---> Physics Paper 4 Past Paper Discussion <---

[blueberryyums]

for the first option work is being done on the system , so there must be an increase in internal energy reight? this is confusing me.. as water is BEING freezed

however, for the third option the water is evaporating , meaning work is done BY the system... so internal energy must decrease... because increae in the internal energy is work done on the system and heat is supplied to the system....

can youclear my confusion?

Alright but in an hour. I was about to sign out and take a nap before seeing this. I'll grab my notes later and explain to you properly. My apologies for not answering immediately. My brain is shutting down already But i'll remember to reply this.

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_4.pdf

Number 4 c

Is this the correct drawing? Just a sketch

Just two points to note (unless it was just the damping you wanted to illustrate, in which case the diagram should be just fine):

i) The graph should start at the same point; the original diagram shows the mass initially displaced by 1.5 cm downwards (-1.5 cm) and the question where you are asked to sketch the graph also says that the initial displacement is 1.5 cm downwards, so the two graphs should start at the same point.

ii) The damping on the situation you've been asked to illustrate is much, much more than the damping in the initial situation, so the oscillating mass loses energy much faster, and thus the velocity decreases faster than the original situation shows. If the velocity decreases and frequency is to remain constant (as it does in your diagram) then the amplitude must decreases faster, so the graph you draw should always be in between the given wave and the x - axis, although your diagram shows one of the lines alternately above and below the other one.

Aside from that, it's all correct.

Good Luck for all your exams!

 

[seneria]

sagar plaese reply to my questions too..

 

[Freaked out]

nov 2 question 5 (a)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_4.pdf
you will have to make two capacitors in series parallel with another two capacitors in series.
as in the series the capacitance will be halved, it will become 24microF and when you add the the overall capacitance with the total capacitance of the capacitors which will be parallel it would be 24+24=48microF.
As the voltage adds up in series meaning the total voltage of the capacitors in series would become 25V+25V=50V
and as the voltages in parallel are same the overall voltage of the circuit would be 50V
Because according to the question we have to make a circuit in which the overall capacitance is 48microF and the voltage is 50V.
I hope it doesn't sound confusing.

 

[seneria]

nov 2 question 5 (a)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_4.pdf
you will have to make two capacitors in series parallel with another two capacitors in series.
as in the series the capacitance will be halved, it will become 24microF and when you add the the overall capacitance with the total capacitance of the capacitors which will be parallel it would be 24+24=48microF.
As the voltage adds up in series meaning the total voltage of the capacitors in series would become 25V+25V=50V
and as the voltages in parallel are same the overall voltage of the circuit would be 50V
Because according to the question we have to make a circuit in which the overall capacitance is 48microF and the voltage is 50V.
I hope it doesn't sound confusing.

okay, you mean that if we want to increar the voltage then we must increse the capacitors in series?

 

[seneria]

nov 2 question 5 (a)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_4.pdf
you will have to make two capacitors in series parallel with another two capacitors in series.
as in the series the capacitance will be halved, it will become 24microF and when you add the the overall capacitance with the total capacitance of the capacitors which will be parallel it would be 24+24=48microF.
As the voltage adds up in series meaning the total voltage of the capacitors in series would become 25V+25V=50V
and as the voltages in parallel are same the overall voltage of the circuit would be 50V
Because according to the question we have to make a circuit in which the overall capacitance is 48microF and the voltage is 50V.
I hope it doesn't sound confusing.

and thank you so much

 

[seneria]

can you explain question 6 part (c)and 9 part a (!!) and (b)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf

 

[sagar65265]

can you explain question 6 part (c)and 9 part a (!!) and (b)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf

6 c) If the r.m.s. current is equal to 2.6 Amperes, the power dissipated in any components in the circuit will be the same, but the actual maximum curent will not be 2.6 Amperes - if the rms current is equal to 2.6 amperes, the maximum possible current will be 2.3 * sqrt(2) = 3.25 Amperes and the minimum current will be -3.25 Amperes, due to the property of alternating current. Since the weight on the scale is modified by the force on the wire (which in turn results in the wire exerting a force on the magnet and thus increasing the reading on the scale), and the force on the wire is given by

F = B * I * L * sin(theta) = B * I * L

in this case, the force is proportional to the change in reading, and increasing the force on the wire by a factor of sqrt(2), the maximum deviation will be
3.25 grams greater than the actual weight of the magnet. However, when the current reverses direction, the force is the other way around, the reading on the scale will decrease and will be 3.25 grams less than the mass of the magnet. Therefore, the total deviation will be 2 * 2.3 * sqrt(2) = 6.5 grams.

