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CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2013 series
9702 PHYSICS
9702/41
Paper 4 (A2 Structured Questions), maximum raw mark 100
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before m

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CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2013 series
9702 PHYSICS
9702/41
Paper 4 (A2 Structured Questions), maximum raw mark 100
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 41
© Cambridge International Examinations 2013
Section A 1 (a)
region of space area / volume B1 where a mass experiences a force B1 [2]
(b) (i)
force proportional to product of two masses M1 force inversely proportional to the square of their separation M1
either
reference to point masses
or
separation >> ‘size’ of masses A1 [3]
(ii)
field strength =
GM
/
x
2
or
field strength
∝
x
C1 ratio = (7.78
×
10
8
)
2
/ (1.5
×
10
8
)
2
C1 = 27 A1 [3]
(c) (i)
either
centripetal force =
mR
ω
2
and
ω
= 2
π
/
T or
centripetal force =
mv
2
/
R
and
v
= 2
π
R
/
T
B1 gravitational force provides the centripetal force B1
either
GMm / R
2
=
mR
ω
2
or
GMm / R
2
=
mv
2
/
R
M1
M
= 4
π
2
R
3
/
GT
2
A0 [3]
(allow working to be given in terms of acceleration)
(ii)
M
= {4
π
2
×
(1.5
×
10
11
)
3
} / {6.67
×
10
–11
×
(3.16
×
10
7
)
2
} C1 = 2.0
×
10
30
kg A1 [2]
2 (a)
obeys the equation
pV
= constant
×
T
or
pV = nRT
M1
p, V
and
T
explained A1 at all values of
p
,
V
and
T
/fixed mass/
n
is constant A1 [3]
(b) (i)
3.4
×
10
5
×
2.5
×
10
3
×
10
–6
=
n
×
8.31
×
300 M1
n
= 0.34
mol A0 [1]
(ii)
for total mass/amount of gas 3.9
×
10
5
×
(2.5 + 1.6)
×
10
3
×
10
–6
= (0.34 + 0.20)
×
8.31
×
T
C1
T
= 360
K A1 [2]
(c)
when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3]
Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 41
© Cambridge International Examinations 2013
3 (a) (i) 1.
amplitude = 1.7
cm A1 [1]
2.
period = 0.36
cm C1 frequency = 1/0.36 frequency = 2.8
Hz A1 [2]
(ii)
a
= (–)
ω
2
x
and
ω
= 2
π
/
T
C1 acceleration = (2
π
/0.36)
2
×
1.7
×
10
–2
M1 = 5.2
m
s
–2
A0 [2]
(b)
graph: straight line, through srcin, with negative gradient M1 from (–1.7
×
10
–2
, 5.2) to (1.7
×
10
–2
, –5.2) A1 [2]
(if scale not reasonable, do not allow second mark)
(c)
either
kinetic energy = ½
m
ω
2
(
x
02
–
x
2
)
or
potential energy = ½
m
ω
2
x
2
and potential energy = kinetic energy B1 ½
m
ω
2
(
x
0
–
x
2
) = ½
×
½
m
ω
2
x
02
or ½
m
ω
2
x
2
= ½
×
½
m
ω
2
x
02
C1
x
02
= 2
x
2
x
=
x
0
/ √2 = 1.7 / √2 = 1.2
cm A1 [3]
4 (a)
work done moving unit positive charge M1 from infinity (to the point) A1 [2]
(b)
(gain in) kinetic energy = change in potential energy B1 ½
mv
2
=
qV
leading to
v
= (2
Vq
/
m
)
½
B1 [2]
(c)
either
(2.5
×
10
5
)
2
= 2
×
V
×
9.58
×
10
7
C1
V
= 330
V M1 this is less than 470
V and so ‘no’ A1 [3]
or
v
= (2
×
470
×
9.58
×
10
7
) (C1)
v
= 3.0
×
10
5
m
s
–1
(M1) this is greater than 2.5
×
10
5
m
s
–1
and so ‘no’ (A1)
or
(2.5
×
10
5
)
2
= 2
×
470
×
(
q
/
m
) (C1) (
q
/
m
) = 6.6
×
10
7
C
kg
–1
(M1) this is less than 9.58
×
10
7
C
kg
–1
and so ‘no’ (A1)
Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 41
© Cambridge International Examinations 2013
5 (a)
(uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1 (creates) force per unit length of 1
N
m
–1
A1 [2]
(b) (i)
flux density = 4
π
×
10
–7
×
1.5
×
10
3
×
3.5 C1 = 6.6
×
10
–3
T A1 [2]
(ii)
flux linkage = 6.6
×
10
–3
×
28
×
10
–4
×
160 C1 = 3.0
×
10
–3
Wb A1 [2]
(c) (i)
(induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2]
(ii)
e.m.f. = (2
×
3.0
×
10
–3
) / 0.80 C1 = 7.4
×
10
–3
V A1 [2]
6 (a) (i)
to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2]
(ii)
either
no power loss in transformer
or
input power = output power B1 [1]
(b)
either
r.m.s. voltage across load = 9.0
×
(8100 / 300) C1 peak voltage across load = √2
×
243 = 340
V A1 [2]
or
peak voltage across primary coil = 9.0
×
√2 (C1) peak voltage across load = 12.7
×
(8100/300) = 340
V (A1)
7 (a) (i)
lowest frequency of e.m. radiation M1 giving rise to emission of electrons (from the surface) A1 [2]
(ii)
E
=
hf
C1 threshold frequency = (9.0
×
10
–19
) / (6.63
×
10
–34
) = 1.4
×
10
15
Hz A1 [2]
(b)
either
300
nm ≡ 10
×
10
15
Hz (and 600
nm ≡ 5.0
×
10
14
Hz)
or
300
nm ≡ 6.6
×
10
–19
J (and 600
nm ≡ 3.3
×
10
–19
J)
or
zinc
λ
0
= 340
nm, platinum
λ
0
= 220
nm (and sodium
λ
0
= 520
nm) M1 emission from sodium and zinc A1 [2]
(c)
each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1 [3]

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