Physics: Post your doubts here!

[Zishi]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w07_qp_1.pdf

Can anyone please explain Q 11 of this paper? Does the answer have anything to do with centripetal force?

No, it isn't related to Centripetal force - you're not expected to know about it in an AS Level paper. B should be the answer, because that is the resultant of normal force and friction force on the tyre. Force of friction is in the same direction as car's motion because the tyre must be rotating in a clockwise direction.

 

[JD REBORN]

In Q5 why cant B and D be correct?
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf

 

[CaptainDanger]

9702_w11_qp_11.pdf
9702_w11_ms_11.pdf

Q 4, 26 & 32?

 

[JD REBORN]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf
Q11
For Q23 answer is B cuz speed of light in vacuum is constant.Is my reasoning correct or is their any other reason?

 

[JD REBORN]

9702_w11_qp_11.pdf
9702_w11_ms_11.pdf

Q 4, 26 & 32?

Q4
For L units are m,for time T s and radius a m.sin theta has no units.Now u put the units ignoring 3:
m(m^2/P)=Qs^2
Now if P has units m^2 and Q has ms^-2 then on both sides there will be m so thats why B is correct.
Q26
The length of cube is l,its area of crosssection is l^2 and force is F.Since young modulus is not given we assume it to be E.Now
E=Fl/delta l*l^2
delta lE=Fl/l^2
delta lE=F/l.Now this is the same as B option
Q32
The force acting on +Q is F=+QE
The force acting on -Q is F=-QE
Now since these forces are equal in magnitude and one has a negative sign so resultant force will be zero
The electric field is orginating from positive charges.So to make it easier assume that each arrow has a negative charge.Now +Q will be attracted to this charge and and -Q will be repelled so left side of rod will move upward(anticlockwise moments)

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
Q11
For Q23 answer is B cuz speed of light in vacuum is constant.Is my reasoning correct or is their any other reason?

Yes,
Because speed of light (red or green) both are same and the wavelenghts different!
Using the formula:
c = f * lamda
We can imply that since wavelenghts are different, frequency must also be different !

 

[JD REBORN]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf
Q11

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
Q11

R1 = R2 because think about it, forget the strings and all, what exactly is R balancing ?
It is balancing the WEIGHT in both cases!
Weight is the same so R is the same!

Now the second thing which is T,
A shorter length of string means that the same weight is to be distributed over a shorter length so more force per unit length is needed in diagram 1
That is why T1 > T2

 

[JD REBORN]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Q5 a(ii,iii)

 

[musa Khan]

Can someone please solve Q8 part B and C? It would be great if you can draw the circuit out and post the image for C! I arrived at the answers for part b but I don't think my reasoning is quite right though..
Thank you..

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_4.pdf

 

[Hussnain]

AS PHYSICS
M/J/2010
Q3(c), Q6(ii)

 

[darknessinme]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_22.pdf
Q5 a(ii,iii)

ii) The trick in this question is that wire BF needs to be split into two separate sections because the current in it changes. The resistance is also R2/2 in each section.
Kirchoff''s second basically law says Sum of EMFs=Sum of PDS
E1=PD across r1 + PD across R1 + PD across (B to J)+ PD across (J to F)
E1=I1 x r1 + I1 x R1 + I2 x R2/2 + I1 x R2/2

iii) Applying same law as before to a different loop.
E1-E2=-(PD across r2) + PD across (J to F) + PD across R1 + PD across r1
E1–E2 =–І3 x r2 + І1 x R1 + І1 x r1 + І1 x R2 / 2

 

[JD REBORN]

ii) The trick in this question is that wire BF needs to be split into two separate sections because the current in it changes. The resistance is also R2/2 in each section.
Kirchoff''s second basically law says Sum of EMFs=Sum of PDS
E1=PD across r1 + PD across R1 + PD across (B to J)+ PD across (J to F)
E1=I1 x r1 + I1 x R1 + I2 x R2/2 + I1 x R2/2

iii) Applying same law as before to a different loop.
E1-E2=-(PD across r2) + PD across (J to F) + PD across R1 + PD across r1
E1–E2 =–І3 x r2 + І1 x R1 + І1 x r1 + І1 x R2 / 2

What i dont understand is that Pd across J to F why do we take current I1 shouldnt it be I2?

 

[JD REBORN]

AS PHYSICS
M/J/2010
Q3(c), Q6(ii)

be more specific like paper and question part

 

[XPFMember]

Assalamoalaikum wr wb!!

PAPER 4 doubts...

