physics credit · pdf filecopper glycerol lead water 0.99 × 105 3.95 × 105 ... a...
TRANSCRIPT
2007 Physics
Standard Grade − Credit
Finalised Marking Instructions Scottish Qualifications Authority 2007 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from the Assessment Materials Team, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre's responsibility to obtain the necessary copyright clearance. SQA's Assessment Materials Team at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.
Physics – Marking Issues The current in a resistor is 1·5 amperes when the potential difference across it is 7·5 volts. Calculate the resistance of the resistor. Answers Mark + Comment Issue 1. V=IR (½) Ideal answer 7·5=1·5R (½) R=5·0 Ω (1) 2. 5·0 Ω (2) Correct answer GMI 1 3. 5·0 (1½) Unit missing GMI 2 (a) 4. 4·0 Ω (0) No evidence/wrong answer GMI 1 5. Ω (0) No final answer GMI 1
6. Ω 04 5·15·7 R ·===
IV
(1½) Arithmetic error GMI 7
7. Ω 04 R ·==IV
(½) Formula only GMI 4 and 1
8. Ω R ==IV
(½) Formula only GMI 4 and 1
9. Ω 5·15·7
R ===IV
(1) Formula + subs/No final answer GMI 4 and 1
10. 04 5·15·7 R ·===
IV
(1) Formula + substitution GMI 2 (a) and 7
11. Ω 05 5·75·1
R ·===IV
(½) Formula but wrong substitution GMI 5
12. Ω 5·0 5·1
75 R ===IV (½) Formula but wrong substitution GMI 5
13. Ω 05 5·15·7 R ·===
VI
(0) Wrong formula GMI 5
14. V = IR 7·5 = 1·5 × R R = 0·2 Ω (1½) Arithmetic error GMI 7 15. V = IR
Ω 20 5·75·1 R ·===
VI
(½) Formula only GMI 20
Page 2
DATA SHEET
Speed of light in materials Speed of sound in materials
Gravitational field strengths Specific heat capacity of materials
Specific latent heat of fusion of materials Melting and boiling points of materials
Specific latent heat of vaporisation of materials SI Prefixes and Multiplication Factors
Page three
Alcohol
Carbon dioxide
Glycerol
Turpentine
Water
Material
Air
Carbon dioxide
Diamond
Glass
Glycerol
Water
3.0 × 108
3.0 × 108
1.2 × 108
2.0 × 108
2.1 × 108
2.3 × 108
Speed in m/s Material Speed in m/s
Aluminium
Air
Bone
Carbon dioxide
Glycerol
Muscle
Steel
Tissue
Water
5200
340
4100
270
1900
1600
5200
1500
1500
Earth
Jupiter
Mars
Mercury
Moon
Neptune
Saturn
Sun
Venus
10
26
4
4
1.6
12
11
270
9
Gravitational field strengthon the surface in N/kg
Material Specific heat capacity in J/kg °C
Alcohol
Aluminium
Copper
Diamond
Glass
Glycerol
Ice
Lead
Water
2350
902
386
530
500
2400
2100
128
4180
Material
Alcohol
Aluminium
Carbon dioxide
Copper
Glycerol
Lead
Water
0.99 × 105
3.95 × 105
1.80 × 105
2.05 × 105
1.81 × 105
0.25 × 105
3.34 × 105
Specific latent heatof fusion in J/kg
Material Melting point in °C
Boilingpoint in °C
Alcohol
Aluminium
Copper
Glycerol
Lead
Turpentine
–98
660
1077
18
328
–10
65
2470
2567
290
1737
156
Material
11.2 × 105
3.77 × 105
8.30 × 105
2.90 × 105
22.6 × 105
Specific latent heatof vaporisation in J/kg
Prefix Symbol Factor
giga
mega
kilo
milli
micro
nano
G
M
k
m
µn
1 000 000 000 = 109
1 000 000 = 106
1000 = 103
0.001 = 10–3
0.000 001 = 10–6
0.000 000 001 = 10–9
Page 4
Question N
os 1(a), (b) and (c)
NO
TE
S
1.A
pu
pil is se
nt e
xam
resu
lts by a
text m
essa
ge o
n a
mob
ile p
hon
e. Th
e
freq
uen
cy o
fth
e sig
nal re
ceiv
ed
by th
e p
hon
e is 1
900
MH
z.
Th
e m
ob
ile p
hon
e re
ceiv
es ra
dio
waves (sig
nals).
(a)
Wh
at is th
e sp
eed
of
rad
io w
aves?
.........................................................................................................
(b)
Calc
ula
te th
e w
avele
ngth
of
the sig
nal.
(c)T
he p
up
il sen
ds a
vid
eo m
essa
ge fro
m th
e m
ob
ile p
hon
e. Th
e m
essa
ge
is tran
smitte
d b
y m
icro
waves. T
he m
essa
ge tra
vels a
tota
l dista
nce o
f
72
000
km
.
Calc
ula
te
the
time
betw
een
th
e
messa
ge
bein
g
tran
smitte
d
an
d
receiv
ed
.
122
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Space for w
orking and answer
3 × 10
8m/s
(1) or (0)
λλ=
= 0 .16
m
t = 0 .24
s ××
8
63
10 =
1900 10
×8
72000000 =
3
10
v fd=
v
OR
v = f λλ
3 × 10
8=
1900 × 10
6×
λλ
λλ=
0 .16m
(½)
(½)
(1)
(a) un
it required
(b) un
it penalty if
no con
version in
to Hz (–½
)
significan
t figure ran
ge:
0 .2 0 .16 0 .158 0 .1579
(c) un
it penalty if
no con
version in
to m (–½
)
Page 5
Question N
os 2(a)
NO
TE
S
2.R
ad
io w
aves h
ave a
wid
e ra
nge o
ffre
qu
en
cie
s.
