teachercol
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OCR Physics A G482 EWP 9th June 2014

Usual disclaimers.
These are just my answers and are in no sense 'official'.
They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.
Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin
so I is not prop to V
so doesn't obey Ohm's law [1]
ii) Line is approx straight so R is approx constant
R = V / I = 0.5v / 50mA = 10ohms [2]
iii) At 6v filament gets hot ( more power dissipated)
Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat
so resistance increases [2]
b) In the daytime Rldr = 1ohm
so Rcombination is approx 1 ohm
so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v
PD across light bulb = PD across LDR is small and doesnt light
At night Rldr = 1000ohm
so R combination = R light bulb approx = 25ohm
so light bulb gets 1/2 of 12v = 6v
so lights up. [5]
Tricky potential divider with that parallel combination.
TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]
b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]
ii) Internal resistance = resistance between the terminals of the cell. [1]
iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it
must have twice the current [2]
iv) PD across top branch = I x total R = 0.16x6 = 0.96v
so 0.96v across bottom branch
Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v
[A lot of work for 2 marks]
v) PD splits 3:3 in top branch and 6:6 in bottom branch
same potential at each point so no difference [2]
vi) Current in bottom branch = 0.08A (half the other one)
so current through cell = 0.16 + 0.08 = 0.24A
V = E - Ir so 0.96 = 1.2 - 0.24 r
so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]
A tough question after the easy definitions
TOTAL 15

Q3 a) resistance = PD / Current [1]
b) i) strange setup which will throw many
R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2
= 224 ohm [3]
ii) no density of electrons = no of atoms per unit vol
= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]
Hard work for 1 mark
iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19
= 3.2E-14A (amazingly small) [2]
iv) Power =I^2 R for each conductor
so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E-16W [3]
c) Electrons undergo many collisions with atoms so mean drift velocity is very small.
EM waves travel at speed of light (ish) so v fast and very short time [2]
Not a straightforward resistivity question.
TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]
ii) PD across LED = 4.0v (from graph)
PD across R = 100 x 20mA = 2.0v
so Terminal PD = 6.0v [3]
b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]
ii) Turn on PD = 2.56v from graph
Photon is emitted when electrons have enough energy to excite atoms in LED
energy lost by electron = Energy of photon [2]
c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]
ii) Total = energy of 1 photon x no of photons /s (possible ecf)
= 1.25E17 x 4.1E-19 = 5.125E-2 [2]
iii) Efficiency = useful power out / power in = power out / VI
= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]
d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]
Not too bad if you know what you're doing but tricky all the same
TOTAL 15

5 a) yes coherent -constant phase difference
b) Superposition principle : At min add displacements together.
Min displacement = 4.0 - 2.0 = 2.0 so not zero
Waves have different amplitudes so cant cancel perfectly [2]
c) i) 90 ii) 135 [2]
d) i ) T = 0.80ms so f = 1250Hz
v = f x lambda so lambda = 340/1250 = 0.272m [3]
ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)
= 7.1m [3]
e) i) Intensity = power per unit area [1]
ii) Min amp = 2 units
Max amp = 6 units
Intensity is prop to amp squared
so max intensity = 9 x min intensity = 9 x 4.0E-6
= 3.6E-5 [3
Finishes with a real tricky bit.
TOTAL 15

Q6 nearest thing to straightforward!
a) Travel at speed of light
through a vacuum [2]
b) i) plane polarised means oscillations are confined to a single plane [2]
ii) Polariser 1 only allows vert components through and absorbs horiz
Polariser 2 only allows horizontal components through and absorbs vert
so all components absorbed and dont see anything [2]
iii) Yes - will see something.
Easiest to draw diagrams showing components - I demo this with microwaves each year.
Hopefully my students remembered it! Difficult if youve never seen it before [3]
TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.
Interference / superposition.
Cancellation - . nodes. Reinforcement -> antinodes etc [3]
b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]
c) Some confusion about scales
Antinode to antinode = 6cm = lambda /2
so lambda = 12cm
so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]
TOTAL 9
Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]
ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]
iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19 [2]
b) i) Energy of photon = hc/lambda = 4.5E-19
so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J
so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]
ii) lambda = h /mv = 1.63E-9m [2]
c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]
ii) 3rd transition means 3 -> 1
Calculate energies of2->1 and 3->2 , add then use E=hf
or add frequencies etc
Ans = 252nm [3]
Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E
They are going to come down from that
Could be 65 and 35.
Might be even lower

Best guess?

