Friday, August 29, 2014

9702 June 2011 Paper 42 43 Worked Solutions | A-Level Physics

  • 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
(a)
A mole is an amount of a substance containing the same number of particles as in 0.012kg of carbon-12.

(b)
2 containers A and B joined by tube of negligible volume.
Containers filled with ideal gas at pressure = 2.3x105Pa. Gas in container A has volume 3.1x103cm3 at temperature 17oC. Gas in container B has volume 4.6x103cm3 at temperature 30oC.
Total amount of gas, in mol, in containers:

pV = nRT
Total amount = (2.3x105 x 3.1x10-3) / (8.31x290) + (2.3x105 x 4.6x10-3) / (8.31x303)
Total amount = 0.296 + 0.420 = 0.716 mol



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:
Amplitude-Frequency Graph for resonance

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 main principles of use of magnetic resonance to obtain diagnostic information about internal body structures:
A strong / large (uniform) magnetic field is applied which causes the nuclei to precess / rotate about the field direction. A (r.f) radio frequency pulse is applied at the Larmor frequency which causes resonance / nuclei absorb energy. On relaxation / de-excitation, the nuclei emit r.f pulse which is detected and processed. Superimposing a non-uniform field on the uniform field allows the position of the resonating nuclei to be determined and for the location of detection to be changed.



Question 11
Use of ionospheric reflection of radio waves for long-distance communication has, to a great extent, been replaced by satellite communication.
(a)
2 reasons why this change has occurred:
Any 2 sensible suggestions:
Examples:
Unreliable communication because the ion layers vary in height / density
Cannot carry all information required since the bandwidth is too narrow
Coverage is limited since reception is poor in hilly areas

(b)
Radio link between a geostationary satellite and Earth may be attenuated by as much as 190dB. Why, as a result of this attenuation, uplink and downlink frequencies must be different:
The signal must be amplified (greatly) before transmission back to Earth. The uplink signal would be swamped by the downlink signal.



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.


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