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AQA A-level Physics Paper 2 Unofficial Mark Scheme
6th June 2025
Unofficial Mark Schemes
Num Question Answer Discussion Mark
1) Topic: Thermal physics Total: 9
1.1) Show that there is 40 mols Pv = nRT Nice start before hell began +3 1
V=1n= U know it was bad when this was the first
T= 20°C P/RT question compared to paper 1
Pressure=105 kPa n=
105000/(8.31x(20+273)) n
= 43 mols n ~ 40 mols
1.2) Density 1.25 kgm^-3, Calculate the root 484m/s using 43mols Is the density not 1.45? Nope 1.25 3
mean square 503m/s-1 using 43.1 mol I said Nm/V is equal to the density so did
3P/density=crms^2, crms=501m/s
1.3) Temperature change when Crms is (Crms)^2 ∝ T Ratio change in T = sqrt2 2
doubled. 3T = 3*(270+23) = 879K Hence change in T=(Sqrt2-1)*T=122.6 K
T ∝ Crms^2 Above is wrong, for c to double T will quadruple.
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1.4) Determine how much heat energy is air mass = dens*vol I got 5.6x10^6? 3
removed per hour by the dehumidifier. = 1.25*960 = 1200 I think i got 8.__x10^8J
i found mass of water before and after and used
Num Question Answer Discussion Mark
Volume = 96 m^-3, Difference in mass of air = the difference in Q=mcΔθ+ml
Initial temp of moist air = 20°C 1200*0.0057 - 1200*0.0037 = 2.4kg Q = ml
Temp of water that collects = 10°C Q = mcΔθ
Temp of dry air leaving = 20°C Energy to change state I omitted the ml as the water isn't changing
Density = 1.25kg^-3 =2.4*2.3x10^6=5.52x10^6J state, just condensing
Energy to change temp
Ratio of mass of water / mass of air =2.4*1860*(20-10)
0.0057 =44640J
0.0037 Total = 5.56x10^6J
SPH of water vapor = 1860Jkg-1K-1
SLH of water = 2.3x10^6
2) Topic: Capacitors Total: 9
2.1) Shows a graph of a voltage of a capacitor C = (Q/V) 2
which is discharging Gradient = (V/t)
C x (V/t) = (Q/t) = I
Explain how a gradient can be used to
determine the initial current I0?
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2.2) Calculate the time constant. It takes 7.5 seconds for V to fall to use V = V0 (1- e^(-t /(RC) 2
36% of initial. i got 7.3 using that so should be fine if they allow
Show that the resistance is 2.4x10^5 a range and it gives correct resistance
Time constant = 7.5s Kp
C = 31uF And use half life instead of the full curve,
7.5=RC also read each box to half What did you take
R=7.5/31x10-6 as V_0
= 2.42x10^5
2.3) Current = 3.6x10-5 V = RI Sub in everything and rearrange for t 3
V = R * I0(1-e^-t/rc) (can someone confirm? idk if the 2nd is true bc
Num Question Answer Discussion Mark
What time is V = 6V 6 = 2.4*10^5 * 3.6*10^-5 (1-e^(t/7.5)) V is at the end and I is at the start)
Then -RC* ln(6/V_0)? Or is it other way around?
t = -7.5* ln(1-6/(2.4*3.6)) - pls check
= 8.98 seconds yes i think ☝️ is right
I got 8.9s <~~~~~ i oncur GOt 8.8s
Same, I used graph on page before though not
the current it gave us
2.4) Ratio of v1/v2 ⅛ ⅛ capacitance is inversely proportional to 2
voltage c1/c2 = v2/v1 Why
is not 8? v1/v2 = c2/c1
na i swear it was 1/16 cuz it’s d^2 and it was 2 d
½/ 4
I got ½ as well
1/2
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3) Topic: EMF magnetic fields Total: 7
3.1) Average emf induced when rod is released Calculate time to fall with suvat Why did they have to have suvat here 🙁 3
from 8m and falls to the ground assuming
Cold ngl
no air resistance. S=8 v² = u² + 2as
U=0 v = √2*9.81*8 = 12.53
Height: 8m
V=X shouldnt u use s=½at² and rearrange for t?
Length: 2m
A=9.81 yes This is another way of doing it. They're all
B = 4.9X10^-5T T=? suvat
Θ = 68
Do you remember the B?
s=ut+0.5at^2 emf = Blvcos64 = B*2*12.53
t=sqrt(2*8/9.81) Divide by sqrt(2) to find average emf
=1.28s Can someone please explain how area is 2, if
that is just the length of the bar? Area would be
EMF= BANcosΘ/t the size of the face of the bar into the field no? I
= (4.9x10^-5)(8*2)cos(68)/(1.28)) felt kinda confused here :(
= 2.29x10^-4V The area is 2*8 it is 2 long and falls 8 meters.
