A-Level physics doubts

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Jian17
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#1
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#1
Hello so I just finished a past paper and I have some doubts I would like some help with.

https://papers.gceguide.xyz/A%20Leve..._w18_qp_42.pdf

This is the paper.

First doubt
So in question 1c, I re-arranged the density of a planet ( density = mass/ volume and density = (4/3)pi*r^3 ) and made M the subject, then I just substituted it in the equation of gravitational field strength.

In which I got g = (4/3)g*pi*r*density.
And then I just calculated it but it's wrong.

I saw in the mark scheme they used the following formula:
https://gyazo.com/37ba6c5feec6c6565b33b4b86350e34c
I would like to know how to find it.

Second doubt

In question 4c)ii) I would like to know how to find that ratio, just got no idea how to do it.

Third doubt

In question 9b I'm not sure if the diagram should be between one way or another:

https://gyazo.com/b5746a367e34f6257b6cf44dd4f89f0b

Last doubt: how to do question 12b)iii), I think I have to use the formula N = NA * n but not sure how to continue.

Thanks
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username3249896
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#2
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#2
First:
They used mass = density x volume, so  M = (4/3 \times \pi R^3) \rho
Then, at a height of R above the surface, the distance from the centre is 2R.
So we place in  g = \dfrac{GM}{r^2} to obtain  g = \dfrac{GM}{(2R)^2}

the distance from the centre is not R, but 2R.

Second:
Total energy of oscillations = Maximum kinetic/potential energy
Maximum kinetic energy,  E_K = 1/2 mv_{\text{max}}^2
 v_{max} = \omega x_0 - in the usual notations
hence, total energy of oscillations =  1/2 m \omega^2 x_0^2
period did not change, hence angular frequency did not change.
mass did not change.
hence we have energy proportional to (amplitude)^2
hence the ratio of energy required is the appropriate (ratio of amplitudes)^2 = (1.3/2)^2

Third:
Formula list has  V_H = \dfrac{BI}{nqt}
So the hall voltage is proportional to B.

Faraday's law: Induced emf is proportional to rate of change of magnetic flux linkage
magnetic flux linkage = magnetic flux density x area x number of turns
so magnetic flux linkage is proportional to magnetic flux density
so rate of change of flux linkage is proportional to magnetic flux density
so induced emf is proportional to magnetic flux density
Lenz's law: Induced current will flow in such a direction so as to produce effects to oppose the change causing it. So E is proportional to -(rate of change of flux linkage)

Hence emf proportional to negative gradient of B-t (magnetic flux density-time) graph
gives emf = 0 for 0 to t1, emf = constant positive value for t1 to t3, emf = 0 for t3 onwards

Fourth:
number of moles of reactant = 1 mol
when 1 deuterium nucleus reacts with 1 tritium nucleus, answer to 12(b)(ii) J of energy is released
for 1 mol, answer to 12(b)(ii) x 6.02 x 10^23 J of energy is released
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Jian17
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#3
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#3
Thank you very much.
I understood all your explanations except for the last part, why is the energy change * avogadro constant = energy released?

(Original post by BobbJo)
First:
They used mass = density x volume, so  M = (4/3 \times \pi R^3) \rho
Then, at a height of R above the surface, the distance from the centre is 2R.
So we place in  g = \dfrac{GM}{r^2} to obtain  g = \dfrac{GM}{(2R)^2}

the distance from the centre is not R, but 2R.

Second:
Total energy of oscillations = Maximum kinetic/potential energy
Maximum kinetic energy,  E_K = 1/2 mv_{\text{max}}^2
 v_{max} = \omega x_0 - in the usual notations
hence, total energy of oscillations =  1/2 m \omega^2 x_0^2
period did not change, hence angular frequency did not change.
mass did not change.
hence we have energy proportional to (amplitude)^2
hence the ratio of energy required is the appropriate (ratio of amplitudes)^2 = (1.3/2)^2

Third:
Formula list has  V_H = \dfrac{BI}{nqt}
So the hall voltage is proportional to B.

