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Reply 80
What was the question about the distance of lens ?
Reply 81
Which resistance did you put circle around?
Reply 82
Original post by walkers38
Am I the only 'girl' here? Now I understand why this test was hard for me. They should give extra 2 marks to any girls sitting this crazy paper. Haha. #


I kid ^_^


I'm a girl to. Lol.
Original post by edexcel
I think that our paper was harder than Jan 10. Anyone else think the same?


I don't think that the paper was particularly difficult (although there were some tricky questions, like the frequency with the n's and stuff), I just think that there was far too much to get through. As a result, when I got to the last question I had barely any time left, and so was rushing answers with not much time too think. :frown:
Reply 84
250 ohms for resistance I put. Does anyone have an unoffical markscheme that someone coul post?
Reply 85
Anyone working on an unofficial mark scheme?


Posted from TSR Mobile
Reply 86
Original post by Fordey
What was the question about the distance of lens ?


The very first question? The waves were coming in with 0 curvature. Curvature =1/u. Since 1/u = 0, u = (object is really far away). I think if you just said the waves had zero curvature, that'd be enough for the mark.
Original post by Fordey
Which resistance did you put circle around?


250Ω250\Omega
Reply 88
Original post by nacho1996
250 ohms for resistance I put. Does anyone have an unoffical markscheme that someone coul post?


I also put that, but it was kind of a guess. How sure are you?
Reply 89
Any one get 500000 Hz or a number that looks similar to that? For the frequency question? The units at the start were piss and the frequency spectrum question was hard. For the last bit I said that to increase the number of bits per sample would reduce quantisation errors then I went on to say something about increasing the sample rate, which actually would have no effect..:frown: Will I not get the marks now?:frown:
Reply 90
Original post by Fordey
Which resistance did you put circle around?


250 ohms

The total resistance of the parallel resistors was 1/100 + 1/100 =1/r so r=50. 200 + 50 = 250 ohms
1. As, JC-1 , AV-1 (3)
2i) Wave-fronts are parallel before the lens (1)
ii) Focusing waves at focus (1) wave-fronts equally spaced by eye (1)
3i.) 250 ohms (1)
ii) 0.048A (2) ecf from i)
4a) 77Hz (1)/ find highest frequency and divide by 5 as uncertainty divided by 5 too (1)
b) Amplitude decreased (1), same frequency (1)/ by same scale factor (1)
5a) 225 MPa (1)
ii) 2.25 x1011 Pa (2)
6a) 3000 bits-1 (3)
b) Analogue signal rounded to nearest voltage level (1)
c) Increase amount of bits/ sample (1) so there are more voltage levels to round to (1)
7. 0.06m (2)
Another part to this question???
8a) 0.0348A (2)
ii) 5,000,000 Hz (1)
ii) 4 (1)
b) 53.3s (2)
ii) Bits/s decreases due to (other file downloading) (1) means takes longer to download desired amount of bits (1)
c) Amount of noise increases with distance (1)/ after 60m noise too much (1) / reference to low Vtotal/Vnoise ratio (1)/ digitising just copies noise (1)/ less waves hit the laptop directly from a distance (1) / interference from other waves (1) Max (4)
9a)i) Planes of atoms at different slants (1) separated by grain boundaries (1)
a)ii Ductile- can be drawn into wires/ can be stretched with plastic deformation (1) carbon atoms pin dislocations into place in steel (1)/ dislocations allow a few planes of atoms to slip in pure iron causing ductility (1)
b)i) Difficult to scratch/dent (1) tip would become blunt (1)
ii) Negative sea of electrons glues positive metal ions together by electrostatic forces (1) covalent bond is sharing of electrons (doubtful mark, certainly wouldn’t need the full covalent bond saga), bonding non-directional in steel/ directional in diamond (1) covalent bonds are (very) strong (1), reference to no slip/ plastic flow/ dislocation in diamond (1) Max (4)
10a) So that conductive layer can bend and complete the circuit (1)
ii) So circuit is not automatically completed (1)
b) i) More conductive than an insulator but less conductive that a conductor (at room temperature) (1)
ii) Reference to doping (1) more free electrons (1) Max (1) for heat up with explanation
c) As X moves to the right, resistance of left resistor increases (1), so Vout increases (1)
d)i) Show 12/60 x 1000 = 20mVmm-1 (1)
ii) 0.25mm (1)
iii) 60/0.25 = 240 (1) , 27 < 240 < 28 (1), so 8 bits needed (1)
(edited 11 years ago)
Reply 92
Original post by ChrissM
I also put that, but it was kind of a guess. How sure are you?


Reasonably it was just a V=IR question. It was one of the only questions on the paper that I thought "I know this one". I found the paper overall really hard.
Reply 93
Original post by ChrissM
The very first question? The waves were coming in with 0 curvature. Curvature =1/u. Since 1/u = 0, u = (object is really far away). I think if you just said the waves had zero curvature, that'd be enough for the mark.

I said the waves were parallel to each other, would this get the mark?

Original post by ChrissM
I also put that, but it was kind of a guess. How sure are you?


If I remember correctly there was 2 100Ω100 \Omega resistors in parallel, and a 200Ω200 \Omega in series with those 2.

Those in parallel can be found using: 1RT=1R1+...\frac{1}{R_T}=\frac{1}{R_1}+...

Then once you've found that you simple add this number to the 200Ω200 \Omega resistor (because of the addition rule of resistors).