Th
e ta
ble
giv
es in
form
atio
n a
bou
t diffe
ren
t waveb
an
ds.
(a)
Coastg
uard
s use
sign
als o
ffre
qu
en
cy 5
00
kH
z.
Wh
at w
aveb
an
d d
o th
ese
sign
als b
elo
ng to
?
.........................................................................................................
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
1
Waveband
Frequency R
angeE
xample
Low
freq
uen
cy
(LF
)30
kH
z –
300
kH
zR
ad
io 4
Med
ium
freq
uen
cy
(MF
)300
kH
z –
3M
Hz
Rad
io S
cotla
nd
Hig
h fre
qu
en
cy
(HF
)3
MH
z –
30
MH
zA
mate
ur ra
dio
Very
hig
h fre
qu
en
cy
(VH
F)
30
MH
z –
300
MH
zR
ad
io 1
FM
Ultra
hig
h fre
qu
en
cy
(UH
F)
300
MH
z –
3G
Hz
BB
C 1
an
d I
TV
Su
per h
igh
freq
uen
cy
(SH
F)
3G
Hz –
30
GH
zS
ate
llite T
V
med
ium
frequen
cy OR
mf
not: R
adio Scotlan
d
Page 6
Question N
os 2(b)
NO
TE
S
1112
2.(con
tinu
ed)
(b)
Th
e d
iagra
m sh
ow
s how
rad
io sig
nals o
fd
iffere
nt w
avele
ngth
s are
sen
t
betw
een
a tra
nsm
itter a
nd
a re
ceiv
er.
(i)W
hic
h o
fth
e w
aves in
the d
iagra
m sh
ow
s diffra
ctio
n?
.................................................................................................
(ii)W
hat d
oes th
is ind
icate
ab
ou
t the w
avele
ngth
of
the d
iffracte
d
wave c
om
pare
d to
the o
ther tw
o w
aves?
.................................................................................................
(iii)T
he
Earth
’s io
no
sph
ere
is
sho
wn
o
n
the
dia
gra
m.
T
he
ion
osp
here
is
a
lay
er
of
ch
arg
ed
p
artic
les
in
the
up
per
atm
osp
here. H
igh
freq
uen
cy w
aves a
re tra
nsm
itted
as sk
y w
aves.
Exp
lain
how
the tra
nsm
itted
waves re
ach
the re
ceiv
er.
.................................................................................................
(iv)
Su
per h
igh
freq
uen
cy (S
HF
) sign
als a
re sh
ow
n a
s space w
aves o
n
the d
iagra
m. A
lthou
gh
they c
an
on
ly tra
vel in
straig
ht lin
es, th
ey
can
be u
sed
for c
om
mu
nic
atio
ns o
n E
arth
betw
een
a tra
nsm
itter
an
d re
ceiv
er.
Desc
ribe h
ow
the S
HF
sign
als g
et to
the re
ceiv
er.
.................................................................................................
.................................................................................................
.................................................................................................
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
space wave
sky wave
receiv
er
Not to
scale
ion
osp
here
tran
smitte
r
surface
wave
Earth
surface (w
aves)
(b) (ii)accept:big
ger/larger/greater/high
er/large/high
not: “w
ider”, an
y answ
er based on
frequen
cy
(b) (iii)accept:reflection
, (total intern
al) reflection
do n
ot accept: “bou
nce (off
ionosph
ere)”, refraction
longer (w
avelength
)
the (rad
io) waves are reflected
by the ion
osphere
men
tion of
satellite (1) +
any valid
fun
ction of
satellite (1)eg
“signals tran
smitted
back to Earth
”“sign
al amplified
/focused
”“sign
al frequen
cy altered”
Page 7
Question N
os 3(a)
NO
TE
S
3.A
door e
ntry
syste
m in
an
offic
e b
lock a
llow
s vid
eo a
nd
au
dio
info
rmatio
n
to b
e se
nt b
etw
een
two p
eop
le.
(a)
Acam
era
at th
e e
ntra
nce u
ses a
len
s to fo
cu
s para
llel ra
ys o
flig
ht o
nto
a d
ete
cto
r.
Part o
fth
e c
am
era
is show
n in
the d
iagra
m b
elo
w.
(i)C
om
ple
te th
e d
iagra
m a
bove b
y:
(A)
dra
win
g th
e le
ns u
sed
;
(B)
com
ple
ting th
e p
ath
of
the lig
ht ra
ys.
(ii)U
sing in
form
atio
n fro
m th
e d
iagra
m, c
alc
ula
te th
e p
ow
er o
fth
e
len
s use
d in
the c
am
era
.
22
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
dete
cto
r
Space for w
orking and answer
cam
era
positio
n
of
len
s
para
llel ra
ys o
f
ligh
t
16
mm
ind
epend
ent m
arks (1) for correct len
s shape
(1) for focused
rays ond
etectorrays m
ust m
eet at mid
-point of
detector
rays draw
n m
ust be reason
ably straight
ignore rays d
rawn
insid
ed
otted box
ifP
= 0 .0625
D as fin
al answ
er – ded
uct (½
) (un
it error)1
P =
1 =
.0016.
=62
5D
f
Page 8
Question N
os 3(b)
NO
TE
S
3.(con
tinu
ed)
(b)T
he d
oor e
ntry
syste
m u
ses a
bla
ck a
nd
wh
ite te
levisio
n sc
reen
.
Desc
ribe h
ow
a m
ovin
g p
ictu
re is se
en
on
the te
levisio
n sc
reen
.