Faraday's law: Induced emf is proportional to rate of change of magnetic flux linkage
magnetic flux linkage = magnetic flux density x area x number of turns
so magnetic flux linkage is proportional to magnetic flux density
so rate of change of flux linkage is proportional to magnetic flux density
so induced emf is proportional to magnetic flux density
Lenz's law: Induced current will flow in such a direction so as to produce effects to oppose the change causing it. So E is proportional to -(rate of change of flux linkage)

Hence emf proportional to negative gradient of B-t (magnetic flux density-time) graph
gives emf = 0 for 0 to t1, emf = constant positive value for t1 to t3, emf = 0 for t3 onwards

Fourth:
number of moles of reactant = 1 mol
when 1 deuterium nucleus reacts with 1 tritium nucleus, answer to 12(b)(ii) J of energy is released
for 1 mol, answer to 12(b)(ii) x 6.02 x 10^23 J of energy is released
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username3249896
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#4
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#4
(Original post by Jian17)
Thank you very much.
I understood all your explanations except for the last part, why is the energy change * avogadro constant = energy released?
The answer to 12(b)(ii) is the energy change when 1 nucleus of tritium reacts with 1 nucleus of deuterium

2 g of deuterium is 1 mol of deuterium which contains 6.02 x 10^23 deuterium nuclei
3 g of tritium is 1 mol of tritium which contains 6.02 x 10^23 tritium nuclei

12(b)(ii) = energy change when 1 nucleus of tritium reacts with 1 nucleus of deuterium

energy change when 2 g of deuterium reacts with 3 g of tritium = energy change when 1 mol of tritium reacts with 1 mol of deuterium = energy change when 6.02 x 10^23 tritium nuclei react with 6.02 x 10^23 deuterium nuclei

energy change when 6.02 x 10^23 tritium nuclei react with 6.02 x 10^23 deuterium nuclei = 6.02 x 10^23 times the energy change when 1 nucleus of tritium reacts with 1 nucleus of deuterium = 6.02 x 10^23 x answer to 12(b)(ii)2.
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Jian17
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#5
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#5
Also, similarly for this question the correct variation would be the one in red right? My teacher told me blue is right. Name:  image-1cfd8dbb-341d-461b-adab-b3e39bbfc09b1365549589-compressed.jpg.jpeg
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username3249896
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(Original post by Jian17)
Also, similarly for this question the correct variation would be the one in red right? My teacher told me blue is right. Name:  image-1cfd8dbb-341d-461b-adab-b3e39bbfc09b1365549589-compressed.jpg.jpeg
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The red one is right. Which paper is this from?

I think according to the ms both answers get full marks. You just need to show them in opposite directions (i.e one negative, one positive).

Also the values on the y-axis are only relative values, not absolute values (unless it is calculated from given values in the question).
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Jian17
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#7
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#7
9702/41/O/N/16 https://papers.gceguide.xyz/A%20Leve..._w16_qp_41.pdf

(Original post by BobbJo)
The red one is right. Which paper is this from?

I think according to the ms both answers get full marks. You just need to show them in opposite directions (i.e one negative, one positive).

Also the values on the y-axis are only relative values, not absolute values (unless it is calculated from given values in the question).
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username3249896
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#8
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ms says:
"E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0s, 1.6 s → 2.0s B1
E = 4 mV for t = 0.3 s → 0.6 s (either polarity) B1
E = 2 mV for t = 1.0 s → 1.6 s B1
with opposite polarity"

so both curves (red and blue) are correct

you would get the opposite polarity if the voltmeter is connected the opposite way
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Jian17
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#9
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Just wondering, how would the voltmeter need to be connected so that the *correct* sketch is blue?
Not too sure where would be opposite, sorry for being a pain
(Original post by BobbJo)
ms says:
"E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0s, 1.6 s → 2.0s B1
E = 4 mV for t = 0.3 s → 0.6 s (either polarity) B1
E = 2 mV for t = 1.0 s → 1.6 s B1
with opposite polarity"

so both curves (red and blue) are correct

you would get the opposite polarity if the voltmeter is connected the opposite way
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username3249896
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#10
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#10
(Original post by Jian17)
Just wondering, how would the voltmeter need to be connected so that the *correct* sketch is blue?
Not too sure where would be opposite, sorry for being a pain
Both sketches are correct because it depends on the way you connect the voltmeter

You will get one of the curves when connecting the voltmeter one way

You will get the other one when reversing the connections, i.e, the one you took at positive is now negative, the one which was negative is now positive

https://www.allaboutcircuits.com/tex...voltage-usage/
This shows a picture of a multimeter
One curve -> red positive, black negative (as supposed to be)
Other curve -> reverse it so that red negative, black positive

Pause at 0:48 https://www.youtube.com/watch?v=ELiTfD6ehuI to see a voltmeter and the actual terminals
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