I will try out question 9, I remember getting that question wrong when we were doing a mock exam at school, so i'll check whatever resources are available to try to get an answer.

Hope this helped!
Good Luck for all your exams!

P.S. really sorry I couldn't reply earlier, power went off here and comp got shut down... sorry!

 

[seneria]

6 c) If the r.m.s. current is equal to 2.6 Amperes, the power dissipated in any components in the circuit will be the same, but the actual maximum curent will not be 2.6 Amperes - if the rms current is equal to 2.6 amperes, the maximum possible current will be 2.3 * sqrt(2) = 3.25 Amperes and the minimum current will be -3.25 Amperes, due to the property of alternating current. Since the weight on the scale is modified by the force on the wire (which in turn results in the wire exerting a force on the magnet and thus increasing the reading on the scale), and the force on the wire is given by

F = B * I * L * sin(theta) = B * I * L

in this case, the force is proportional to the change in reading, and increasing the force on the wire by a factor of sqrt(2), the maximum deviation will be
3.25 grams greater than the actual weight of the magnet. However, when the current reverses direction, the force is the other way around, the reading on the scale will decrease and will be 3.25 grams less than the mass of the magnet. Therefore, the total deviation will be 2 * 2.3 * sqrt(2) = 6.5 grams.

I will try out question 9, I remember getting that question wrong when we were doing a mock exam at school, so i'll check whatever resources are available to try to get an answer.

Hope this helped!
Good Luck for all your exams!

P.S. really sorry I couldn't reply earlier, power went off here and comp got shut down... sorry!

i still didnt get that 2.3? are u assuming it to be 2.3?? how did u deduce this value?

and no problem.... thanks for the explanation btw

 

[sagar65265]

and also question 9part b(1) of nov 12 41
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_41.pdf

The question mentions an ideal op amp is being used - one of the main characteristics of an ideal op amp is that it has an infinite open loop gain (this means that when there is no feedback, like in this question, the op amp will always be saturated unless the input voltage is zero), so this op amp will be saturated throughout the question and it will behave like a comparator.

A comparator basically sees which input (the non inverting input (+) or the inverting input (-)) is greater than the other, and produces a corresponding output, If, for example, the non inverting input is at a greater potential than the inverting input, then the output will be positive - it will flow from the op amp to the rest of the external circuit. If the inverting input is greater than the non inverting input, then there will be a flow of current from the Earth connection to the op amp, i.e. a negative current.

So, they have told you that the potential at the non inverting input, V2, is constant at 1.5 V and that the potential at the inverting input, V1, is constantly varying. So, when the potential V2 is greater than the potential V1 (when the line is above the curve on the graph), the output will be a positive voltage & current. When it is not so, the output will be a negative voltage(at the op amp output) & current.

But since the op amp is saturated, the output will either be zero, +5V or -5V, depending on which input is greater.

So, when the line is above the curve, the output voltage will be positive and constant at +5V and thus the graph will be a straight line until the time at which the line and the curve meet. At this point, the difference in potential is zero and thus the output will also decrease sharply towards zero and then go towards -5V since the curve will then be above the line.

A file is attached to this post, I hope it helped clear up the description and the concept (the output line is the bold red one).

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

i still didnt get that 2.3? are u assuming it to be 2.3?? how did u deduce this value?

and no problem.... thanks for the explanation btw

It is given in the question that turning on the direct current of 2.6 Amperes increases the reading on the scale by 2.3 grams, here's the exact paragraph:

The wire XY is horizontal and normal to the magnetic field. The length of wire between
the poles is 4.4 cm.
A direct current of magnitude 2.6 A is passed through the wire in the direction from X
to Y.
The reading on the top-pan balance increases by 2.3 g.

So because the force increases linearly with current, so does the reading on the scale and increasing the current by a factor of sqrt(2) will increase the present changed reading on the scale (2.3 grams change ==> 2.3 * sqrt(2) = 3.25 grams) by the same factor, sqrt(2). Similarly for the negative current, the same change from the original value will occur, except that the reading will change in the other "direction" (when the current is positive, the reading increased; when the current turns negative, the reading should decrease).

Hope this helped!
Good Luck for all your exams!

