A2 Physics | Post your doubts here

[user]

Nah, I don't have any reason for that one. I just know that if anything like that comes up in the test, I will answer that the half-life is too large

Do u have any idea abt this question:



answer is : http://www.xtremepapers.com/communi...st-your-doubts-here.13317/page-63#post-236268

 

[angelicsuccubus]

But the internal energy will increase right?

only the potential energy of the molecules will increase.. not k.e

 

[Mobeen]

Do u have any idea abt this question:

View attachment 9415

answer is : http://www.xtremepapers.com/communi...st-your-doubts-here.13317/page-63#post-236268

according to the definition E would be opposite to the direction of current hence the direction of E. in the first interval I is zero so E zero . in the second interval I is changing so E induced but I is reducing because gradient is decreasing so E would also decrease hence the curve for E with reducing gradient.. in the next interval I is constant so no E induced . in the last interval I has a 'constant' change i.e the gradient is constant so the E would be constant but not zero ..
basically E is the gradient of the first graph but in opposite direction .

 

[smzimran]

Do u have any idea abt this question:

View attachment 9415

answer is : http://www.xtremepapers.com/communi...st-your-doubts-here.13317/page-63#post-236268

For the first part, there is an increasing current in the primary coil that means flux is being cut , but the rate is not constant because rate of increase of current in primary coil is not constant!
Also, the e.m.f is not in phase with current of primary coil so the emf has to be negative
The emf will increase from a negative value (not uniform increase so not a straight line) to zero
Zero because current eventually becomes constant that means there is no change in flux!


For the second part, as the current is constant that means there is no change in flux linkage and hence no emf is induced!


For the third part, the current decreases uniformaly (uniform because it is a striaight line with constant gradient)
So there is a constant rate of change of flux linkage and therefore constant emf is induced!
+ve value of emf because current is decreasing in primary coil!

I hope you understood!

 

[user]

For the first part, there is an increasing current in the primary coil that means flux is being cut , but the rate is not constant because rate of increase of current in primary coil is not constant!
Also, the e.m.f is not in phase with current of primary coil so the emf has to be negative
The emf will increase from a negative value (not uniform increase so not a straight line) to zero
Zero because current eventually becomes constant that means there is no change in flux!


For the second part, as the current is constant that means there is no change in flux linkage and hence no emf is induced!


For the third part, the current decreases uniformaly (uniform because it is a striaight line with constant gradient)
So there is a constant rate of change of flux linkage and therefore constant emf is induced!
+ve value of emf because current is decreasing in primary coil!

I hope you understood!

my problem is only in the second interval..
graph is actually like this:

​

but my answer:​

my answer..the solid line..

why so, is what i'm confused abt..​

 

[angelicsuccubus]

considering I'm only going to have enough time to look at 5 papers or so.. which years/seasons should I look at?

 

[alphabravocharlie]

my problem is only in the second interval..
graph is actually like this:

​

but my answer:​

my answer..the solid line..​

​

why so, is what i'm confused abt..​

Dude its a law, negative gradient of a sinusoidal graph WILL NEVER give you a graph like the! Negative gradient of a sinusoidal graph is a sinusoidal graph!

 

[smzimran]

my problem is only in the second interval..
graph is actually like this:

​

but my answer:​

my answer..the solid line..​

​

why so, is what i'm confused abt..​

The current in the question is changing with decreasing gradient (See the gradient at the start of the line and then see it at the end!)
That means flux is changing with decreasing rate
Whereas
your graph suggests that the flux is changing with increasing rate and suddenly drops to zero, which is wrong

 

[user]

Dude its a law, negative gradient of a sinusoidal graph WILL NEVER give you a graph like the! Negative gradient of a sinusoidal graph is a sinusoidal graph!

The current in the question is changing with decreasing gradient (See the gradient at the start of the line and then see it at the end!)
That means flux is changing with decreasing rate
Whereas
your graph suggests that the flux is changing with increasing rate and suddenly drops to zero, which is wrong

contradicting explanations

the way i drew..was actually cuz i thought what alpha said..

