9702 June 2011 Paper 42 & 43 Worked Solutions | A-Level Physics
Paper 42 & 43
SECTION A
Question 1
(a)
A field of force is a region (of
space) where a particle / body experiences a force.
(b)
Gravitational fields and electric
fields: 2 examples of fields of force. 1 similarity and 1 difference between
these 2 fields of force:
Similarity: Both are inversely proportional to the separation. E.g force ∝
1/r2 and potential ∝ 1/r
Difference: Gravitational force is always attractive while the electric force
can be both attractive and repulsive.
(c)
2 protons isolated in space. Centres
separated by distance R. Each proton considered to be point mass with point
charge. Ratio of force between protons due to electric field to force between
due to gravitational field:
Either
Ratio = Q1Q2 /
4πϵom1m2G
Ratio = (1.6x10-19)2
/ 4π (8.85x10-12) (1.67x10-27)2
(6.67x10-11)
Ratio = 1.2x1036
Or
FE = 2.30x10-28
x R-2
FG = 1.86x10-64
x R-2
FE / FG =
1.2x1036
Question 2
Two containers A and B are joined by a tube of negligible volume, as illustrated in Fig. 2.1.
Question 3
Capacitor consists of 2 metal plates
separated by insulator. Pd between plates is V. Variation with V of magnitude
of charge Q on one plate shown.
(a)
Why capacitor stores energy but not
charge:
The charges on the plates are equal
and opposite. So, there is no resultant charge. Energy is stored because of the
charge separation.
(b)
Use Fig to determine:
(i)
Capacitance of capacitor:
Capacitance = Q / V = (18x10-3)
/ 10 = 1800μF
(ii)
Loss in energy stored in capacitor
when the pd V is reduced from 10.0V to 7.5V:
Either
Use area under graph line in the
region required
Energy = 2.5x 15.7x10-3 =
39mJ
Or
Energy = ½ CV2 = ½ (1800x10-6)
(102 – 7.52) = 39mJ.
(c)
3 capacitors X, Y and Z, each of
capacitance 10μF, are connected as shown.
Initially, capacitors are uncharged.
Pd of 12V applied between points A and B. Magnitude of charge on 1 plate of
capacitor X:
Either
The combined capacitance of Y and Z
= 20μF
Pd across capacitor X = 8V
Charge = 10x10-6 x 8 = 80μC
Or
Total capacitance = 6.67μF
Pd across combination = 12V
Charge = 6.67x10-6 x 12 =
80μC
Question 4
{Detailed explanations for this question is
available as Solution 547 at Physics 9702 Doubts | Help Page 107 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-107.html}
Question 5
Bar magnet suspended vertically from
free end of helical spring as shown. 1 pole of magnet situated in a coil. Coil connected
in series with high-resistance voltmeter. Magnet displaced vertically and then
released. Variation with time t of reading V of voltmeter shown.
(a)
(i)
Faraday’s law of electromagnetic induction
states that the (induced) e.m.f is proportional to the rate of change of (magnetic)
flux (linkage) / rate of flux cutting.
(ii)
Use Faradays’ law to explain why
1.
There is a reading on voltmeter:
The moving magnet causes a change of
the flux linkage.
2.
Reading varies in magnitude:
The speed of the magnet varies, so
there is a varying rate of change of flux.
3.
Reading has both +ve and –ve values:
The magnet changes direction of
motion (so the current also changes direction).
(b)
Use Fig to determine frequency fo of
oscillations of magnet:
Period, T = 0.75s
Frequency, fo = 1/ T = 1
/ 0.75 = 1.33Hz.
(c)
Magnet now brought to rest and
voltmeter replaced by variable frequency alternating current supply that
produces a constant r.m.s current in the coil. Frequency of supply gradually
increased from 0.7fo to 1.3fo where fo is
frequency calculated in (b).
Sketch a graph to show variation
with frequency f of amplitude A of new oscillations of bar magnet:
Smooth, correctly shaped curve with
peak at fo. A never reaches zero.
(d)
(i)
Phenomenon illustrated from graph:
Resonance
(ii)
Situation where phenomenon in (i) is
useful:
Example:
Quartz crystal for timing /
production of ultrasound
Question 6
Alternating current supply connected
in series with resistor R as shown. Variation with time t (in seconds) of
current I (in amps) on resistor given by I = 9.9sin(380t)
(a)
For current in resistor R:
(i)
Frequency:
2πf
= 380
Frequency, f = 60Hz
(ii)
r.m.s current, Irms:
Irms x √2 =
Io = 9.9A
Irms = 9.9 / √2 =
7.0A
(b)
To prevent over-heating, mean power
dissipated in resistor R must not exceed 400W.
