• 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 / 4πϵ_om₁m₂G

Ratio = (1.6x10^-19)² / 4π (8.85x10^-12) (1.67x10^-27)² (6.67x10^-11)

Ratio = 1.2x10³⁶  

Or

F_E = 2.30x10^-28 x R^-2

F_G = 1.86x10^-64 x R^-2

F_E / F_G = 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 f_o of oscillations of magnet:

Period, T = 0.75s

Frequency, f_o = 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 0.7f_o to 1.3f_o where f_o 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 f_o. 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, I_rms:

I_rms x √2 = I_o = 9.9A

I_rms = 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 E_k = 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 V_OUT of circuit given by

V_OUT = (R_F/R) x (V₂ – V₁)

(i)

Name given to ratio R_F/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 V_OUT = 0:

For V_OUT = 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 R_F/R = 12. Magnitude of V_OUT:

Now, V₂ = (1000 / [1000+128]) x 4.5 = 3.99V

V_OUT = 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.