You
r desc
riptio
n m
ust in
clud
e th
e te
rms:
line bu
ild u
pim
age retention
brightn
ess variation.
.........................................................................................................
.........................................................................................................
.........................................................................................................
.........................................................................................................3
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
award
marks:
2 × (½
) for describin
g line bu
ild u
p
2 × (½
) for describin
g image reten
tion
(½) for d
escribing brigh
tness variation
(½) for m
ention
ofall 3 term
s in con
text w
ith correct explan
ation
eg•
line bu
ild u
p is wh
en electron
s (½) scan
across screen (½
)•
image reten
tion is w
hen
brain/eye retain
s (½)
each pictu
re wh
ile next is prod
uced
(½) (or pictu
re isprod
uced
25 times per secon
d)
•brigh
tness variation
is by chan
ging n
um
ber/inten
sity ofelectron
beam (½
)+
all 3 terms correctly u
sed in
context (½
)
Page 9
Question N
os 4(a)
NO
TE
S
4.T
he c
on
sum
er u
nit in
a h
ou
se c
on
tain
s a m
ain
s switc
h a
nd
circ
uit b
reakers
for d
iffere
nt c
ircu
its.
(a)
(i)W
hat is th
e p
urp
ose
of
the m
ain
s switc
h?
.................................................................................................
(ii)T
wo o
fth
e c
ircu
its have n
ot b
een
lab
elle
d.
Wh
ich
circ
uit is:
the rin
g c
ircu
it?.............................
the lig
htin
g c
ircu
it?.............................
(iii)T
he c
urre
nt ra
tings fo
r the rin
g c
ircu
it an
d th
e lig
htin
g c
ircu
it are
diffe
ren
t.
Sta
te a
noth
er d
iffere
nce b
etw
een
the rin
g c
ircu
it an
d th
e lig
htin
g
circ
uit.
.................................................................................................
111
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
cooker
A45
Am
ain
s
switc
h
30
A20
A15
A5
A
B
show
er
C
wate
rh
eate
r
DE
to switch
offall circu
its(a) (i)
acce pt:to tu
rn on
/offall
circuits/electrical applian
cesto sw
itch off: electricity su
pply/power/cu
rrent
to isolate main
s
do n
ot accept:“to sw
itch off
main
s”, “to chan
ge fuse”
(a) (iii)acce pt:w
ire thickn
esscostch
eaper/thin
ner w
ire for lightin
g circuit
B(½
)
(½)
E
thicker w
ire in rin
g circuit O
R tw
o paths (for cu
rrent) in
ring circu
it
Page 1
0Q
uestion Nos 4(b)
NO
TE
S
32
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
4.(con
tinu
ed)
(b)
(i)A
25
W la
mp
is desig
ned
to b
e u
sed
with
main
s volta
ge.
Calc
ula
te th
e re
sistan
ce o
fth
e la
mp.
(ii)F
ou
r of
these
lam
ps a
re c
on
necte
d in
para
llel.
Calc
ula
te th
e total
resista
nce o
fth
e la
mp
s.
Space for w
orking and answer
Space for w
orking and answer
V =
230V
(1)
Ω
P25
.I =
=
= 0
109V
230
V R
= I
230=
.0109
=2110
Ω 22
VR
=
P
230=
25
=2116
OR
T1
23
4
T 11
11
1 =
+
++
RR
RR
R
11
11
=
+
+
+
21162116
21162116
4 =
2116
2116R
=
= 529
4ΩΩ
(½)
(½)
(1)
(½)
(½)
(1)s.f. ran
ge: 2000 2100 2120 2116
OR
P =
4 × 25 =
100W
⇒⇒
(½)
(½)
(1)
2
2
VP
=R
230100
=R
R=
529ΩΩ
(½)
(½)
(1)
(b) (i)(½
) max if
any voltage oth
er than
230V
or 240V
used
if240
V u
sed th
en d
edu
ct (1)if
230V
is left as the fin
al answ
er then
mu
st have u
nit for (1)
sig.fig. range: if
I = 0 .109
A ⇒⇒
R =
2000 2100 2110
(b) (ii)if
less than
4 “compon
ents”
eg
then
award
(½) on
ly
if:aw
ard 0 m
arks
accept:
T1
23
T1
23
4
T 11
11
RR
RR
11
11
RR
RR
R
14
2116529
R2116
4
==++
++⇒⇒
==++
++++
====
==ΩΩ
accept
Page 1
1Q
uestion Nos 5(a), (b) and (c)
NO
TE
S
5.T
wo g
rou
ps o
fp
up
ils are
investig
atin
g th
e e
lectric
al p
rop
ertie
s of
a la
mp.
(a)
Gro
up
1 is g
iven
the fo
llow
ing e
qu
ipm
en
t:
am
mete
r; voltm
ete
r; 12
V d
.c. sup
ply
; lam
p; c
on
nectin
g le
ad
s.
Com
ple
te th
e c
ircu
it dia
gra
m to
show
how
this e
qu
ipm
en
t is use
d to
measu
re th
e c
urre
nt th
rou
gh
, an
d th
e v
olta
ge a
cro
ss, the la
mp.
(b)
Gro
up
2
u
ses
the
sam
e
lam
p
an
d
is o
nly
g
iven
th
e
follo
win
g
eq
uip
men
t:
lam
p; o
hm
mete
r; con
nectin
g le
ad
s.
Wh
at p
rop
erty
of
the la
mp
is measu
red
by th
e o
hm
mete
r?
.........................................................................................................
(c)T
he re
sults o
fb
oth
gro
up
s are
com
bin
ed
an
d re
cord
ed
in th
e ta
ble
belo
w.
