Physics paper 2 Discuss and Share

[Jinkglex]

Since its Physics paper 22 tomorrow, i thought we could all share our queries and notes in this thread. Feel free to post notes and ask questions!

 

[MaxStudentALevel]

Hey! My sister is giving P2 tommorow and she's fully prepped just going through final revision! She has a problem with
ON 2010/ 22 with Q6 Part b could you give her an explanation of the formula used in the marking scheme?

 

[Jinkglex]

Thats strange...the only way the marking scheme makes sense is if we take this into really basic terms, meaning

if voltage drop across 1200 is 5
voltage drop across x is 4

1200/5 = 240/V

4x240= 960.

But Im pretty sure that cant be right...its just breaking down the whole thing into a really basic form...

 

[MaxStudentALevel]

Hmmm. How would you do it your way? Like solving it just not the MS way?

Thats strange...the only way the marking scheme makes sense is if we take this into really basic terms, meaning

if voltage drop across 1200 is 5
voltage drop across x is 4

1200/5 = 240/V

4x240= 960.

But Im pretty sure that cant be right...its just breaking down the whole thing into a really basic form...

Hmmm. How would you do it your way? Like solving it just not the MS way?

 

[Gimmick]

Hmmm. How would you do it your way? Like solving it just not the MS way?

Hmmm. How would you do it your way? Like solving it just not the MS way?

The way I do it is:

The e.m.f of the cell is = 9V
There are 2 total resistors = 1,200 ohms and x ohms
x ohms is split up into 2 resistors in parallel: 1,600 and the thermistor value Rt.
therefore 1 / ( (1/1600) + (1 / Rt) ) = x
the voltmeter reading is 4 volts
voltage is the same in a parallel circuit
therefore 1200 ohms resistor "drops" 5 volts
9 - 5 volts = remaining volts for the resistors.
= 4 volts for x

circuit current is I = V / R
potential drop acros 1200 ohm is 5
therefore I = 5 / 1200
I = 4.166666[..]7e-3


therefore

9 = (1200 * [4.1666666...7e-3]) + ([4.166666...7e-3] * R)
4 = [4.166666...7e-3] * R
4 / [4.166666...7e-3] = R = 960

(funnily enough I just finished this paper a half-hour back )

I've uploaded a picture too...forgive my handwriting and the poor quality, heh.


Also, how do you do 4) b) i) and 4) b) ii)? The MS just says "correct area", which I don't know.

 

[Ariel Robert]

Hmmm. How would you do it your way? Like solving it just not the MS way?

Hmmm. How would you do it your way? Like solving it just not the MS way?

I would say find the current in the main circuit as the resistance and voltage is given, then use that current in the parallel area to find the resistance...

 

[axetreme.O]

No one..? ^-^

 

[Gimmick]

No one..? ^-^

Counting squares? That's all I could do.

 

[Eng Minhal]

Tha paper was easy right.

 

[Beaconite007]

Too easy to be true D:

 

[aslam.pervez]

I have my exams coming in the following session. What kind of questions came in today's phy p2?

 

[pakiboy]

it was halwa, and i ate it !

 

[aslam.pervez]

damn, nice. seemed as though it was pretty easy. i should've sat for it

 

[saadgujjar]

it was halwa, and i ate it !

except q2

 

[pakiboy]

except q2

the graph one?

 

[Aamer Faiyaz]

i screwed up on the superposition question

 

[saadgujjar]

the graph one?

yup

 

[saadgujjar]

what is expected GT???i think it will be 30_35

 

[pakiboy]

i

what is expected GT???i think it will be 30_35

think its going to be above 40!

 

[Aamer Faiyaz]

wayy too low probably 44ish since alot of people are saying it was easy :l