Minimum resistance of R:
(Maximum) Power dissipated in
resistor = I2R = 400W
Minimum resistance, R = 400 / 7.02
= 8.2Ω
Question 7
(a)
The de Broglie wavelength is the
wavelength of the wave associated with a particle that is moving.
(b)
Electron accelerated in vacuum from
rest through potential difference of 850V.
(i)
Show final momentum of electron =
1.6x10-23Ns:
Energy of the electron = VQ = 850 x
1.6x10-19 = 1.36x10-16J.
Energy = p2/2m or p =mv and Ek = ½
mv2
Momentum, p = [1.36x10-15
x 2(9.11x10-31)]0.5 = 1.6x10-23Ns
(ii)
de Broglie wavelength of this
electron:
λ = h / p
Wavelength, λ = (6.63x10-34)
/ (1.6x10-23) = 4.1x10-11m
(c)
Experiment to demonstrate wave
nature of electrons:
A diagram or description showing:
An electron beam in vacuum is
incident on a thin metal target / carbon film. A fluorescent
screen is placed at a distance behind the target. A pattern of concentric rings
are observed on the screen. This pattern is similar to the diffraction pattern
observed with visible light.
Question 8
(a)
The binding energy of a nucleus is
the energy required to separate nucleons in a nucleus to infinity.
(b)
Show energy equivalence of 1.0u is
930MeV:
1u = 1.66x10-27kg
E = mc2 = (1.66x10-27)
(3.0x108)2 = 1.49x10-10J
E = (1.49x10-10) / (1.6x10-13)
= 930MeV
(c)
Data for masses of some particles
and nuclei given: proton: 1.0073u,
neutron: 1.0087u, deuterium (21H): 2.0141u and zirconium
(9740Zr): 97.0980u.
Use data and information from (b) to
determine, in MeV,
(i)
Binding energy of deuterium:
Δm = 2.0141u – (1.0073 + 1.0087)u = –
1.9x10-3u
Binding energy = 1.9x10-3
x 930 = 1.8MeV
(ii)
Binding energy per nucleon of
zirconium:
Δm = (57 x 1.0087u) + (40 x 1.0073u)
– 97.0980u = (-)0.69u
Binding energy per nucleon = (0.69x930)
/ 97 = 6.61MeV
SECTION B
Question 9
(a)
Structure of metal wire strain
gauge:
It consists of a thin / fine metal,
having the shape of a grid, encased in plastic.
(b)
Strain gauge S connected into circuit
shown. Operational amplifier (op-amp) is ideal. Output potential VOUT
of circuit given by
VOUT = (RF/R)
x (V2 – V1)
(i)
Name given to ratio RF/R:
Gain (of the amplifier)
(ii)
Strain gauge S has resistance 125Ω when not under strain. Magnitude of V1
such that, when strain gauge S not strained, output VOUT = 0:
For VOUT = 0, then V1
= V2 or V+
= V-
V1 = (1000 / [1000+125])
x 4.5 = 4.0V
(iii)
In a particular test, resistance of
S increases to 128 Ω. V1 is unchanged. Ratio RF/R = 12.
Magnitude of VOUT:
Now, V2 = (1000 / [1000+128])
x 4.5 = 3.99V
VOUT = 12 x (3.99 – 4.00)
= (-)0.12V
Question 10
Explain briefly the main principles of the use of magnetic resonance to obtain diagnostic information about internal body structures.
Question 11
The use of ionospheric reflection of radio waves for long-distance communication has, to a great extent, been replaced by satellite communication.
Question 12
(a)
Signal-to-noise ratio in an optic
fibre must not fall below 24dB. Average noise power in fibre = 5.6x10-19W.
(i)
Minimum effective signal power in
optic fibre:
Ratio / dB = 10 log (P1/P2)
24 = 10 log (P1/{5.6x10-19})
P1 = 1.4x10-16W
(ii)
Fibre has attenuation per unit
length of 1.9dBkm-1. Maximum uninterrupted length of fibre for input
signal of power 3.5mW:
Either
Attenuation per unit length = (1/L)
x 10 log (P1/P2)
1.9 = (1/L) x 10 log ({3.5x10-3}/{1.4x10-16})
L = 71km
Or
Attenuation = 10 log({3.5x10-3}/{5.6x10-19})
= 158dB
Attenuation along fibre = (158 - 24)
L = (158 - 24) / 1.9 = 71km
(b)
Why infra-red radiation, rather than
ultraviolet radiation, is used for long-distance communication using optic
fibres:
For infra-red radiation, there is
less attenuation (per unit length) / longer uninterrupted length of fibre.