A 65
B 58
C 50
D 43
E 35

Good Luck.
Everyone is going to find this paper hard.


(Edited for various typos)
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Red Fox
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(Original post by teachercol)
.
For 7c, I converted the frequency to a period and did wavelength/ period, 12x10^-2/whatever it was and got 3x10^8. No idea why I did this but it's the same thing so will I get the marks?
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gothmog827
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thank you for this. I wish I had you for a teacher!
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Elcor
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This might be wrong, but for 3.b.iv shouldn't you x I by 10^9 and then square that value for V=I^2R (and x R by 10^9 too as you have)? Getting approx. 230W.
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L'Evil Fish
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In the I = neva

I used n = 2 x 10^28 because my own value for n wasn't right (calc error, i did 80/vol)

Will I get the marks?

Because I wrote

If n = 2 x 10^28 m^-3 then carried on calculating? And had 5.5 I think instead of 5.2
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Flauta
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Yeah looks mostly in line with what I got, thanks so much for this!
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teachercol
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(Original post by Elcor)
This might be wrong, but for 3.b.iv shouldn't you x I by 10^9 and then square that value for V=I^2R (and x R by 10^9 too as you have)? Getting approx. 230W.
I don't think so.
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teachercol
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(Original post by L'Evil Fish)
In the I = neva

I used n = 2 x 10^28 because my own value for n wasn't right (calc error, i did 80/vol)

Will I get the marks?

Because I wrote

If n = 2 x 10^28 m^-3 then carried on calculating? And had 5.5 I think instead of 5.2
I'd expect ecf there . so yes you should get the marks
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SH0405
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(Original post by teachercol)
OCR Physics A G482 EWP 9th June 2014

Usual disclaimers.
These are just my answers and are in no sense 'official'.
They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.
Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin
so I is not prop to V
so doesn't obey Ohm's law [2]
ii) Line is approx straight so R is approx constant
R = V / I = 0.5v / 50mA = 10ohms [2]
iii) At 6v filament gets hot ( more power dissipated)
Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat
so resistance increases [2]
b) In the daytime Rldr = 1ohm
so Rcombination is approx 1 ohm
so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v
PD across light bulb = PD across LDR is small and doesnt light
At night Rldr = 1000ohm
so R combination = R light bulb approx = 25ohm
so light bulb gets 1/2 of 12v = 6v
so lights up. [5]
Tricky potential divider with that parallel combination.
TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]
b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]
ii) Internal resistance = resistance between the terminals of the cell. [1]
iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it
must have twice the current [2]
iv) PD across top branch = I x total R = 0.16x6 = 0.96v
so 0.96v across bottom branch
Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v
[A lot of work for 2 marks]
v) PD splits 3:3 in top branch and 6:6 in bottom branch
same potential at each point so no difference [2]
vi) Current in bottom branch = 0.08A (half the other one)
so current through cell = 0.16 + 0.08 = 0.24A
V = E - Ir so 0.96 = 1.2 - 0.24 r
so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]
A tough question after the easy definitions
TOTAL 15

Q3 a) resistance = PD / Current [1]
b) i) strange setup which will throw many
R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2
= 224 ohm [3]
ii) no density of electrons = no of atoms per unit vol
= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]
Hard work for 1 mark
iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19
= 3.2E-14A (amazingly small) [2]
iv) Power =I^2 R for each conductor
so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E16W [3]
c) Electrons undergo many collisions with atoms so mean drift velocity is very small.
EM waves travel at speed of light (ish) so v fast and very short time [2]
Not a straightforward resistivity question.
TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]
ii) PD across LED = 4.0v (from graph)
PD across R = 100 x 20mA = 2.0v
so Terminal PD = 6.0v [3]
b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]
ii) Turn on PD = 2.56v from graph
Photon is emitted when electrons have enough energy to excite atoms in LED
energy lost by electron = Energy of photon [2]
c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]
ii) Total = energy of 1 photon x no of photons /s (possible ecf)
= 1.25E17 x 4.1E-19 = 5.125E-2 [2]
iii) Efficiency = useful power out / power in = power out / VI
= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]
d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]
Not too bad if you know what you're doing but tricky all the same
TOTAL 15

