• 9702 November 2010 Paper 41 & 42 Worked Solutions | A-Level Physics

Paper 41 & 42

SECTION A

Question 1

{Detailed explanations for this question is available as Solution 504 at Physics 9702 Doubts | Help Page 98 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-98.html}

Question 2

{Detailed explanations for this question is available as Solution 915 at Physics 9702 Doubts | Help Page 187 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-187.html}

Question 3

{Detailed explanations for this question is available as Solution 934 at Physics 9702 Doubts | Help Page 192 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-192.html}

Question 4

(a)

Capacitance is defined as the ratio of charge to potential (difference)

(b)

Isolated metal sphere has radius r. When charged to potential V, charge on sphere is q. Charge may be considered to act as point charge at centre of sphere.

(i)

Expression, in terms of r and q, for potential V of sphere:

V = q / 4πϵ_or

(ii)

Isolated sphere has capacitance.  Use answers from (a) and (b)(i) to show that capacitance of sphere is proportional to its radius:

C = q / V = 4πϵ_or         and      4πϵ_o is constant

So, the capacitance, C is proportional to the radius, r.

(c)

Sphere has capacitance of 6.8pF and is charged to potential of 220V.

(i)

Radius of sphere:

r = C / 4πϵ_o = (6.8x10^-12) / (4π (8.85x10^-12) ) = 6.1x10^-2m

(ii)

Charge, in coulomb, on sphere:

q = CV = (6.8x10^-12) (220) = 1.5x10^-9C

(d)

A 2^nd uncharged metal sphere is brought up to sphere in (c) so that they touch. Combined capacitance of 2 spheres = 18pF.

(i)

Potential of the 2 spheres:

V = q / C = (1.5x10^-9) / (18x10^-12) = 83V

(ii)

Change in total energy stored on spheres when they touch:

Either

Energy = ½ CV²

Change in energy, ΔE = [ ½ (6.8x10^-12) (220²) ] – [ ½ (18x10^-12) (83²) ]

ΔE = (1.65x10^-7) – (6.2x10^-8) = 1.03x10^-7J

Or

Energy = ½ QV

Change in energy, ΔE = [ ½ (1.5x10^-9) (220) ] – [ ½ (1.5x10^-9) (83) ]

ΔE = 1.03x10^-7J  

Question 5

Positive ions are travelling through a vacuum in a narrow beam. The ions enter a region of uniform magnetic field of flux density B and are deflected in a semi-circular arc, as shown in Fig. 5.1.
 

Question 6

Simple iron-cored transformer illustrated.

(a)

(i)

Primary and secondary coils are wounds on core made of iron:

E.g. It prevents flux losses / It improves the flux linkage

(ii)

Thermal energy is generated in core when transformer is in use:

Since the flux in the core is changing, there is an e.m.f / current (induced) in the core. This induced current in the core causes heating.

(b)

Root-mean-square (r.m.s) voltage and current in primary coil are V_p and I_p respectively. r.m.s voltage and current in secondary coil are V_s and I_s respectively.

(i)

By reference to direct current, what is meant by root-mean-square value of an alternating current:

The root-mean-square value of an alternating current is that value of the direct current producing the same (mean) power / heating in a resistor.

(ii)

Show that, for ideal transformer, V_s / V_p = I_p / I_s :

Power in primary coil = Power in secondary coil

V_pI_p = V_sI_s

So, V_s / V_p = I_p / I_s

Question 7

{Detailed explanations for this question is available as Solution 623 at Physics 9702 Doubts | Help Page 123 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-123.html}

Question 8

In some power stations, nuclear fission used as source of energy.

(a)

Nuclear fission is the splitting of a heavy nucleus into two (lighter) nuclei of approximately the same mass.

(b)

Nuclear fission reaction produces neutrons. In power station, neutrons may be absorbed by rods made of boron-10.

Nuclear equation for absorption of single neutron by boron-10 nucleus with emission of an alpha-particle:

¹⁰₅B      +          ¹₀n        - - - >   ⁷₃Li      +          ⁴₂He

¹₀n

⁴₂He

⁷₃Li

(c)

Why, when neutrons are absorbed in boron rods, the rods become hot as a result of this nuclear reaction:

The emitted particles have kinetic energy. The range of particles in the control rods is short / the particles are stopped in the rods / the particles lose kinetic energy in the rods. The kinetic energy of the particles is converted to thermal energy.  

