Physics: Post your doubts here!

[abdul moeed rana]

https://www.xtremepapers.com/community/members/syed1995.17088/.
in our second mock our chm teacher gave us the specimen paper of year 14.since syllabi has not changed then how come thers a specimen paper.there should be of physics as well.i asked him keh sir kia specimen paper saaare subjects ke aate ha each year n he said yes.but he is not giving us the papers.dnt know why!so if u or any one of ur friends has access to the teachers website jo cie ke teachers ke official website ha ,us se password le k yaha attachment k toar par post kar do.there is a specimen paper n it comes every year,thats for sure

 

[Faaiz Haque]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf ' Q6(d)(ii)'
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf 'Q6(a)(i)'

Help required in these two please

6 d ii
Since increasing the voltage increases the current (V = IR)
Resistance also increases as voltage increases, in this case current is increasing non linearly , so resistance increases as well.
I'm not actually sure why current increases non linearly but when that does happen , resistance increases.


6 a i

R = V/I at any point of the graph (basically the reciprocal of its gradient)
Since gradient increases, it's reciprocal obviously decreases. Meaning resistance decreases with increase in voltage.
I'm not 100% , these were hard. I tried, sorry.

 

[meerul264]

Help me with 9702 Physics June 2013 Paper 23 (9702/23/M/J/13).

Q 6 (c)(ii) please, and thanks!
http://www.examtestprep.com/CIE/A-A...013.html#pdfsection_354d1e97-page_12-locus_32

 

[meerul264]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

MCQ - Question 11 , 17 Please

With explanation

Thanks

Assalamualaikum.

Q 11.
When doing questions regarding momentum you have to bear in mind that momentum is a vector quantity.
Whenever vector quantity is involved, you have to fix a positive direction. In this case I'd let the right direction be positive.

So velocity of the lorry would be 20.0 ms-1, while the velocity of the car would be -30.0 ms-1 (note that the velocity of car is negative since its in the opposite direction).

Total momentum = momentum of lorry + momentum of car

Q 17.

Whenever a question compares one quantity to another, like "velocity of A is twice that of B", its wise to do the working which I'm going to show you below.
In the case of this question, it compares K.E, so:

K.E = 1/2 m v^2

Working:

K.E of X = 1/2 (2m) (0.5v)^2
= (1/4) m v^2
K.E of Y = 1/2 m v^2

ratio of K.E of X to K.E of Y = (K.E of X) / (K.E of Y) = 1/2

This implies,

2(K.E of X) = (K.E of Y)

K.E of Y is twice that of K.E of X, or
K.E of X is half that K.E of Y.

Hence K.E of X is half the K.E of Y.

 

[syed1995]

https://www.xtremepapers.com/community/members/syed1995.17088/.
in our second mock our chm teacher gave us the specimen paper of year 14.since syllabi has not changed then how come thers a specimen paper.there should be of physics as well.i asked him keh sir kia specimen paper saaare subjects ke aate ha each year n he said yes.but he is not giving us the papers.dnt know why!so if u or any one of ur friends has access to the teachers website jo cie ke teachers ke official website ha ,us se password le k yaha attachment k toar par post kar do.there is a specimen paper n it comes every year,thats for sure

Ask her about it. I am pretty sure there were no specimen papers for Phy and Chem in 2014. Can you post one of the questions from the specimen paper of Chemistry?

The only specimen paper of Chemistry has been in 2007.. http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_y07_sp_4.pdf

Because our teacher told us that Specimen papers come out as a sample of the new pattern of paper.. So I am pretty sure there are no specimen papers of Phy in 2014.

 

[midha.ch]

Can anyone help me with Physics applications? I need good notes especially for amplifier
I have a copy of the CIE provided booklet. But I need a much "easy to understand" explanation.
And also of Medical Physics

 

[midha.ch]

Help me with 9702 Physics June 2013 Paper 23 (9702/23/M/J/13).

