Physics specimen papers set 1 paper 2 A LEVEL

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za19
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Is anyone finding that the answers for paper 2 are wrong? I don't know if i'm being stupid but I cannot get the answers for question 2 or question 4. If you have managed to get the right answer could you please share the solution
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Starmock99
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Physics A? Are they multiple choice questions? Make it clearer or post a pic of the questions !
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za19
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(Original post by Starmock99)
Physics A? Are they multiple choice questions? Make it clearer or post a pic of the questions !
sorry! it's aqa board and here is a link to the paper
http://filestore.aqa.org.uk/resource...-74081-SQP.PDF
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Starmock99
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(Original post by tasha1898)
sorry! it's aqa board and here is a link to the paper
http://filestore.aqa.org.uk/resource...-74081-SQP.PDF
What question don't you get? 1.2? Any particular one in 2 or 4????
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za19
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(Original post by Starmock99)
What question don't you get? 1.2? Any particular one in 2 or 4????
i must've copied the wrong link its this one
http://filestore.aqa.org.uk/resource...-74082-SQP.PDF
and essentially 2.2 and 2.3 and the whole of 4 because the first answer is used in the other questions.
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Eimmanuel
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(Original post by tasha1898)
i must've copied the wrong link its this one
http://filestore.aqa.org.uk/resource...-74082-SQP.PDF
and essentially 2.2 and 2.3 .......

For 2.2, make use of the capacitance formula that depends on the geometry of the setup to compute the capacitance when the insulator is pulled out:

  C =\epsilon_0  \dfrac{ A}{ d }

Then use the definition of capacitance to compute the p.d:

  C = \dfrac{ Q}{ V }

For 2.3, the energy stored by the capacitor is “defined” by the following formula:

  C = \frac{ 1}{ 2 }CV^2 = \frac{ 1}{ 2 }QV = \dfrac{ Q^2}{ 2C }
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Eimmanuel
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(Original post by tasha1898)
....the whole of 4 ....

For 4.1:

Compute the number of moles of sodium-24 using

 \text{mol.} = \dfrac{\text{ amount of Na-24 in g}}{ A_r(\text{Na}) }

Then use Avogadro constant to find the number of atoms in 15 cm3:

 \text{Number} = \text{mol.} \times N_A \times \dfrac{15}{ 1500 }

For 4.2, you need to find the decay constant first but note that you need to convert the half-life in hours to seconds:

 \lambda = \dfrac{\ln 2}{ t_{1/2} }

Then use the number in 4.1 in the formula:


A = λ N


For 4.3, compute the total activity in the flask say AT after 3.5 h.

 A_T = A_0 e^{- \lambda t}

 A_0 is from 4.2

Then use the following ratio:

For 15 cm3, the activity is 3600 Bq. If the activity is AT, then what is the volume?


For 4.4, the estimated mass of the liquid = 5.2 – 1.5 = 3.7 kg.

If the liquid is water, the density of water is 1 g/cm3, so the volume of the liquid is 3700 cm3.

See the MS for the deduction. They should make senses. If they don’t make sense, you can always post your query.
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za19
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(Original post by Eimmanuel)
For 2.2, make use of the capacitance formula that depends on the geometry of the setup to compute the capacitance when the insulator is pulled out:

  C =\epsilon_0  \dfrac{ A}{ d }

Then use the definition of capacitance to compute the p.d:

  C = \dfrac{ Q}{ V }

For 2.3, the energy stored by the capacitor is “defined” by the following formula:

  C = \frac{ 1}{ 2 }CV^2 = \frac{ 1}{ 2 }QV = \dfrac{ Q^2}{ 2C }
thanks! i used those and i was still getting it wrong but it was just a number error. question 4 is wrong though according to my teacher so no worries i guess
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Eimmanuel
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(Original post by tasha1898)
thanks! i used those and i was still getting it wrong but it was just a number error. question 4 is wrong though according to my teacher so no worries i guess
Could you explain why question 4 is incorrect? Thanks.
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za19
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(Original post by Eimmanuel)
Could you explain why question 4 is incorrect? Thanks.
the calculation they give on the mark scheme actually comes out as 7.5*10^12 not 10^10 so they have made a mistake i think, the numbers are right it's just the order of magnitude
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Eimmanuel
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(Original post by tasha1898)
the calculation they give on the mark scheme actually comes out as 7.5*10^12 not 10^10 so they have made a mistake i think, the numbers are right it's just the order of magnitude
I think that the answer is correct. The method suggested by the MS indeed gives 7.5E12 because it lacks the terms 15/1500, not that the question is wrong.
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za19
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(Original post by Eimmanuel)
I think that the answer is correct. The method suggested by the MS indeed gives 7.5E12 because it lacks the terms 15/1500, not that the question is wrong.
oh yes! that makes sense thanks. just an error in their method then
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