SECTION B

Question 9

{Detailed explanations for this question is available as Solution 617 at Physics 9702 Doubts | Help Page 122 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-122.html}

Question 10

(a)

(i)

The acoustic impedance of a medium is the product of the density of the medium and the speed of the wave (in the medium).

(ii)

Data for some media given.

Medium           Speed of ultrasound / ms^-1      acoustic impedance / kgm^-2s^-1

Air                               330                                          4.3x10²

Gel                              1500                                        1.5x10⁶

Soft tissue                   1600                                        1.6x10⁶

Bone                            4100                                        7.0x10⁶

Use data to calculate value for density of bone:

Density, ρ = (7.0x10⁶) / 4100 = 1700kgm^-3

(b)

Parallel beam of ultrasound has intensity I. It is incident at right-angles to boundary between 2 media, as shown. Media have acoustic impedance Z₁ and Z₂. Transmitted intensity of ultrasound beam is I_T and reflected intensity is I_R.

(i)

Relation between I, I_T and I_R:

I = I_T + I_R

(ii)

Reflection coefficient α given by α = [(Z₂ – Z₁)²] / [(Z₂ + Z₁)²]

Use data to determine reflection coefficient α for boundary between

1.

Gel and soft tissue:

(α = ([1.6 – 1.5]x10⁶)² / ([1.6 + 1.5]x10⁶)² )

α = (0.1x10⁶)² / (3.1x10⁶)² = 0.001

2.

Air and soft tissue:

α ≈ 1

(c)

Explain use of gel on surface of skin during ultrasound diagnosis:

Either

There is very little transmission at an air-skin boundary while (almost) complete transmission occurs at a gel-skin boundary when the wave travels in or out of the body.

Or

When there is no gel, there is a majority reflection while with the gel, little reflection occurs when the wave travels in or out of the body.

Question 11

(a)

Wire pairs provide one means of communication but are subject to high levels of noise and attenuation.

(i)

Noise is the unwanted random power / signal / energy.

(ii)

Attenuation is the loss of (signal) power / energy.

(b)

Microphone connected to receiver using wire pair, as shown. Wire pair has attenuation per unit length of 12dBkm^-1. Noise power in wire pair is 3.4x10^-9W. Microphone produces signal power of 2.9μW.

(i)

Maximum length of wire so that minimum signal-to-noise ratio is 24dB:

Either

Signal-to-noise ratio at microphone = 10 log(P₂/P₁)

Signal-to-noise ratio = 10 log({2.9x10^-6}/{3.4x10^-9}) = 29dB

Maximum length = (29 – 24) / 12 = 0.42km = 420m

Or

Signal-to-noise ratio at receiver = 10 log(P₂/P₁)

Power at receiver, P:

24 = 10 log(P/{3.4x10^-9})

P = 8.54x10^-7W

Power loss in cable = 10 log({2.9x10^-6}/{8.54x10^-7}) = 5.3dB

Length = 5.3 / 12 = 440m.

(ii)

Communication over distances greater than that calculated in (i) required. Modification to circuit of Fig so that the minimum signal-to-noise ratio at receiver not reduced:

The use of an amplifier coupled to the microphone.

Question 12

(a)

Principles of use of a geostationary satellite for communication on Earth:

A (carrier wave) is transmitted from Earth to the satellite. The satellite receives a signal which is greatly attenuated. This signal is then amplified and transmitted back to Earth at a different (carrier) frequency. The different frequencies prevent swamping of the uplink signal. Examples of frequencies used are (6/4 GHz, 14/11 GHz, 30/20 GHz).

(b)

Polar-orbiting satellites also used for communication on Earth. 1 advantage and 1 disadvantage of polar-orbiting satellites as compared with geostationary satellites:

Advantage:

Examples:

There is a much shorter time delay because the orbits are much lower.

The whole Earth may be covered in several orbits / with networks.

Disadvantage:

Examples:

Either It must be tracked        Or there is limited use in any one orbit

As more satellites are required for continuous operation.