(i)U
se th
ese
resu
lts to c
om
ple
te th
e la
st two c
olu
mn
s of
the ta
ble.
(ii)W
hat
qu
an
tity is
rep
rese
nte
d b
y th
e la
st tw
o colu
mn
s of
the
tab
le?
.................................................................................................
(iii)W
hat is th
e u
nit fo
r this q
uan
tity?
.................................................................................................
31211
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
12
V d
.c.
I(A
)V
(V)
R(Ω
)IV
I2R
212
624
24
Space for w
orking
V
A
(½) each
for correct symbols
(½) for correct position
ing of
amm
eter(½
) for correct positionin
g ofvoltm
eter(½
) for correct “workin
g circuit”
IV =
2 × 12 =
24 I 2R =
22
× 6 =
24
Pow
er (not P
)
Watt / W
Resistan
ce / R
(1) for each correct en
tryd
edu
ct (½) if
answ
ers not in
serted
into table
un
it not n
ecessary
(a) for wron
g positionin
g ofam
meter or voltm
eter lose “w
orking circu
it”m
ark
(b) Not
“ohm
s”
(c) (i) ifan
swers d
o not appear in
table ded
uct (½
)bu
t the an
swers m
ust
be iden
tifiable as IV an
d I 2R
ifou
twith
the table
Page 1
2Q
uestion Nos 6(a) and (b)
NO
TE
S
6.T
he th
yro
id g
lan
d, lo
cate
d in
the n
eck, is e
ssen
tial fo
r main
tain
ing g
ood
health
.
(a)
(i)A
rad
ioactiv
e so
urc
e, w
hic
h is a
gam
ma ra
dia
tion
em
itter, is u
sed
as
a
rad
ioactiv
e
tracer
for
the
dia
gn
osis
of
thy
roid
gla
nd
diso
rders.
A sm
all q
uan
tity o
fth
is tracer, w
ith a
n a
ctiv
ity o
f20
MB
q, is
inje
cte
d in
to a
patie
nt’s b
od
y. Afte
r 52 h
ou
rs, the a
ctiv
ity o
fth
e
tracer is m
easu
red
at 1
.25
MB
q.
Calc
ula
te th
e h
alf
life o
fth
e tra
cer.
(ii)A
noth
er ra
dio
activ
e so
urc
e is u
sed
to treat
can
cer o
fth
e th
yro
id
gla
nd
. Th
is sou
rce e
mits o
nly
beta
rad
iatio
n.
Wh
y is th
is sou
rce u
nsu
itab
le a
s a tracer
?
.................................................................................................
.................................................................................................
(iii)T
he e
qu
ivale
nt d
ose
is mu
ch
hig
her fo
r the b
eta
em
itter th
an
for
the g
am
ma e
mitte
r.
Wh
y is th
is hig
her d
ose
necessa
ry?
.................................................................................................
(b)
Wh
at a
re th
e u
nits o
feq
uiv
ale
nt d
ose
?
.........................................................................................................
2111
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Th
yro
id
gla
nd
20 →→10 →→
5 →→2 .5 →→
1 .25M
Bq
(½) for h
alving
4 half
lives=
52 hou
rs(½
) for correct no of
½ lives
half
life=
13 hou
rs(1) for an
swer
(un
it required
)
it is a beta emitter, absorbed
with
in th
e body
OR
gamm
a emitter requ
ired, to pass th
rough
body
(larger dose requ
ired) to kill th
e (cancerou
s) cells
Sievert (Sv)
(a) (ii)accept:
any an
swer in
dicatin
g the absorption
ofor
penetration
by
do n
ot accept:an
y answ
er involvin
g ½ life,
“m
ore ionisin
g”, “is w
eaker”
(a) (iii)n
ot :w
eaker/stronger/sh
rink an
swer O
R “to be m
ore effective”
ββγγ
ββββ
Page 1
3Q
uestion Nos 7(a)(i)
NO
TE
S
7.A
n
ew
born
b
ab
y is
giv
en
a h
earin
g te
st. A
sm
all
devic
e,
con
tain
ing a
lou
dsp
eaker a
nd
a m
icro
ph
on
e, is p
laced
in th
e b
ab
y’s e
ar.
(a)
A p
ulse
of
au
dib
le so
un
d la
sting 10
µs
is tra
nsm
itted
th
rou
gh
th
e
lou
dsp
eaker. T
he so
un
d is p
layed
at a
level o
f80
dB
.
(i)G
ive a
reaso
n w
hy th
is pu
lse o
fso
un
d d
oes n
ot c
au
se d
am
age to
the b
ab
y’s h
earin
g.
.................................................................................................
.................................................................................................1
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Th
e du
ration of
the pu
lse, 10 microsecon
ds, is very sm
all.O
R O
nly prolon
ged exposu
re at this level w
ill cause
dam
ageO
R 80
dB
is thresh
old level for d
amage
OR
90d
B is th
reshold
level for dam
age
Page 1
4Q
uestion Nos 7(a)(ii), (iii) and (b)
NO
TE
S7.
(a)(con
tinu
ed)
(ii)T
he tra
nsm
itted
pu
lse o
fso
un
d m
akes th
e in
ner e
ar v
ibra
te to
pro
du
ce a
new
sou
nd
, wh
ich
is receiv
ed
by th
e m
icro
ph
on
e.
Sig
nals fro
m th
e tra
nsm
itted
an
d re
ceiv
ed
sou
nd
s are
vie
wed
on
an
osc
illosc
op
e sc
reen
, as sh
ow
n b
elo
w.