June 2002

· Q:4 b & c​

· Q:6a how should we draw that...? :s​

Nov 2002

· Q:3 b(ii)​

Nov 2003

· Q:5 b (ii) need explanation, I don’t get it... :s​

· Q:4 c Can you show, how is the less ripple represented, please.. :s​

· Q:2 b (ii) –ve sign?​

June 2004

· Q:4 c reason for the answer...​

· Q:6 I’m totally blank abt this...i don’t get how they worked this out..​

· Q:8 iii who said we didn’t use them? :s​

​

June 2005

· Q:5b​

· Q:6b​

Nov 2005

· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?​

· Q:4 c Why speed max when displacement is zero? :s​

· Q:5b(ii) why when speed is reduced, deflection is larger? :s​

· Q:6 b I don’t understand​

·Q:6 c (ii) which way do we do...negative of the upper graph gradient? ​

June 2006

· Q:2 b (ii) how to do?? :s​

· Q:4 c can you show, plz?​

· Q:6 a what I understand is the field should be circle..right? I don’t get this :s​

· Q:6 c what do I write for this?​

· Q:7 Can you answer the complete question, I couldn’t do this one... ​

· Q:8 a For such questions where exactly are we supposed to draw the arrow? Like beside the paper shown..or inside that region? Where do we show....?​

​

Nov 2006

· Q:3 c what to write? :s​

· Q:4 b ans. Is 5.99 x 10^24 how does ms say 6.00 :s​

· Q:5 a (ii) what’s eddy current?​

· Q:6 a (ii) How do the diode works then, normally...isn’t that the way we usually connect? :s​

​

June 2007

· Q:3 d (i) Why is the answer zero? Just to confirm, is it because E=F/Q and F = ma so E is proportional to a...and hence a will be max. when E is max? is it?​

· Q:4 c (ii) How’ll the diagram be? :s​

· Q:7 b I don’t get why the damping is reduced...i took it as, resistance is increased, more power dissioated, hence more energy losses...so oscillations die off quicker..but it say damping is less..:s confused..​

· Q:7 c (iii) can you give some examples? :s​

​

I'm sorry for so many doubts, but please can you answer them by few at a time..​

​

P.S. A2 doubts to be shifted here..So post the answers in that thread..​

Jazak Allahu Khairen..​

 

[darknessinme]

What i dont understand is that Pd across J to F why do we take current I1 shouldnt it be I2?

The current flowing from J to F has to be the same as the current flowing from J all the way to B. That's because there are no junctions between J to B for the current to split up. If I2 were flowing from J to F, then we'd be saying I2=I1 which is wrong as we know from part i).

What you have to do is apply Kirchoff's first law to the junction at J where the ammeter is connected to the resistance wire. This will once again end up being I2=I1+I3, with I1 flowing from J to F.

 

[Hussnain]

be more specific like paper and question part

PAPER 2

 

[JD REBORN]

PAPER 2

22 23 or 21?

 

[unique840]

Assalamoalaikum wr wb!!

PAPER 4 doubts...

June 2002

· Q:4 b & c​

· Q:6a how should we draw that...? :s​

Nov 2002

· Q:3 b(ii)​

Nov 2003

· Q:5 b (ii) need explanation, I don’t get it... :s​

· Q:4 c Can you show, how is the less ripple represented, please.. :s​

· Q:2 b (ii) –ve sign?​

June 2004

· Q:4 c reason for the answer...​

· Q:6 I’m totally blank abt this...i don’t get how they worked this out..​

· Q:8 iii who said we didn’t use them? :s​

​

June 2005

· Q:5b​

· Q:6b​

Nov 2005

· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?​

· Q:4 c Why speed max when displacement is zero? :s​

· Q:5b(ii) why when speed is reduced, deflection is larger? :s​

· Q:6 b I don’t understand​

·Q:6 c (ii) which way do we do...negative of the upper graph gradient? ​

June 2006

· Q:2 b (ii) how to do?? :s​

· Q:4 c can you show, plz?​

· Q:6 a what I understand is the field should be circle..right? I don’t get this :s​

· Q:6 c what do I write for this?​

· Q:7 Can you answer the complete question, I couldn’t do this one... ​

· Q:8 a For such questions where exactly are we supposed to draw the arrow? Like beside the paper shown..or inside that region? Where do we show....?​

​

Nov 2006

· Q:3 c what to write? :s​

· Q:4 b ans. Is 5.99 x 10^24 how does ms say 6.00 :s​

· Q:5 a (ii) what’s eddy current?​

· Q:6 a (ii) How do the diode works then, normally...isn’t that the way we usually connect? :s​

​

June 2007

· Q:3 d (i) Why is the answer zero? Just to confirm, is it because E=F/Q and F = ma so E is proportional to a...and hence a will be max. when E is max? is it?​

· Q:4 c (ii) How’ll the diagram be? :s​

· Q:7 b I don’t get why the damping is reduced...i took it as, resistance is increased, more power dissioated, hence more energy losses...so oscillations die off quicker..but it say damping is less..:s confused..​

· Q:7 c (iii) can you give some examples? :s​

​

I'm sorry for so many doubts, but please can you answer them by few at a time..​

​

P.S. A2 doubts to be shifted here..So post the answers in that thread..​

Jazak Allahu Khairen..​

nov 2002 q3bii) the freq is 1/2fo so time will be 2T. means after twice of time period the oscillation will be forced. so every alternate oscillation is energized. and every pulse increases the amplitude. it is a resonance like process

didnt i give u the ans before when u posted the same ques?

 

[XPFMember]

nov 2002 q3bii) the freq is 1/2fo so time will be 2T. means after twice of time period the oscillation will be forced. so every alternate oscillation is energized. and every pulse increases the amplitude. it is a resonance like process

didnt i give u the ans before when u posted the same ques?

sorry....my mistake..i updated the list ..

P.S. A2 doubts to be shifted here....