 = 0.67 A
Voltmeter reading = - (2 + 0.67) x 2 + 8
= 2.66 V
( or any logical method )
1
1
1
1
1
1
12 (b)(i) χ c =
rms
rms
I
V
=2
oV .rms I
1
=2
6.
6.1
1
= 2.7 Ω
(ii) χ c = fC 2
1
C =7.22
01.0
= 5.9 x 10-4 F
1
1
1
1
1
1
I1
I2
I1 + I2
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9
(iii)
Show cosine graph
Show value 2.3 and – 2.3
Show value 0.01
(0 mark if axes not label and no unit shown)
1
11
13 (a) (i) Threshold frequency is the minimum frequency required to emit an electron
from the surface of a metal.
(ii) Work function is the minimum work required to emit an electron from the
surface of a metal
1
1
(b) - Photoelectrons can only be emitted if the frequency of the radiation is greaterthan the threshold frequency of the metal
- Photoelectric emission is an instantaneous process
- The kinetic energy of the photoelectron emitted is independent of the intensity
of the radiation
** any 2 answers 2
(c) (i) maximum kinetic energy, ½ mvmax = hc/ λ - W
= (6.63 x 10-34) )1013.4
100.3(
7
8
- 20(1.6 x 10-19)
= 1.62 x 10-19 J
(ii) ½ mvmax2 = 1.62 x 10-19 J
mvmax = 3119 1011.91062.12
de Broglie wavelength, λ =max
mv
h =
3119
34
1011.91062.12
1063.6
= 1.22 x 10-9 m
(iii) If n is the number of electrons per second falling on the caesium surface,
n
hc = 5.0 x 25 x 10-4
n =
834
47
100.31063.6
10250.51013.4
1
1
1
1
1
0.01 0.02
I/A
t/s
2.3
-2.3
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10
the number of electrons emitted =100
60 x
834
47
100.31063.6
10250.51013.4
= 1.56 x 1016
1
1
(d)
(i) electric potential energy = r
e
0
2
4 =
1112
219
1031.51085.84
1060.1
= -4.34 x 10-18 J
(ii) from2
0
22
4 r
e
r
mv
kinetic energy, ½ mv2 =r
e
0
2
8
=
1112
219
1031.51085.88
1060.1
= 2.17 x 10-18 J
1
1
1
1
14 (a)
14 (b)
14 (c)
(i) the difference between the total mass of the constituent nucleons and the
mass of the nucleus .
(ii) the energy required to separate completely all the nucleons in the nucleus
E = Δm c2
E = binding energy
Δm = mass defect
c = speed of light in vacuum
( i ) N 14
7 + He4
2 O17
8 + H 1
1
(ii) Δm = [ (2.82282 + 0.16735) x 10-26 ] – [ ( 2.32500 + 0.66466) x 10-26 ]
= 5.1 x 10-30 kg
Minimum kinetic energy of the alpha particle = Δm c2
= 5.1 x 10-30 x (3 x 108)2
= 4.59 x 10-13 J
(i) Usable energy per fission = 200 x 106 x 1.6 x 10-19 x100
20
= 6.4 x 10-12 J
1
1
1
1
2
1
1
1
1
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8/12/2019 [Edu.joshuatly.com] Kedah STPM Trial 2010 Physics [w Ans] [0453A76A]
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11
Number of fission required per second =12
8
104.6
107
x
x
= 1.094 x 1020
Number of uranium atom used in one day = 1.094 x 1020 x 24 x 60 x 60
= 9.45 x 1024
(ii) Mass of uranium used in each day =23
24
1002.6
1045.9
x
x (0.235)
= 3.68 kg
1
1
1
1