Th
e a
vera
ge sp
eed
of
sou
nd
insid
e th
e e
ar is 1
500
m/s.
Calc
ula
te th
e d
istan
ce b
etw
een
the d
evic
e a
nd
the in
ner e
ar.
(iii)S
uggest a
freq
uen
cy th
at c
ou
ld b
e u
sed
for th
e h
earin
g te
st.
.................................................................................................
(b)
An
ultra
sou
nd
scan
can
be u
sed
to p
rod
uce a
n im
age o
fan
un
born
bab
y.
Exp
lain
h
ow
th
e
image
of
an
u
nb
orn
b
ab
y
is fo
rmed
b
y
ultra
sou
nd
.
Space for w
orking and answer
312
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
tran
smitte
d so
un
d
receiv
ed
sou
nd
90
80
70
60
50
40
30
20
100
05
10
15
20
25
30
35
40
45
50
55
60
time in
µs
sou
nd
level in
dB
×
×
-6
-6
3010
t =
2
= 15
10=
1500 × 15 ×
10–6
= 0 .0225
m
(½)
(½)
(½)
(1)
any stated
value betw
een 20 – 20000
Hz in
clusive (u
nit requ
ired)
(1) mark for in
dicatin
g reflectioneg
“ultrasou
nd
reflects ofbaby in
wom
b”
+ (1) m
ark for anoth
er relevant poin
teg
“ultrasou
nd
takes differen
t times to reflect from
differen
td
epths of
tissue”
“reflected u
ltrasoun
d is d
etected by receiver/com
puter”
“apply jelly + valid
description
”
total time =
30 × 10
–6(½
)d
= v ×
t
(a) (ii)d
ivide total t by 2 can
appear at any stage in
answ
er if
wron
g time selected
from graph
then
:
(½) for d
= v ×
t, (½) for d
ivide by 2 on
ly
mu
st use v =
1500m
/s
ded
uct (½
) ifn
o conversion
oft in
to second
s
(a) (iii)ran
ge ofvalu
es not
acceptable
(b)n
ot :“u
ltrasoun
d bou
nces back”
“soun
d rays”
Page 1
5Q
uestion Nos 8(a)
NO
TE
S
8.A
hig
h
inte
nsity
L
ED
is
use
d
as
a
gard
en
lig
ht.
T
he
ligh
t tu
rns
on
au
tom
atic
ally
wh
en
it becom
es d
ark
.
Th
e lig
ht a
lso c
on
tain
s a so
lar c
ell w
hic
h c
harg
es a
rech
arg
eab
le b
atte
ry
du
ring d
aylig
ht h
ou
rs.
(a)
Part o
fth
e c
ircu
it is show
n b
elo
w.
(i)S
tate
the e
nerg
y tra
nsfo
rmatio
n in
a so
lar c
ell.
.................................................................................................
(ii)A
t a p
artic
ula
r ligh
t level, th
e v
olta
ge g
en
era
ted
by th
e so
lar c
ell
is 1.5
V.
Calc
ula
te th
e v
olta
ge a
cro
ss the re
ch
arg
eab
le b
atte
ry a
t this lig
ht
level.
12
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
rech
arg
eab
le
batte
ry
600
Ω
2400
Ω
sola
r cell
Space for w
orking and answer
light to electrical (en
ergy)RT
= R
1+
R2
= 600 +
2400 = 3000
( ΩΩ) (½
)
(½)
(1)
(½)
(½)
2B
S1
2
RV
=×
VR
+ R
2400.
=
15
3000××
= 1 .2V
(1)
OR
cell
T
B
V1
5I
00005(A
)R
3000
VIR0
00052400
12V
⋅⋅==
====
⋅⋅
====⋅⋅
××==
⋅⋅
accept“electric”
do n
ot accept“electricity”
Page 1
6Q
uestion Nos 8(b)
NO
TE
S8.
(contin
ued
)
(b)
Th
e L
ED
is switc
hed
on
usin
g th
e fo
llow
ing c
ircu
it.
(i)N
am
e c
om
pon
en
t X.
.................................................................................................
Th
e g
rap
h b
elo
w sh
ow
s the v
olta
ge a
cro
ss the L
DR
in th
is circ
uit
for d
iffere
nt lig
ht le
vels.
Lig
ht le
vel is m
easu
red
in lu
x.
(ii)F
or th
e L
ED
to b
e lit, th
e v
olta
ge a
cro
ss the L
DR
mu
st be a
t least
0.7
V.
Wh
at is th
e m
axim
um
ligh
t level fo
r the L
ED
to b
e lit?
.................................................................................................
(iii)E
xp
lain
the p
urp
ose
of
resisto
r R.
.................................................................................................
111
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
rech
arg
eab
le
batte
ry
R
X
ligh
t level
in lu
x
00.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
100
80
60
40
200
volta
ge a
cro
ss LD
R in
volts
transistor (sw
itch)
20 (lux)
to protect the L
ED
OR
to limit th
e curren
t OR
to redu
ceth
e voltage across the L
ED
(b) (i)ign
ore: “pnp”, “n
pn”
not :
phototran
sistor,m
osfet transistor,
switch
(b) (iii)d
o not acce pt:
“to redu
ce voltage”on
ly“to stop L
ED
blowin
g”“to red
uce th
e voltage throu
gh/in
the L
ED
”“to red
uce ch
arge/power to th
e LE
D”
Page 1
7Q
uestion Nos 9(a) and (b)
NO
TE
S
9.A
n e
lectro
nic
tun
er fo
r a g
uita
r con
tain
s a m
icro
ph
on
e a
nd
an
am
plifie
r.