Therefore:
RT=11100+1100+200=250R_T=\frac{1}{\frac{1}{100}+\frac{1}{100}}+200 = 250

I can't remember if these were the resistor values, but this is definitely the method I used
(edited 11 years ago)
Reply 95
Original post by metaltron
1. As, JC-1 , AV-1 (3)
2i) Wave-fronts are parallel before the lens (1)
ii) Focusing waves at focus (1) wave-fronts equally spaced by eye (1)
3i.) 250 ohms (1)
ii) 0.048A (2) ecf from i)
4a) 77Hz (1)/ find highest frequency and divide by 5 as uncertainty divided by 5 too (1)
5a) 225 MPa (1)
ii) 2.25 x1011 Pa (2)
b) Amplitude decreased (1), same frequency (1)/ by same scale factor (1)
6a) 3000 bits-1 (3)
b) Analogue signal rounded to nearest voltage level (1)
c) Increase amount of bits/ sample (1) so there are more voltage levels to round to (1)
7. 0.06m (2)
Another part to this question???
8a) 0.0348A (2)
ii) 5,000,000 Hz (1)
ii) 4 (1)
b) 53.3s (2)
ii) Bits/s decreases due to (other file downloading) (1) means takes longer to download desired amount of bits (1)
c) Amount of noise increases with distance (1)/ after 60m noise too much (1) / reference to low Vtotal/Vnoise ratio (1)/ digitising just copies noise (1)/ less waves hit the laptop directly from a distance (1) / interference from other waves (1) Max (4)
9a) Ductile- can be drawn into wires/ can be stretched with plastic deformation (1) carbon atoms pin dislocations into place in steel (1)/ dislocations allow a few planes of atoms to slip in pure iron causing ductility (1)
b)i) Difficult to scratch/dent (1) tip would become blunt (1)
ii) Negative sea of electrons glues positive metal ions together by electrostatic forces (1) covalent bond is sharing of electrons (doubtful mark, certainly wouldn’t need the full covalent bond saga), bonding non-directional in steel/ directional in diamond (1) covalent bonds are (very) strong (1), reference to no slip/ plastic flow/ dislocation in diamond (1) Max (4)
10a) So that conductive layer can bend and complete the circuit (1)
ii) So circuit is not automatically completed (1)
b) i) More conductive than an insulator but less conductive that a conductor (at room temperature) (1)
ii) Reference to doping (1) more free electrons (1) Max (1) for heat up with explanation
c) As X moves to the right, resistance of left resistor increases (1), so Vout increases (1)
d)i) Show 12/60 x 1000 = 20mVmm-1 (1)
ii) 0.25mm (1)
iii) 60/0.25 = 240 (1) , 27 < 240 < 28 (1), so 8 bits needed (1)


Thank you very very very very much!
Reply 96
Looks like I've failed though. :-(
Reply 97
Original post by edexcel
I said the waves were parallel to each other, would this get the mark?


Yes I think so
Reply 98
Original post by metaltron
1. As, JC-1 , AV-1 (3)
2i) Wave-fronts are parallel before the lens (1)
ii) Focusing waves at focus (1) wave-fronts equally spaced by eye (1)
3i.) 250 ohms (1)
ii) 0.048A (2) ecf from i)
4a) 77Hz (1)/ find highest frequency and divide by 5 as uncertainty divided by 5 too (1)
5a) 225 MPa (1)
ii) 2.25 x1011 Pa (2)
b) Amplitude decreased (1), same frequency (1)/ by same scale factor (1)
6a) 3000 bits-1 (3)
b) Analogue signal rounded to nearest voltage level (1)
c) Increase amount of bits/ sample (1) so there are more voltage levels to round to (1)
7. 0.06m (2)
Another part to this question???
8a) 0.0348A (2)
ii) 5,000,000 Hz (1)
ii) 4 (1)
b) 53.3s (2)
ii) Bits/s decreases due to (other file downloading) (1) means takes longer to download desired amount of bits (1)
c) Amount of noise increases with distance (1)/ after 60m noise too much (1) / reference to low Vtotal/Vnoise ratio (1)/ digitising just copies noise (1)/ less waves hit the laptop directly from a distance (1) / interference from other waves (1) Max (4)
9a) Ductile- can be drawn into wires/ can be stretched with plastic deformation (1) carbon atoms pin dislocations into place in steel (1)/ dislocations allow a few planes of atoms to slip in pure iron causing ductility (1)
b)i) Difficult to scratch/dent (1) tip would become blunt (1)
ii) Negative sea of electrons glues positive metal ions together by electrostatic forces (1) covalent bond is sharing of electrons (doubtful mark, certainly wouldn’t need the full covalent bond saga), bonding non-directional in steel/ directional in diamond (1) covalent bonds are (very) strong (1), reference to no slip/ plastic flow/ dislocation in diamond (1) Max (4)
10a) So that conductive layer can bend and complete the circuit (1)
ii) So circuit is not automatically completed (1)
b) i) More conductive than an insulator but less conductive that a conductor (at room temperature) (1)
ii) Reference to doping (1) more free electrons (1) Max (1) for heat up with explanation
c) As X moves to the right, resistance of left resistor increases (1), so Vout increases (1)
d)i) Show 12/60 x 1000 = 20mVmm-1 (1)
ii) 0.25mm (1)
iii) 60/0.25 = 240 (1) , 27 < 240 < 28 (1), so 8 bits needed (1)


Thanks! What happened to question 7 though?
Reply 99
I know this may be a pain but has anyone got a copy of the paper they could upload so I could look at the markscheme and the paper and work out how well I did?

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