Wednesday, August 13, 2014

9702 June 2012 Paper 42 Worked Solutions | A-Level Physics

  • 9702 June 2012 Paper 42 Worked Solutions | A-Level Physics




SECTION A
Question 1
{Detailed explanations for this question is available as Solution 711 at Physics 9702 Doubts | Help Page 144 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html}





Question 2
{Detailed explanations for this question is available as Solution 510 at Physics 9702 Doubts | Help Page 99 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-99.html}





Question 3
(a)
The internal energy of a system is the sum of the potential energy and the kinetic energy of the atoms / molecules / particles which are randomly distributed in the system.

(b)
Explain change in internal energy of following systems:
(i)
Lump of ice at 0oC melts to form liquid water at 0oC:
As the lattice structure is ‘broken’ / the bonds are broken / the forces between the molecules are reduced, the potential energy of the molecules increases while no change occur in the kinetic energy. So, the internal energy of the system increases.

(ii)
Cylinder containing gas at constant volume in sunlight so that its temperature rises from 25oC to 35oC:
The molecules / atoms / particles move faster / the mean squared speed, <c2> is increasing  /  Kinetic energy of the gas molecules increases with an increase in temperature. So, no change occurs to the potential energy of the molecules while their kinetic energy increases. Thus, internal energy increases.





Question 4
{Detailed explanations for this question is available as Solution 994 at Physics 9702 Doubts | Help Page 206 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-206.html}





Question 5
(a)
(long) straight conductor carrying current of 1A                               
Current/wire normal to magnetic field                                            
(for a flux density 1T,) force per unit length is Nm-1                        

(b)
Horseshoe magnet placed on balance. Stiff metal wire clamped horizontally between poles. Magnetic flux density in space between poles of magnet is uniform and zero outside.
Length of metal wire normal to magnetic field = 6.4cm
When current switched on, reading on balance increases by 2.4g.
Current in wire = 5.6A.
(i)
Direction of force on wire due to current:
there is originally a downward force on the horseshoe magnet (due to the current). From Newton’s third law, there will be an upward force on the wire.

(ii)
Magnitude of magnetic flux density between poles of magnet:
F = BIL
(2.4x10-3)(9.8) = B(5.6)(6.4x10-2)
B = 0.066T

(c)
Low frequency AC now passed through wire. R.m.s value of current = 5.6A.
Variation of reading seen on balance:
The new reading on the balance is now 2.4√2 = 3.4g.
Either The reading changes between +3.4g and -3.4g
Or Total change is 6.8g





Question 6
{Detailed explanations for this question is available as Solution 727 at Physics 9702 Doubts | Help Page 147 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-147.html}
 





Question 7
(a)
Work function energy is the minimum energy required to remove an electron from the metal/surface.

(b)
Variation with frequency f of maximum kinetic energy Ek of photoelectrons emitted from surface of sodium metal.
Use gradient to determine value of Planck constant h:
Gradient = 4.17x10-15 (allow 4.1 – 4.3)
h = (4.15x10-15)(1.6x10-19)                  or h = (4.1 to 4.3) x10-15 eVs
h = 6.6x10-34Js

(c)
Sodium metal (Work function = 2.4eV) replaced by Calcium (Work function = 2.9eV).
Draw line to show variation with f of max Ek of photoelectrons emitted from surface of calcium:

Graph: 
a straight line parallel to the given line with intercept at frequency between 6.9x1014Hz and 7.1x1014Hz



Question 8
Strontium has at least 16 isotopes. One is strontium-89, with half-life = 52days
(a)
Isotopes of an element are nuclei of that element having the same number of protons(atomic number) but different number of neutrons

(b)
Probability per second of decay of a nucleus of strontium-89:
The probability of decay per unit time is the decay constant.
λ = ln2 / t½ = 0.693 / (52x24x3600) = 1.54x10-7s-1    

(c)
Activity of strontium-89 source measured 21 days after preparation of source and found to be = 7.4x106Bq.
At time it was prepared,
(i)
Activity:
A = Aoexp(-λt)
7.4x106 = Aoexp(-1.54x10-7(21x24x3600))
Ao = 9.8x106Bq