Th
e o
utp
ut v
olta
ge fro
m th
e a
mp
lifier is 9
V.
(a)
Th
e v
olta
ge g
ain
of
the a
mp
lifier is 1
50.
Calc
ula
te th
e in
pu
t volta
ge to
the a
mp
lifier.
(b)
Th
e tu
ner is u
sed
to m
easu
re th
e fre
qu
en
cy o
fsix
gu
itar strin
gs.
Th
e n
um
ber a
nd
freq
uen
cy o
feach
string is g
iven
in th
e ta
ble
belo
w.
Th
e
tun
er
has
an
ou
tpu
t so
cket
wh
ich
h
as
been
con
necte
d
to
an
osc
illosc
op
e. Th
e tra
ce fo
r string 5
is show
n in
Fig
ure
1.
(i)T
he c
on
trols o
fth
e o
scillo
scop
e a
re n
otalte
red
.
In
Fig
ure
2, d
raw
the tra
ce o
bta
ined
ifstrin
g 1
is pla
yed
loud
erth
an
string 5
.
(ii)S
tring 3
is plu
cked
.
Wh
at is th
e fre
qu
en
cy o
fth
e o
utp
ut sig
nal fro
m th
e a
mp
lifier?
.................................................................................................
221
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arks
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T
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IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Num
ber ofstring
Frequency
(Hz)
1330.0
2247.0
3196.0
4147.0
5110.0
682.5
Fig
ure
1F
igu
re 2
out
inin
VG
ain =
V9 150 =
V
Vin
= 0 .06
V
196H
z
(½)
(½)
(1)Ind
epend
ent
marks
(1) for show
ing
increased
amplitu
de
(1) for show
ing
three w
aves
(b) (i)m
ust sh
ow 3 w
aves only
(b) (ii)d
o not acce pt: “th
e same”
(1) or (0) marks u
nit requ
ired
Page 1
8Q
uestion Nos 10(a) and (b)
NO
TE
S
10.C
am
era
s pla
ced
at 5
km
inte
rvals a
lon
g a
stretc
h o
fro
ad
are
use
d to
record
the a
vera
ge sp
eed
of
a c
ar.
Th
e c
ar is tra
vellin
g o
n a
road
wh
ich
has a
speed
limit o
f100
km
/h. T
he
car tra
vels a
dista
nce o
f5
km
in 2
.5 m
inu
tes.
(a)
Does th
e a
vera
ge sp
eed
of
the c
ar sta
y w
ithin
the sp
eed
limit?
You
mu
st justify
you
r an
swer w
ith a
calc
ula
tion
.
(b)
At o
ne p
oin
t in th
e jo
urn
ey, th
e c
ar sp
eed
om
ete
r record
s 90
km
/h.
Exp
lain
wh
y th
e a
vera
ge sp
eed
for th
e e
ntire
jou
rney is n
ot a
lways th
e
sam
e a
s the sp
eed
record
ed
on
the c
ar sp
eed
om
ete
r.
.........................................................................................................
.........................................................................................................
.........................................................................................................
.........................................................................................................
32
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UP
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arks
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T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
()
d v =
t5
=.25 ÷
60
= 120
km/h
No (1) m
ust attem
pt to justify to
obtain fin
al mark
Fin
al answ
er mu
st be consisten
tw
ith calcu
lated valu
e and
un
it(or im
plied u
nit)
Explan
ation m
ust in
clud
e the follow
ing:
•car speed
ometer m
easures in
stantan
eous speed
(½)
•in
stantan
eous speed
will ch
ange d
urin
g journ
ey (½)
•average speed
is measu
red over/greater tim
e/wh
ole journ
ey (1)
(½)
(½)
(1)
Note: if
t roun
ded
to 0 .04 hou
rs ⇒⇒v ==
125 km/h
then
accept
Page 1
9Q
uestion Nos 11(a)
NO
TE
S
221
11.A
n a
ero
pla
ne o
n a
n a
ircra
ft carrie
r mu
st reach
a m
inim
um
speed
of
70
m/s
to sa
fely
take o
ff. Th
e m
ass o
fth
e a
ero
pla
ne is 2
8000
kg.
(a)
Th
e a
ero
pla
ne a
ccele
rate
s from
rest to
its min
imu
m ta
ke o
ffsp
eed
in 2
s.
(i)C
alc
ula
te th
e a
ccele
ratio
n o
fth
e a
ero
pla
ne.
(ii)C
alc
ula
te th
e fo
rce re
qu
ired
to p
rod
uce th
is accele
ratio
n.
(iii)T
he a
ero
pla
ne’s e
ngin
es p
rovid
e a
tota
l thru
st of
240
kN
. An
ad
ditio
nal
forc
e
is su
pp
lied
b
y
a
cata
pu
lt to
p
rod
uce
the
accele
ratio
n re
qu
ired
.
Calc
ula
te th
e fo
rce su
pp
lied
by th
e c
ata
pu
lt.
K&
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SM
arks
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NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Space for w
orking and answer
Space for w
orking and answer
v-u
a =
t
()
70-
0 =
2=
35m
/s2
F=
ma
= 28
000 × 35
= 980
000N
add
itional force requ
ired=
total force – aircraft thru
st
= 980
000 – 240000
= 740
000N
(a) (i)n
ot:
(a) (iii)if
980000 – 240 th
en u
nit pen
alty (–½)
va =
t
Page 2
0Q
uestion Nos 11(b)
NO
TE
S
22
11.(con
tinu
ed)
(b)
Late
r, the sa
me a
ero
pla
ne tra
vellin
g a
t a sp
eed
of
65
m/s, to
uch
es d
ow
n
on
the c
arrie
r.
