AS Physics P1 MCQs Preparation Thread.

[kinglynx]

https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/282891_10150999944406535_846620522_n.jpg

third time im asking guys ,come on pleaseee!!!

 

[Oliveme]

questions 10, 15, 23, 26, 37 and 38. thank you very much.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

 

[kinglynx]

why is the answer C ? that makes no sense at all!!!!

 

[Gémeaux]

https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/282891_10150999944406535_846620522_n.jpg

third time im asking guys ,come on pleaseee!!!

The answer is A, right?
What i understand of this is that because pure metals are more ductile than alloys, aluminium is the answer. Pure aluminium has sheets of molecules that can easily slide over each other. Steel has in it atoms between the layers making it difficult for them to slide over.

 

[ChrisRedfield]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf
I have problems with 13,14 and 29 !

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf
questions 4,8,9,12,14,27
Can''t find the solution,help would be appreciated!
I'm horrible when it comes to P1

 

[Gémeaux]

questions 10, 15, 23, 26, 37 and 38. thank you very much.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

10) I didn't get that either. tell me too when you understand this one... n also 26th and 35th
15)The initial K.E is 4J in each case. for the force F, the change in K.E is 8-4= 4J
also, K.E = work done
so, 4J = (F x s)
Then, for the next part, with 2F and 2s,
the w.d, i.e the K.E = 2F x 2s = 4Fs which means four times (F x s)
i.e, 4 x 4 J = 16 J
now this is the *change* in K.E., therefore the new K.E would be 4 + 16 = 20J

 

[Gémeaux]

questions 23, 26, 37 and 38. thank you very much.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

23) In a force-extension graph the E.P.E is the area under that graph. For this question select the option that has a maximum area under the graph, which is clearly B.

37) the contact X would be adjusted around the resistor with 4 kΩ, so the max. n min. limits of p.d. across PQ wud be when its complete resistance is considered and when none is considered. at R = 4kΩ, the resistance across it is 20V
[ 4 /(4+1)] x 25 = 20V
when there is no 4kΩ resistor, p.d. across PQ becomes zero. therefore the answr's B.

38) Consider the voltmeter as a resistor with 200kΩ resistance, connected in parallel to the other resistor of 200kΩ.
with this the effective resistance of these two becomes 100kΩ.
the p.d. across resistors in parallel is equal therefore, by unitary method calculate V across voltmeter.
(100/500) x 60 = 12V

 

[Abdullah syed]

Need some help
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s03_qp_1.pdf #30
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s03_qp_1.pdf #28
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s03_qp_1.pdf

#18,14,12,5,3​

 

[Gémeaux]

Kindly help me with these two questions.
Q 13 and Q 36
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
the answers are A for 13th, and B for 36th.
Thankyou.

 

[XPFMember]

Kindly help me with these two questions.
Q 13 and Q 36
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
the answers are A for 13th, and B for 36th.
Thankyou.

AsSalamoAlaikum Wr Wb!

13. They said it's in equilibrium, so the vector triangle should be a closed one...that's only the case in A... this might not be the right reason...if you come across a better explanation, plz let me know too [tag me]

36. I too need an explanation...but one thing i noticed, whenever such a quesstion comes..the answer is always 'due to the pd across internal resistance..'

 

[XPFMember]

AsSalamoAlaikum Wr Wb

need help with these questions, plz. Answers have been marked.

JazakAllahu khairen

 

[Peter Check]

Guys for finding out the intensity, is the time period given here of any importance??

 

[Peter Check]

This is the third time Im posting this, please anyone who is good in moments show me hot to to this!! plssssss

 

[XPFMember]

View attachment 12096Guys for finding out the intensity, is the time period given here of any importance??

yup, it is.

cuz intensity is proportional to square of amplitude AND square of frequency. Frequency is 1/time period.

 

[Peter Check]

How is the answer D? If the strain increases wont the length decrease(strain=x/L)??? And hence the resistance decrease??

 

[Peter Check]

yup, it is.

cuz intensity is proportional to square of amplitude AND square of frequency. Frequency is 1/time period.

okay thnx, can u take a look at my next post? about the tensile strain..

 

[USMAN Sheikh]

mates i am posting dis another time plssssssssssssssssssss help me on these can anyone pls help me out on these question plssss 6 24 30 39 pls

 

[leosco1995]

okay thnx, can u take a look at my next post? about the tensile strain..

This one is a fact. As you stretch a wire, its length increases and the cross-sectional area decreases. Since R = ρL/A, and ρ is a constant, the resistance increases.

 

[leosco1995]

because full diameter is only used when we take couple/torque thingy here as the lower side of belt is slag... so only upwards force acts on one side.... creating moment so maybe thats why we use half diameter = the radius

I see, thanks.

 

[shan5674]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf
can someone explain why 40 is C? how do we know?