 

[histephenson007]

The current in the question is changing with decreasing gradient (See the gradient at the start of the line and then see it at the end!)
That means flux is changing with decreasing rate
Whereas
your graph suggests that the flux is changing with increasing rate and suddenly drops to zero, which is wrong

Complete

 

[smzimran]

contradicting explanations

the way i drew..was actually cuz i thought what alpha said..

My explanation is correct, trust me...

 

[smzimran]

QP: http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_43.pdf
MS: http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_43.pdf

Q1(c)(i)

After doing (b) if we use the value stored in our calculator of the speed and use the exact value to calculate the temperature, the temperature comes out to be 2022 K
This, when rounded off, should become 2020 K

However the mark scheme uses the rounded off value from (b) and its answer comes out to be 2030 K !!!
Does this mean we should not use stored values for next parts of a question ??

 

[angelicsuccubus]

I hate magnetic fields and forces....

 

[angelicsuccubus]

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_43.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_43.pdf

Q1(c)(i)

After doing (b) if we use the value stored in our calculator of the speed and use the exact value to calculate the temperature, the temperature comes out to be 2022 K
This, when rounded off, should become 2020 K

However the mark scheme uses the rounded off value from (b) and its answer comes out to be 2030 K !!!
Does this mean we should not use stored values for next parts of a question ??

I've noticed this in all cie papers.. if the question is in parts, always use the value from your last answer, so the rounded off version unless it's like angles in mechanics or something

 

[histephenson007]

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_43.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_43.pdf

Q1(c)(i)

After doing (b) if we use the value stored in our calculator of the speed and use the exact value to calculate the temperature, the temperature comes out to be 2022 K
This, when rounded off, should become 2020 K

However the mark scheme uses the rounded off value from (b) and its answer comes out to be 2030 K !!!
Does this mean we should not use stored values for next parts of a question ??

I had the exact same problem. But, I suppose they would allow for small mistakes like that.

 

[smzimran]

I hate magnetic fields and forces....

I love them!

I've noticed this in all cie papers.. if the question is in parts, always use the value from your last answer, so the rounded off version unless it's like angles in mechanics or something

 

[smzimran]

I had the exact same problem. But, I suppose they would allow for small mistakes like that.

An error of 10 K is not a small mistake.

 

[user]

I've noticed this in all cie papers.. if the question is in parts, always use the value from your last answer, so the rounded off version unless it's like angles in mechanics or something

but this thing doesnt apply in maths atleast...and even in physics there are questions in which they use the exact values :/

 

[smzimran]

but this thing doesnt apply in maths atleast...and even in physics there are questions in which they use the exact values :/

Exactly!
I have seen this thing first time, elsewhere I have seen that it is written in brackets that allow the bla bla bla answer if candidate uses stored value!

 

[SkyPilotage]

The current in the question is changing with decreasing gradient (See the gradient at the start of the line and then see it at the end!)
That means flux is changing with decreasing rate
Whereas
your graph suggests that the flux is changing with increasing rate and suddenly drops to zero, which is wrong

I know the answer drawn is correct but
Let me comment on something,
1-How do you know if the rate of decrease is decreasing or constant or increasing? -Please note that emf is the Numerical value of the rate of change of flux, so in both graphs the rate of change of flux is decreasing. But what I dont get is how to know whether the rate of decrease is decreasing or constant or increasing?
2-When you have alternating current , why isnt the graph of the induced emf same like the one drawn in this question?

I think there are 2 mistakes in that graph,
-First thing, the emf induced in the last part should not be greater than the first part.
-Second thing, at the very beginning the gradient is 99% vertical, so I think the emf induced at that point is almost at infinity, hence a line should not be connecting the x-axis to the graph.
[ This is one of the reasons I believe the second graph is wrong ]
Another reason is that the gradient of the current tends to become zero slightly before the end mark, so the emf graph should touch the x-axis.