(i)C
alc
ula
te th
e k
inetic
en
erg
y o
fth
e a
ero
pla
ne a
t this sp
eed
.
(ii)T
he gra
ph
sh
ow
s th
e m
otio
n of
the aero
pla
ne fro
m th
e p
oin
t
wh
en
it tou
ch
es d
ow
n o
n th
e c
arrie
r un
til it stop
s.
Calc
ula
te th
e d
istan
ce tra
velle
d b
y th
e a
ero
pla
ne o
n th
e c
arrie
r.
K&
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SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
70
60
50
40
30
20
100
speed
in
m/s
01
23
4
time in
s
Space for w
orking and answer
= 59 .2
MJ
2
2
k1
E =
m
v21
=
28000 65
2×
×
distan
ce = area u
nd
er speed tim
e graph
= 128 .75
m
()
⎛⎞
⎛⎞
⎜⎟
⎜⎟
⎝⎠
⎝⎠
11
=
b × h +
l × b +
b × h2
2
11
.=
× 1 ×10
+ 1 × 55
+
× 25 × 55
22
(½)
(½)
(1)
(b) (i)acce pt:
59200
000J
59150
000J
59 .15M
J
(b) (ii)if
any portion
s missin
g then
(½) m
ax for (implied
) relationsh
ipif
wron
g value(s) extracted
from graph
treat as su
bstitution
error (½) m
ax
significan
t figure ran
ge 100 130 129 128 .8acce pt
128 .75m
Page 2
1Q
uestion Nos 12(a) and (b)
NO
TE
S
12.T
he a
dvertise
men
t belo
w is fo
r a n
ew
torc
h.
(a)
(i)E
xp
lain
how
a v
olta
ge is in
du
ced
in th
e c
oil.
.................................................................................................
.................................................................................................
(ii)W
hat is th
e e
ffect o
fsh
akin
g th
e to
rch
faste
r?
.................................................................................................
(iii)D
raw
the c
ircu
it sym
bol fo
r a c
ap
acito
r.
(b)
Wh
en
lit, the c
urre
nt in
the L
ED
is 20
mA
.
Calc
ula
te h
ow
mu
ch
ch
arg
e flo
ws th
rou
gh
the L
ED
in 1
2 m
inu
tes.
2112
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Space for sym
bol
coil
LE
D
sprin
gs
magn
et
movem
en
t of
magn
et
Kinetic T
orchN
o batteries needed – magnet pow
ered!B
right white LE
D w
on’t burn out!30-40 seconds of gentle shaking produces 10-15 m
inutes of light!C
apacitor holds the charge generated by passing the magnet through the
(shakin
g the torch
) makes th
e magn
et move (1) in
and
out of
the coils (1)
OR
chan
ging th
e magn
etic field at coils (1) cau
sed by sh
aking th
e magn
et (1)
the (in
du
ced)
voltage will in
crease
Q=
It
= 0 .02 ×
12 × 60
= 14 .4
C
(a) (i)an
swer sh
ould
ind
icate movem
ent
eg magn
et is moved
(1) in/ou
t ofcoils/w
ires (1)
(a) (ii)accept:
“(ind
uced
) curren
t will in
crease”“w
ill charge m
ore quickly”
“produ
ce more electrical en
ergy/power”
“more ch
arge in th
e capacitor”
do n
ot accept: “lamp w
ill shin
e brighter”
(b) ifn
o conversion
into am
peres and
/or second
s then
un
it error (–½)
mu
st show
conn
ecting w
ires
Page 2
2Q
uestion Nos 12(c)
NO
TE
S
12.(con
tinu
ed)
(c)T
he to
rch
pro
du
ces a
beam
of
ligh
t.
Th
e d
iagra
m sh
ow
s th
e L
ED
p
ositio
ned
at
the fo
cu
s of
the to
rch
refle
cto
r.
Com
ple
te th
e d
iagra
m b
y d
raw
ing lig
ht ra
ys to
show
how
the b
eam
of
ligh
t is pro
du
ced
.2
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
LE
Dre
flecto
r(1) for ligh
t ray paths
(1) for direction
ind
epend
ent m
arks
min
imu
m of
2 rays draw
nreason
ably straight/parallel
Page 2
3Q
uestion Nos 13(a) and (b)
NO
TE
S
13.A
n e
lectric
kettle
is use
d to
heat 0
.4kg o
fw
ate
r.
(a)
Th
e in
itial te
mp
era
ture
of
the w
ate
r is 15
°C.
Calc
ula
te h
ow
mu
ch
heat e
nerg
y is re
qu
ired
to b
ring th
is wate
r to its
boilin
g p
oin
t of
100
°C.
(b)
Th
e au
tom
atic
sw
itch
on
th
e kettle
is
not
work
ing.
T
he kettle
is
switc
hed
off
5 m
inu
tes a
fter it h
ad
been
switc
hed
on
.
Th
e p
ow
er ra
ting o
fth
e k
ettle
is 2000
W.
(i)C
alc
ula
te
how
m
uch
ele
ctric
al
en
erg
y
is con
verte
d
into
h
eat
en
erg
y in
this tim
e.
(ii)C
alc
ula
te th
e m
ass o
fw
ate
r ch
an
ged
into
steam
in th
is time.
