Wednesday, September 10, 2014

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

  • 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 / 4πϵor

(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 = 4πϵor         and      4πϵo 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 / 4πϵo = (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 2nd 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
+ve ions travelling through a vacuum in narrow beam. Ions enter region of uniform magnetic field of flux density B and are defected in semi-circular arc, as shown.
Ions, travelling with speed 1.40x105ms-1, are detected at fixed detector when diameter of arc in magnetic field = 12.8cm.
(a)
Reference to Fig, direction of magnetic field:
The field is into (the plane of) the paper
                                            
(b)
Ions have mass 20u and charge +1.6x10-19C. Show magnetic flux density = 0.454T.
The force due to the magnetic field provides the centripetal force.
mv2/r = Bqv
B = [(20x1.66x10-27) (1.40x105)] / [(1.6x10-19) (6.4x10-2)] = 0.454T

(c)
Ions of mass 22u with same charge and speed as those in (b) also present in beam.
(i)
Sketch path of these ions in the magnetic field of magnetic flux density 0.454T:
Path is a semicircle with the diameter greater than 12.8cm

(ii)
In order to detect those ions at fixed detector, magnetic flux density is changed.
New magnetic flux density:
New magnetic flux density, B = (22/20) x 0.454 = 0.499T  




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 Vp and Ip respectively. r.m.s voltage and current in secondary coil are Vs and Is 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, Vs / Vp = Ip / Is :
Power in primary coil = Power in secondary coil
VpIp = VsIs
So, Vs / Vp = Ip / Is



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 IT and reflected intensity is IR.
(i)
Relation between I, IT and IR:
I = IT + IR

(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.


14 comments:

  1. for 1 cii, h=why did you take 60RE – x for the moon? why couldnt you apply 60RE – x to earth?is it because earth has more mass?

    ReplyDelete
    Replies
    1. It's because because I have taken the point to be at a distance x from the Earth. The separation between the Earth and the Moon is 60RE. So, the point is at a distance of 60RE - x from the Moon.

      The question asks fro the distance from the centre of the Earth.

      If I had taken the object to be at a distance x from the Moon, the distance from Earth would then be 60RE - x. But I would have to perform an additional calculation (60RE - x), after finding x, to obtain the distance from Earth

      Delete
  2. im getting 59.29 RE if i make moon the x. i know that the point isnt equidistant from the earth and moon. i subrated 60- .73 to get the distance from earth but its wrong, so could you explain how to find it if i were to make moon x.

    ReplyDelete
    Replies
    1. Details for the explanations of Q1 has been updated. Check it again

      Delete
  3. In Q7 (b) (i), shouldn't there only be 3 lines because the wavelength of the photon emitted in the smaller transitions is outside the visible light range? Like from -0.87 x 10^-19 J to -1.36 x 10^-19 J, the wavelength is 4060nm.

    ReplyDelete
    Replies
    1. Details have been updated for the question

      Delete
  4. And also, for Q9 b ii, I got 18V, not 9V? Can you show the working?

    ReplyDelete
    Replies
    1. More explanations have been added for the question

      Delete
  5. 43w10 3b please!
    jazzakAllah khair

    ReplyDelete
    Replies
    1. See solution 570 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-112.html

      Delete
    2. 43w 10 question 7b a.please explain.
      Thanks

      Delete
    3. Check solution 1101 at
      http://physics-ref.blogspot.com/2016/02/physics-9702-doubts-help-page-235.html

      Delete
  6. For question November 2010 paper 43 2(c)(ii) why is the kinetic energy is halved ? Why do they have the same equal kinetic energies? Thanks.

    ReplyDelete
    Replies
    1. Check solution 730 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-148.html

      Delete

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