5 a) yes coherent -constant phase difference
b) Superposition principle : At min add displacements together.
Min displacement = 4.0 - 2.0 = 2.0 so not zero
Waves have different amplitudes so cant cancel perfectly [2]
c) i) 90 ii) 135 [2]
d) i ) T = 0.80ms so f = 1250Hz
v = f x lambda so lambda = 340/1250 = 0.272m [3]
ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)
= 7.1m [3]
e) i) Intensity = power per unit area [1]
ii) Min amp = 2 units
Max amp = 6 units
Intensity is prop to amp squared
so max intensity = 9 x min intensity = 9 x 4.0E-6
= 3.6E-5 [3
Finishes with a real tricky bit.
TOTAL 15

Q6 nearest thing to straightforward!
a) Travel at speed of light
through a vacuum [2]
b) i) plane polarised means oscillations are confined to a single plane [2]
ii) Polariser 1 only allows vert components through and absorbs horiz
Polariser 2 only allows horizontal components through and absorbs vert
so all components absorbed and dont see anything [2]
iii) Yes - will see something.
Easiest to draw diagrams showing components - I demo this with microwaves each year.
Hopefully my students remembered it! Difficult if youve never seen it before [3]
TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.
Interference / superposition.
Cancellation - . nodes. Reinforcement -> antinodes etc [3]
b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]
c) Some confusion about scales
Antinode to antinode = 6cm = lambda /2
so lambda = 12cm
so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]
TOTAL 9
Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]
ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]
iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19
b) i) Energy of photon = hc/lambda = 4.5E-19
so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J
so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]
ii) lambda = h /mv = 1.63E-9m [2]
c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]
ii) 3rd transition means 3 -> 1
Calculate energies of2->1 and 3->2 , add then use E=hf
or add frequencies etc
Ans = 252nm [3]
Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E
They are going to come down from that
Could be 65 and 35.
Might be even lower

Best guess?

A 65
B 58
C 50
D 43
E 35

Good Luck.
Everyone is going to find this paper hard.
Was question one 10 or 11 marks?
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teachercol
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(Original post by Red Fox)
For 7c, I converted the frequency to a period and did wavelength/ period, 12x10^-2/whatever it was and got 3x10^8. No idea why I did this but it's the same thing so will I get the marks?
Odd but should be Ok - unless they explicitly have v = f lambda in the mark scheme
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L'Evil Fish
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(Original post by teachercol)
I'd expect ecf there . so yes you should get the marks
Thanks, hopefully 85+ then.
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Elcor
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(Original post by teachercol)
I don't think so.
But if you think about it, 2.3E16 W, the value you got, is a massive power for a small chip used in a computer.

My logic was to look at the chip as a whole, so the current will be 10^9 x larger and so will R. Then to P=I^2R on the chip as a whole, getting ~230W. Is that wrong?
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teachercol
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(Original post by SH0405)
Was question one 10 or 11 marks?
Its 10 marks.

Typo in a) its only 1 mark. I'll edit it.

Thanks

Edited
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MollyMcFly1
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Thank you for this! Feeling a lot more confident now
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SH0405
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(Original post by teachercol)
Its 10 marks.

Typo in a) its only 1 mark. I'll edit it.

Thanks

It wont let me .....bah

Ok - also for question 6.iii), I believe I explained it quite well. I brought in Malus' Law - is this right? However I forgot to square cos(theta) when explaining the Law. Assuming I explained the polarisation theory correctly, how many marks would you say this is?
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SH0405
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How many marks was 8aiii?
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teachercol
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(Original post by Elcor)
But if you think about it, 2.3E16 W, the value you got, is a massive power for a small chip used in a computer.

My logic was to look at the chip as a whole, so the current will be 10^9 x larger and so will R. Then to P=I^2R on the chip as a whole, getting ~230W. Is that wrong?
I got 2.3E-16 which is a very small amount.

The current wont be 1E9 times higher.
Youre told each chip has the same current (3.2E-14A) calculated in iii
You just multiply the number of chips by the power for one chip.
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teachercol
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(Original post by SH0405)
How many marks was 8aiii?
2 marks
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Red Fox
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Somewhere between 60-72, I pray I did enough for an A
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teachercol
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(Original post by SH0405)
Ok - also for question 6.iii), I believe I explained it quite well. I brought in Malus' Law - is this right? However I forgot to square cos(theta) when explaining the Law. Assuming I explained the polarisation theory correctly, how many marks would you say this is?
Should be an Ok approach.
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