323
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Space for w
orking and answer
Space for w
orking and answer
E=
mc ∆∆
T(½
)(1) d
ata mark
= 4180 ×
0 .4 × 85
(½)
= 142
120J
(1)
E=
P ×
t
= 2000 ×
5 × 60
= 600
000J
E=
600000 – 142
120 = 457
880J
(½)
(1) data m
ark
EH
= m
× l
(½)
457880
= m
× 22 .6 ×
105
m
= 0 .2
Kg
(a) if
wron
g value of
c selected from
specific heat capacity of
materials d
atatable th
en can
get (2) marks m
ax
ifc =
4200 or any oth
er value th
en (½
) mark m
ax
significan
t figure ran
ge: 100000 140
000 142000
(b) (i)if
E =
100000
J (no con
version in
to second
s then
ded
uct (½
) un
it error)
(b) (ii)if
wron
g value of
Vselected
from V
data table th
en can
get (2) m
arks max
any oth
er value of
Vth
en (½
) mark m
ax
C =
4180(JK
g–1°C
–1)
V=
22 .6 × 10
5(J/K
g)
Page 2
4Q
uestion Nos 14(a) and (b)
NO
TE
S
212
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Gam
ma
rays
PU
ltravio
let
QIn
frare
dR
TV
an
d
Rad
io
Ele
ctro
magn
etic
Sp
ectru
m
incre
asin
g w
avele
ngth
14.T
he
dia
gra
m
rep
rese
nts
the
ele
ctro
mag
netic
sp
ectru
m
in
ord
er
of
incre
asin
g w
avele
ngth
. Som
e o
fth
e ra
dia
tion
s have n
ot b
een
nam
ed
.
(a)
(i)N
am
e ra
dia
tion
:P
..............................................................
Q..............................................................
R..............................................................
(ii)W
hic
h ra
dia
tion
in th
e e
lectro
magn
etic
spectru
m h
as th
e h
igh
est
freq
uen
cy?
.................................................................................................
(b)
Sta
rs em
it ultraviolet
an
d in
fraredra
dia
tion
.
Nam
e a
dete
cto
r for each
of
these
two ra
dia
tion
s.
Infrared
...........................................................................................
Ultraviolet.......................................................................................
X-rays
visible light
microw
aves
gamm
a rays
(black bulb) th
ermom
eter OR
photod
iode
OR
phototran
sistorflu
orescent pain
t/material/U
V film
(½) for each
nam
e
(½) for correct
order
(a) (i)d
o not accept:
Q—
spectrum
/RO
YG
BIV
/laser radiation
/visible radiation
R —
“micro”
or “µ”
ifan
y entries are blan
k/wron
g then
lose “order”
(½) m
ark
(b) accept:
IR—
therm
ofilm/th
ermistor/th
ermopile/th
ermocou
ple/th
ermograph
ic film/h
eat sensitive paper/IR
film
do n
ot accept:IR
—skin
/photograph
ic filmIR
camera
UV
—ph
otographic film
Page 2
5Q
uestion Nos 15(a)
NO
TE
S
15.In
Jun
e 2
005, a
space v
eh
icle
calle
d M
ars L
an
der w
as se
nt to
the p
lan
et
Mars.
(a)
Th
e g
rap
h sh
ow
s the g
ravita
tion
al fie
ld stre
ngth
at d
iffere
nt h
eig
hts
ab
ove th
e su
rface o
fM
ars.
(i)T
he M
ars L
an
der o
rbite
d M
ars a
t a h
eig
ht o
f200
km
ab
ove th
e
pla
net’s su
rface.
Wh
at is th
e v
alu
e o
fth
e g
ravita
tion
al fie
ld stre
ngth
at th
is heig
ht?
.................................................................................................
(ii)T
he M
ars L
an
der, o
fm
ass 5
30
kg, th
en
lan
ded
.
Calc
ula
te th
e w
eig
ht o
fth
e M
ars L
an
der o
n th
e su
rface.
13
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
heig
ht a
bove M
ars su
rface in
km
gra
vita
tion
al
field
stren
gth
in N
/kg
Space for w
orking and answer
4.543210
3.5
2.5
1.5
0.5
0100
200
300
400
500
600
700
800
3 .6m
/s2
OR
N/kg u
nit requ
ired
W=
mg
(½)
= 530 ×
4(½
)
= 2120
N(1)
(1) for 4N
/kg
(a) (i)3 .6 valu
e only
—n
o range of
answ
ers
(a) (ii)if
any oth
er value of
g from gravitation
al field stren
gth d
ata table isu
sed th
en (2) m
axif
any oth
er value of
g is used
then
(½) m
ax
Page 2
6Q
uestion Nos 15(b)
NO
TE
S
15.(con
tinu
ed)
(b)
Th
e M
ars L
an
der re
lease
d a
rover e
xp
lora
tion
veh
icle
on
to th
e su
rface
of
Mars.
To c
olle
ct d
ata
from
the b
otto
m o
fa la
rge c
rate
r, the ro
ver la
un
ch
ed
a
pro
be h
oriz
on
tally
at 3
0m
/s. Th
e p
rob
e to
ok 6
s to re
ach
the b
otto
m
of
the c
rate
r.
(i)C
alc
ula
te th
e h
oriz
on
tal d
istan
ce tra
velle
d b
y th
e p
rob
e.
(ii)C
alc
ula
te th
e v
ertic
al sp
eed
of
the p
rob
e a
s it reach
ed
the b
otto
m
of
the c
rate
r.
[EN
D O
F M
AR
KIN
G IN
ST
RU
CT
ION
S]
22
K&
UP
SM
arks
DO
NO
T
WR
IT
E I
N
TH
IS
MA
RG
IN
Space for w
orking and answer
Space for w
orking and answer
d=
v × t
(½)
= 30 ×
6(½
)
= 180
m(1)
v-u
a =
t()
v-0
4 =
6v =
24m
/s
(½)
(½)
(1)
(b) (ii)n
ot
ifth
en (½
) max
va =
ta
≠≠