Q 6 (c)(ii) please, and thanks!
http://www.examtestprep.com/CIE/A-A...013.html#pdfsection_354d1e97-page_12-locus_32

For minimum pd across R2 take maximum resistance of the variable resistor. In series circuit, V(total) = V1 + V2
and V = IR
so if you increase the pd across R1 by increasing it's resistance, you'll get minimum value for pd across R2 to keep V(total) constant.
Your current can be found by I = [V(total)]/(maxR1 + R2)
then simply find the min pd across R2 by using the formula
pd across R2 = I x R2

 

[Snackbox86]

Hey can i get help with MCQ questions from 2013 may paper 11
questions
5 and 8

 

[midha.ch]

Hey can i get help with MCQ questions from 2013 may paper 11
questions
5 and 8

For Question 8 equate the distance travelled by the two trains considering the time taken to overtake = t

for Question 5
it's given,
(ΔT/T) x 100 = 1%

ΔT = (1/100) x T

ΔT/T = 0.01

Using the same concept,

Δl/l = 0.04

Δg/g = (2ΔT/T) + (Δl/l)

{ Δg/g = 0.06 } x 100

(Δg/g) x 100 = 6%

 

[meerul264]

For minimum pd across R2 take maximum resistance of the variable resistor. In series circuit, V(total) = V1 + V2
and V = IR
so if you increase the pd across R1 by increasing it's resistance, you'll get minimum value for pd across R2 to keep V(total) constant.
Your current can be found by I = [V(total)]/(maxR1 + R2)
then simply find the min pd across R2 by using the formula
pd across R2 = I x R2

Sorry, but I think you're answering for question 6 (b) (ii) instead of 6 (c) (ii) :S
I need help with 6 (c) (ii).

http://www.examtestprep.com/CIE/A-A...013.html#pdfsection_354d1e97-page_15-locus_26

 

[midha.ch]

Sorry, but I think you're answering for question 6 (b) (ii) instead of 6 (c) (ii) :S
I need help with 6 (c) (ii).

http://www.examtestprep.com/CIE/A-A...013.html#pdfsection_354d1e97-page_15-locus_26

Oh well my mistake.
In circuit 1 resistance for R2 was 600 ohm
In circuit 2 resistance Rp (R2 + LDR) is less than that in both high and low intensity of light.
since resistance of that component circuit 2 (considering R2 and LDR as a combined component in the circuit) is less than that of in circuit 1 the pd across R2 will also be less in circuit 2. (In parallel combination the pd remains same)

This is just theory and although the question says you don't need to show any numerical values, a bit of calculation actually will give you a better idea about this!
(this is only for understanding!!)

For circuit 1, minimum pd across R2 was 12V

For circuit 2
In maximum intensity of light
Rp = 400ohm
pd across R2 = (Rp/Rtotal) x V
= 10V
In minimu intensity of light
Rp = 535 ohm (approximately)
pd across R2 = 11.5V

Hope this cleared your doubt (though it's bit lengthy but it'll give you a clear concept )

 

[meerul264]

Assalamualaikum.

I have some doubt.

(9702/22/O/N/09)
Q 4 (c)(ii)(3)
Q 5 (a) and (b)

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_22.pdf

Please explain to me the concepts behind these questions. I don't mind lengthy ones

 

[Faaiz Haque]

Assalamualaikum.

Q 11.
When doing questions regarding momentum you have to bear in mind that momentum is a vector quantity.
Whenever vector quantity is involved, you have to fix a positive direction. In this case I'd let the right direction be positive.

So velocity of the lorry would be 20.0 ms-1, while the velocity of the car would be -30.0 ms-1 (note that the velocity of car is negative since its in the opposite direction).

Total momentum = momentum of lorry + momentum of car

Q 17.

Whenever a question compares one quantity to another, like "velocity of A is twice that of B", its wise to do the working which I'm going to show you below.
In the case of this question, it compares K.E, so:

K.E = 1/2 m v^2

Working:

K.E of X = 1/2 (2m) (0.5v)^2
= (1/4) m v^2
K.E of Y = 1/2 m v^2

ratio of K.E of X to K.E of Y = (K.E of X) / (K.E of Y) = 1/2

This implies,

2(K.E of X) = (K.E of Y)

K.E of Y is twice that of K.E of X, or
K.E of X is half that K.E of Y.

Hence K.E of X is half the K.E of Y.

Thanks!!!