(ii)
Mass of strontium-89:
o = λN                       and      mass = N x 89/N­A
mass = (9.8x106 x 89) / (1.54x10-7 x 6.02x1023) = 9.4x10-9g  





SECTION B

Question 9
(a)
4 properties of an ideal operational amplifier (op-amp):
Any 4:
e.g Infinite input impedance/resistance
Zero output impedance/resistance
Infinite (open loop) gain
Infinite bandwidth
Infinite slew rate

(b)
Comparator: VOUT = Ao (V+ - V-). Here V+ = 0V and V- = VIN
Graph: square wave with a 180o phase change and amplitude 5.0V

(c)
Connections of the 2 LEDs to output of op-amp – red when Vout +ve and green when Vout -ve:
The correct symbol of LED
Diodes are connected correctly between Vout and earth
Diodes are identified



Question 10
(a)
Aluminum block placed near small source of X-ray radiation. Reasons why intensity of X-ray beam at point B not as great as the intensity at point A:
Any 2 sensible suggestions:
The beam is divergent / obeys an inverse square law
Due to absorption (in the block)
Due to scattering (of the beam in the block)
Reflection (occurs at the boundaries)

(b)
Thickness of model = 2.4cm. Thickness of bone in model = 1.1cm
Linear attenuation (absorption) coefficient for bone = 3.0cm-1, for soft tissue = 0.27cm-1
Ratio of intensity of X-ray beam incident on model to intensity of X-ray beam emergent from model:
(i)
For beam AB:
I = Ioexp(-μx)
Ratio = Io/I = exp(0.27x2.4)  = 1.9

(ii)
For beam CD:
Ratio = Io/I = exp(0.27x1.3) x exp(3.0x1.1)
Ratio = 1.42 x 27.1 = 38.5
(A common error was to calculate the ratios for the bone and the soft tissue and then to add these two quantities, rather than find the product.)

(c)
Why, for this model, X-ray image with good image contrast may be obtained:
Either There is much greater absorption in the bone than in the soft tissue
Or Io/I is much greater for the bone than for the soft tissue



Question 11
Signal transmitted over long distance will be attenuated and it will pick up noise.
(a)
(i)
Attenuation is the loss of (signal) power.

(ii)
Noise is the random unwanted power that gets picked up on the signal.

(b)
Why regenerator amplifiers do not amplify noise that has been picked up on digital signals:
For a digital signal, only the ‘high’ and the ‘low’ \ 1 and 0 are necessary. So, any variation between ‘highs’ and ‘lows’ which are caused by noise is not required.

(c)
Transmitter on Earth produces signal of power 2.4kW. Signal, when received by satellite, is attenuated by 195dB.
Signal power received by satellite:
Attenuation = 10log(P2/P1)
Either 195 = 10log({2.4x103}/P1)
Or –195 = 10log(P1/{2.4x103})
P1 = 7.6x10-17W  



Question 12
Simplified block diagram of circuitry for a mobile-phone handset.
(a)
(i)
X = modulator

(ii)
Y = serial-to-parallel converter

(b)
Explain purpose of
(i)
Switch:
The switch enables one aerial to be used for transmission and receipt of signals.

(ii)
Parallel-to-serial converter:
All the bits for one number arrive at the same time. The parallel-to-serial converter allows the bits to be sent out one after another.

7 comments:

  1. in question 2b) why did you use (a2 – x2) ?

    ReplyDelete
  2. for 2 c iii) why is it (-2.0,3.5) &(2.0,3.5)? shouldn't it be (-2.8,0) &(+2.8,0)?

    ReplyDelete
    Replies
    1. Check the updated details and see if you understand now

      Delete
  3. in 6b why not 1.6 * 10^-19 like other questions

    ReplyDelete
    Replies
    1. Details have been updated for this part of the question

      Delete
  4. Please explain 2012 mj paper 41 question 7 last part

    ReplyDelete
    Replies
    1. See solution 1072 at
      http://physics-ref.blogspot.com/2015/12/physics-9702-doubts-help-page-226.html

      Delete

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