Friday, September 19, 2014

9702 November 2008 Paper 4 Worked Solutions | A-Level Physics

  • 9702 November 2008 Paper 4 Worked Solutions | A-Level Physics


Paper 4


SECTION A

Question 1
{Detailed explanations for this question is available as Solution 709 at Physics 9702 Doubts | Help Page 143 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-143.html}





Question 2
{Detailed explanations for this question is available as Solution 1049 at Physics 9702 Doubts | Help Page 220 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-220.html}




Question 3

{Detailed explanations for this question is available as Solution 582 at Physics 9702 Doubts | Help Page 114 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-114.html}






Question 4
(a)
Equation to represent first law of thermodynamics in terms of heating q of a system, work w done on system and increase ΔU in internal energy:
ΔU = q + w

(b)
Pressure of ideal gas decreased at constant temperature. What change, if any, occurs in internal energy of gas:
EITHER
The kinetic energy is constant because the temperature is constant. The potential energy is also constant because there are no intermolecular forces (in an ideal gas). So, there is no change in the internal energy.
OR
The kinetic energy and potential energy are both constant. So there is no change in the internal energy. Reason for either constant KE or constant PE given



Question 5
{Detailed explanations for this question is available as Solution 559 at Physics 9702 Doubts | Help Page 109 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-109.html}




Question 6
Simple iron-cored transformer illustrated.
(a)
Why core is
(i)
a continuous loop:
EITHER To prevent loss of magnetic flux
OR It improves flux linkage with the secondary coil

(ii)
laminated:
It reduces eddy current (losses). It also reduces losses of energy (in the core)

(b)
(i)
Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to / equal to the rate of change of (magnetic) flux (linkage).

(ii)
Use law to explain the operation of the transformer:
The changing current in the primary coil gives rise to a changing flux in the core. The flux links with the secondary coil, changing the flux in the secondary coil inducing an e.m.f.

(c)
2 advantages of use of alternating voltages for transmission and use of electrical energy:
It can change the voltage easily / efficiently
High voltage transmission reduces power losses





Question 7
{Detailed explanations for this question is available as Solution 988 at Physics 9702 Doubts | Help Page 205 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-205.html}





Question 8
{Detailed explanations for this question is available as Solution 965 at Physics 9702 Doubts | Help Page 199 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-199.html}






SECTION B


Question 9
Different frequencies and wavelengths used in different channels of communication.
Why
(a)
Infra-red radiation rather than visible light usually used with optic fibres:
Infra-red radiation has less attenuation (per unit length). So, fewer (repeater) amplifiers are needed / There is a longer uninterrupted length.

(b)
Base stations in mobile networks operate on UHF:
Either UHF has a limited range. So, the cells do not overlap (appreciably)
OR UHF has a short wavelength. So, a convenient length for the aerial can be used (on the mobile phone)

(c)
For satellite communication, frequencies of order of GHz used, with uplink having different frequency to downlink:
Frequencies are of the order of GHz have a large bandwidth / large information carrying capacity. The frequencies are different so that the uplink signal is not swamped by the downlink signal.



Question 10
(a)
Circuit for amplifier incorporating an ideal operational amplifier (op-amp) shown.
(i)
1.
Name of this type of amplifier circuit:
Inverting (amplifier)

2.
Why point P is referred to as virtual earth:
The gain of the operational amplifier (op-amp) is very large / infinite. The non-inverting input is at earth / 0V. For the amplifier not to saturate, P must be about earth / 0V.

(ii)
Show that gain of amplifier circuit given by expression         G = - R2 / R1:
The input resistance is very large. So, the current in R1 = current in R2.
I = VIN / R1
And I = - VOUT / R2  
Hence, gain, G = VOUT / VIN = - R2 / R1

(b)
Circuit of Fig modified by connecting light-dependent resistor (LDR) as shown. Resistance R1 and R2 are 5.0kΩ and 50kΩ respectively. Input voltage VIN = +1.2V. High-resistance voltmeter measures output VOUT. Circuit used to monitor low light intensities.
(i)
Voltmeter for light intensities such that LDR has resistance of
1.
100kΩ:
Feedback resistance (= [1/50 + 1/100]-1) = 33.3kΩ
Gain (= (-) RF / RIN = 33.3 / 5) = 6.66
VOUT (= Gain x VIN= 6.66 x 1.2) = 8.0V

2.
10kΩ:
Feedback resistance (= [1/50 + 1/10]-1) = 8.33kΩ
(Gain = 8.33 / 5 = 1.67 )
VOUT (= 1.67 x 1.2) = 2.0V

(ii)
Light incident on LDR provided by single lamp. Use answers in (i) to describe and explain qualitatively variation of voltmeter reading as lamp is moved away from LDR:
(An increase in the lamp-LDR distance gives) a decrease in intensity. The feedback / LDR resistance increases. So, the voltmeter reading increases / becomes more negative.



Question 11
(a)
Distinguish images produced by CT scanning and X-ray imaging:
The CT image is a (thin) slice (through the structure) + any further detail e.g. it is built from many ‘slices’ / 3-D image
The X-ray image is a ‘shadow’ image (of the whole structure) / 2-D image

(b)
Reference to principles of CT scanning; suggest why CT scanning could not be developed before powerful computers were available:
An X-ray image of a slice is taken from many different angles. These images are combined (and processed). This is repeated for many different slices in order to build up a 3-D image which can be rotated. (Powerful) computers are required to store and process huge quantity of data.




16 comments:

  1. In Question 5 part (a), why do we multiply m by 2, and square the charge q?

    ReplyDelete
    Replies
    1. I have updated the explanations.

      Delete
    2. where can i find the question 5? when i press the link its display calibration chapter which is one of the AS topic

      Delete
    3. As said above, its solution 559 on that link

      Delete
    4. thank you very much.this blogspot is very useful.keep it up .cheers

      Delete
  2. Plz explain qs 3 part c ii and in question 2 part b iii why can't we convert 64.7 g to 10 minutes?

    ReplyDelete
    Replies
    1. Some details have been added for question 3.

      I'll try to see the other one later

      Delete
  3. Can you explain question 1 a (iii) and b (i)?

    ReplyDelete
  4. can you plz explain question no.8 last part

    ReplyDelete
    Replies
    1. Some details have been added. See if it helps

      Delete
    2. i got it alhamduillah
      jazakallah khair.

      Delete
  5. june 08 Q10, thanks!

    ReplyDelete
    Replies
    1. Check solution 1040 at
      http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-218.html

      Delete
  6. question 2 biii, why did you times half of 16.6?

    ReplyDelete

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