Num Question Answer Discussion Mark
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AQA A-level Physics Paper 2 Unofficial Mark Scheme
6th June 2025
Unofficial Mark Schemes
Num Question Answer Discussion Mark
1) Topic: Thermal physics Total: 9
1.1) Show that there is 40 mols Pv = nRT Nice start before hell began +3 1
V=1n= U know it was bad when this was the first
T= 20°C P/RT question compared to paper 1
Pressure=105 kPa n=
105000/(8.31x(20+273)) n
= 43 mols n ~ 40 mols
1.2) Density 1.25 kgm^-3, Calculate the root 484m/s using 43mols Is the density not 1.45? Nope 1.25 3
mean square 503m/s-1 using 43.1 mol I said Nm/V is equal to the density so did
3P/density=crms^2, crms=501m/s
1.3) Temperature change when Crms is (Crms)^2 ∝ T Ratio change in T = sqrt2 2
doubled. 3T = 3*(270+23) = 879K Hence change in T=(Sqrt2-1)*T=122.6 K
T ∝ Crms^2 Above is wrong, for c to double T will quadruple.
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1.4) Determine how much heat energy is air mass = dens*vol I got 5.6x10^6? 3
removed per hour by the dehumidifier. = 1.25*960 = 1200 I think i got 8.__x10^8J
i found mass of water before and after and used
Num Question Answer Discussion Mark
Volume = 96 m^-3, Difference in mass of air = the difference in Q=mcΔθ+ml
Initial temp of moist air = 20°C 1200*0.0057 - 1200*0.0037 = 2.4kg Q = ml
Temp of water that collects = 10°C Q = mcΔθ
Temp of dry air leaving = 20°C Energy to change state I omitted the ml as the water isn't changing
Density = 1.25kg^-3 =2.4*2.3x10^6=5.52x10^6J state, just condensing
Energy to change temp
Ratio of mass of water / mass of air =2.4*1860*(20-10)
0.0057 =44640J
0.0037 Total = 5.56x10^6J
SPH of water vapor = 1860Jkg-1K-1
SLH of water = 2.3x10^6
2) Topic: Capacitors Total: 9
2.1) Shows a graph of a voltage of a capacitor C = (Q/V) 2
which is discharging Gradient = (V/t)
C x (V/t) = (Q/t) = I
Explain how a gradient can be used to
determine the initial current I0?
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2.2) Calculate the time constant. It takes 7.5 seconds for V to fall to use V = V0 (1- e^(-t /(RC) 2
36% of initial. i got 7.3 using that so should be fine if they allow
Show that the resistance is 2.4x10^5 a range and it gives correct resistance
Time constant = 7.5s Kp
C = 31uF And use half life instead of the full curve,
7.5=RC also read each box to half What did you take
R=7.5/31x10-6 as V_0
= 2.42x10^5
2.3) Current = 3.6x10-5 V = RI Sub in everything and rearrange for t 3
V = R * I0(1-e^-t/rc) (can someone confirm? idk if the 2nd is true bc
Num Question Answer Discussion Mark
What time is V = 6V 6 = 2.4*10^5 * 3.6*10^-5 (1-e^(t/7.5)) V is at the end and I is at the start)
Then -RC* ln(6/V_0)? Or is it other way around?
t = -7.5* ln(1-6/(2.4*3.6)) - pls check
= 8.98 seconds yes i think ☝️ is right
I got 8.9s <~~~~~ i oncur GOt 8.8s
Same, I used graph on page before though not
the current it gave us
2.4) Ratio of v1/v2 ⅛ ⅛ capacitance is inversely proportional to 2
voltage c1/c2 = v2/v1 Why
is not 8? v1/v2 = c2/c1
na i swear it was 1/16 cuz it’s d^2 and it was 2 d
½/ 4
I got ½ as well
1/2
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3) Topic: EMF magnetic fields Total: 7
3.1) Average emf induced when rod is released Calculate time to fall with suvat Why did they have to have suvat here 🙁 3
from 8m and falls to the ground assuming
Cold ngl
no air resistance. S=8 v² = u² + 2as
U=0 v = √2*9.81*8 = 12.53
Height: 8m
V=X shouldnt u use s=½at² and rearrange for t?
Length: 2m
A=9.81 yes This is another way of doing it. They're all
B = 4.9X10^-5T T=? suvat
Θ = 68
Do you remember the B?
s=ut+0.5at^2 emf = Blvcos64 = B*2*12.53
t=sqrt(2*8/9.81) Divide by sqrt(2) to find average emf
=1.28s Can someone please explain how area is 2, if
that is just the length of the bar? Area would be
EMF= BANcosΘ/t the size of the face of the bar into the field no? I
= (4.9x10^-5)(8*2)cos(68)/(1.28)) felt kinda confused here :(
= 2.29x10^-4V The area is 2*8 it is 2 long and falls 8 meters.
Num Question Answer Discussion Mark
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