 

[seneria]

need help in question june 9Q11b(11)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf

 

[seneria]

It is given in the question that turning on the direct current of 2.6 Amperes increases the reading on the scale by 2.3 grams, here's the exact paragraph:

The wire XY is horizontal and normal to the magnetic field. The length of wire between
the poles is 4.4 cm.
A direct current of magnitude 2.6 A is passed through the wire in the direction from X
to Y.
The reading on the top-pan balance increases by 2.3 g.

So because the force increases linearly with current, so does the reading on the scale and increasing the current by a factor of sqrt(2) will increase the present changed reading on the scale (2.3 grams change ==> 2.3 * sqrt(2) = 3.25 grams) by the same factor, sqrt(2). Similarly for the negative current, the same change from the original value will occur, except that the reading will change in the other "direction" (when the current is positive, the reading increased; when the current turns negative, the reading should decrease).

Hope this helped!
Good Luck for all your exams!

thank you so much....

 

[seneria]

The question mentions an ideal op amp is being used - one of the main characteristics of an ideal op amp is that it has an infinite open loop gain (this means that when there is no feedback, like in this question, the op amp will always be saturated unless the input voltage is zero), so this op amp will be saturated throughout the question and it will behave like a comparator.

A comparator basically sees which input (the non inverting input (+) or the inverting input (-)) is greater than the other, and produces a corresponding output, If, for example, the non inverting input is at a greater potential than the inverting input, then the output will be positive - it will flow from the op amp to the rest of the external circuit. If the inverting input is greater than the non inverting input, then there will be a flow of current from the Earth connection to the op amp, i.e. a negative current.

So, they have told you that the potential at the non inverting input, V2, is constant at 1.5 V and that the potential at the inverting input, V1, is constantly varying. So, when the potential V2 is greater than the potential V1 (when the line is above the curve on the graph), the output will be a positive voltage & current. When it is not so, the output will be a negative voltage(at the op amp output) & current.

But since the op amp is saturated, the output will either be zero, +5V or -5V, depending on which input is greater.

So, when the line is above the curve, the output voltage will be positive and constant at +5V and thus the graph will be a straight line until the time at which the line and the curve meet. At this point, the difference in potential is zero and thus the output will also decrease sharply towards zero and then go towards -5V since the curve will then be above the line.

A file is attached to this post, I hope it helped clear up the description and the concept (the output line is the bold red one).

Hope this helped!
Good Luck for all your exams!

i didnt get the point where the curve and line meet why do the vout decreases?

 

[sagar65265]

i didnt get the point where the curve and line meet why do the vout decreases?

If the gain is infinite, as is expected of an ideal op amp, then even a difference of 10^-15 (it's just a random value!) will result in an infinite output, since anything (except zero) times infinity is equal to infinity, and since an infinite output voltage is impossible, the op amp will be saturated for any difference in the voltages. However, when the line meets the curve, there is a single point when the difference between the inverting input potential is equal to the non inverting input potential, and so the difference is zero, and thus the output signal reduces to zero.

I'm not really sure about the following, but just see if it makes sense to you:
The reason the graph just doesn't have 1 instantaneous point at zero and has to decrease from +5V to -5V without any time passing by may be that as the line comes close to the graph, there may be a VERY, VERY short time period when the difference is so small that even with the ideal op amp, the output reduces, and that the output touches zero only exactly when they meet - it might be decreases from +5V to -5V over a period of 10^-20 seconds or some really small value, but since the graph doesn't go to that accuracy, we can't really display that time duration on the output line.

I'm again, not really sure about the above paragraph, so you may have to confirm it with someone else.

Hope this helped!

Good Luck for all your exams!

 

[seneria]

OHHHHHH THATS WHERE 2.1 COMES FROM!!!!
All i was doing wrong was missing out the 3R! Kept using 2R -_______-

I hope I dont get a silly confusion like this in the paper :'(

Anyways THANKYOUSOMUCH!!! I feel like im under so much oressure and every little bit helps!!! <3

2.1 is not coming ... can you show the calculation please?

 

[Jiyad Ahsan]

right ok umm
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf

Q9 (b), can someone explain the working please.. in words.. if possible
pretty pleaaaaaaaaaaaassseee ​

 

[seneria]

right ok umm​

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf​

​

Q9 (b), can someone explain the working please.. in words.. if possible​

pretty pleaaaaaaaaaaaassseee ​

use the formulae A=A. (exp)-lamda t
A is 3.5*10^5 -8%of 3.5*^105 and then determine t by applying log...

 

[seneria]

can someone solve question 4 C(!!)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_42.pdf

 

[seneria]

and Q1 c part 111 and Q 8 part C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_41.pdf