 

[usama321]

Assalamualaikum.

I have some doubt.

(9702/22/O/N/09)
Q 4 (c)(ii)(3)
Q 5 (a) and (b)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf

Please explain to me the concepts behind these questions. I don't mind lengthy ones

For the first question, work done always equals some of all the energy changes of the system. There was a decrease in GP and increase in EP. Thus WD equals their sum.

For the second question, both points are at the same distance from nodes. Considering that this is a stationary wave, the particles will only move up and down in the same place. I don't think there is any question of different frequencies as none of the two points are nodes.As the points are at similar distances from the nodes, there amplitude should be the same.

About the second part, I solved it wrong myself . I think it is 180 degrees because its a standing wave and not a progressive one. X is at its highest possible point, and Y at its lowest. Therefore there is a phase difference of 180.

 

[midha.ch]

For the first question, work done always equals some of all the energy changes of the system. There was a decrease in GP and increase in EP. Thus WD equals their sum.

For the second question, both points are at the same distance from nodes. Considering that this is a stationary wave, the particles will only move up and down in the same place. I don't think there is any question of different frequencies as none of the two points are nodes.As the points are at similar distances from the nodes, there amplitude should be the same.

About the second part, I solved it wrong myself . I think it is 180 degrees because its a standing wave and not a progressive one. X is at its highest possible point, and Y at its lowest. Therefore there is a phase difference of 180.

can you explain the answer for 5a more clearly?? I didn't understand it

 

[usama321]

can you explain the answer for 5a more clearly?? I didn't understand it

http://www.physicsclassroom.com/Class/waves/h4.gif
See this maybe you will understand it better.

Both points will move the same distance up and down, as they are the same distance from nodes.

 

[Hassan Ali Abid]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf Q#8 (C)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_ms_41.pdf

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_ms_4.pdf

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_ms_4.pdf

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_4.pdf Q#6 (C)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_ms_4.pdf

I need help in these parts. Can anyone plz explain me in detail how to solve these parts ?

 

[midha.ch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_41.pdf Q#8 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_41.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_4.pdf Q#6 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_4.pdf

I need help in these parts. Can anyone plz explain me in detail how to solve these parts ?

for the 1st question, lets say initially we had N1 number of Phosphorus nuclei
after 30 days, due to radioactive decay the number of P nuclei has decreased to N2
it's given in question that P nuclei forms Sulfur when it undergoes radioactive decay
that means the decrase in the number of P nuclei after 30 days is actually the number of S nuclei formed in 30 days
the difference in number of P nuclei after 30 days = N1-N2 = number of S nuclei formed after 30 days
thus we can say the ratio is {N1/(N1-N2)}

2nd question
Uranium splits to produce equal number of both Ba and Kr nuclie
so initially Number of Kr nuclei = Number of Ba nuclei = N1
Let after time t, N of Ba/N of Kr =8

N of Ba = N1 x e^(-λ1 x t)

λ2 = ln2/T1/2 of Ba

λ1 = 6.41x10-4

N of Kr = N1 x e^(-λ2 x t)

λ2 = ln2/T1/2 of Kr

λ2 = 0.231

N of Ba/N of Kr = 8

N of Ba = 8xNKr

Now put the values on both sides and equate them

N1 gets cancelled as it’s there on both sides

Remains an equation of the form e^a = e^b

Take ln both sides

ln(e^y) = y

using this formula get the value of t by solving the equation

 

[midha.ch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_41.pdf Q#8 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_41.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_4.pdf Q#6 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_4.pdf

I need help in these parts. Can anyone plz explain me in detail how to solve these parts ?

your 3rd question is similar to 1st question. Here instead of the ratio, you are asked to find the time taken to reach that ratio
If you have understood my answer to first question it's easy to find the result for third question. But if you still have confuson then let me know.

I'll skip numerical calculations for your 4th question
Consider Initial activity is A1
after time t, Activity reduces to (A1 x 1/10)
put the values in the equation of activity
you have already found decay constant in previous parts
now just put the values and solve it.

If you are still unsure or confused about any of the answers, then let me know I'll be glad to help

 

[papajohn]

Question 7 part b...Please help usama321 and others.
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf