AS Physics P1 MCQs Preparation Thread.

Hassi123 · Jun 7, 2012

MAY/JUNE 2003: Q10 ans D and Q40 ans D
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_1.pdf
OCT/NOV 2003 : Q15 ans B Q16 ans C Q27 ans D

PLEASE DO NOT IGNORE IT, I SERIOUSLY NEED HELP OR IM SCREWED :/

1357913579 · Jun 7, 2012

i have a doubt in electricity
in this question in redspot they have got the answer 12volt as shown in the picture below but as we kno9w the voltage in series is divided so voltage acorss the first resistor plus voltage across the second resistor is equal to 12 whereas in the redspot they have mentioned 12 volts as the mximum voltage which is wrong right?
so we have do to take 1/2 multiplied by 12 which will be 6 so answer should be c whereas the asnwer in the redsopt is d am i right?
i didnt find this question in the past papers till the poibt i have done 2010.
thanks for helping

00tanveer · Jun 7, 2012

How do you do this people??

xyz! · Jun 7, 2012

00tanveer said:
View attachment 12276
How do you do this people??

is the answer B?
i guess its dne lyk dis:-
atm=P0
so inside the tube the pressure will be 10% lesser, dat is 90% of the atm
so pressure inside=9P0/10
pressure=rho*g*h
so h=P/(rho*g)
=9P0/10 divided by rho*g
=9P0/(10*rho*g)

leosco1995 · Jun 7, 2012

1357913579 said:
i have a doubt in electricity
in this question in redspot they have got the answer 12volt as shown in the picture below but as we kno9w the voltage in series is divided so voltage acorss the first resistor plus voltage across the second resistor is equal to 12 whereas in the redspot they have mentioned 12 volts as the mximum voltage which is wrong right?
so we have do to take 1/2 multiplied by 12 which will be 6 so answer should be c whereas the asnwer in the redsopt is d am i right?
i didnt find this question in the past papers till the poibt i have done 2010.
thanks for helping

I think you're right. The answer is C.

xyz! · Jun 7, 2012

00tanveer said:
View attachment 12276
How do you do this people??

how do u upload such screen shots btw?

leosco1995 · Jun 7, 2012

xyz! said:
how do u upload such screen shots btw?

There's an 'Upload a File' button next to the 'Post Reply' one at the bottom right corner of your posts. Just select an image of your MCQ and then post.

xyz! · Jun 7, 2012

leosco1995 said:
There's an 'Upload a File' button next to the 'Post Reply' one at the bottom right corner of your posts. Just select an image of your MCQ and then post.

oh ya....thnx! i dint c dat only!

Ahmed Ali Akbar · Jun 7, 2012

1357913579 said:
i have a doubt in electricity
in this question in redspot they have got the answer 12volt as shown in the picture below but as we kno9w the voltage in series is divided so voltage acorss the first resistor plus voltage across the second resistor is equal to 12 whereas in the redspot they have mentioned 12 volts as the mximum voltage which is wrong right?
so we have do to take 1/2 multiplied by 12 which will be 6 so answer should be c whereas the asnwer in the redsopt is d am i right?
i didnt find this question in the past papers till the poibt i have done 2010.
thanks for helping

mark scheme is correct..and this one is good question...when variable resistor would have zero resistance all current will flow through it not through resistor of 1.0k ohm PARALLEL TO IT, because current PREFER TO take shorter route and route in which there is no resistance....all current flow through other 1.0k ohm resistor...so all voltage will drop accross it..AND VOLTAGE WILL BE 12 ACCROSS IT...therefor max voltage is 12v...and when variable resistor have max value of 1.0k ohms,total voltage accross it and 1.0k ohm parallel resistor will be 4v...and ofcourse voltage accross other resistor would be 8vv....

HOPE IT WILL HELP.....

00tanveer · Jun 7, 2012

Use the Snipping Tool if you have Windows 7.

Great app right!

1357913579 · Jun 7, 2012

Ahmed Ali Akbar said:
mark scheme is correct..and this one is good question...when variable resistor would have zero resistance all current will flow through it not through resistor of 1.0k ohm PARALLEL TO IT, because current PREFER TO take shorter route and route in which there is no resistance....all current flow through other 1.0k ohm resistor...so all voltage will drop accross it..AND VOLTAGE WILL BE 12 ACCROSS IT...therefor max voltage is 12v...and when variable resistor have max value of 1.0k ohms,total voltage accross it and 1.0k ohm parallel resistor will be 4v...and ofcourse voltage accross other resistor would be 8vv....HOPE IT WILL HELP.....

ASSALAMOALIKUM
Thanls alot brother i never knew this concept that current flows from the path where there,s less resistance but it does makes sense as resistance opposes current so current would want to move from a place which opposes it less
thanks alot brother JAZAKALLAH KHAIR.

00tanveer · Jun 7, 2012

Answer's A pal....

Ahmed Ali Akbar · Jun 7, 2012

1357913579 said:
ASSALAMOALIKUM
Thanls alot brother i never knew this concept that current flows from the path where there,s less resistance but it does makes sense as resistance opposes current so current would want to move from a place which opposes it less
thanks alot brother JAZAKALLAH KHAIR.[/quotE
WALAIKUMOSLAM...AND YOUR WELCOMEE..

1357913579 · Jun 7, 2012

ASSALAMOALIKUM,
I have another doubt in the paper link youll find below number-33 in this like i can find the v1 pd between p and s but for p2 for pd between s and p iam confused like i did this way answer is correct but i want to confirm thaat is my tecnique corrct this is how i did v1=5/10 multiplied by 2 =1 for v2 resistor between pand s is connected first to 3 ohm resistor so from s to q only there will be 3 ohm resistance and as we know voltage in parallel combination is same so it means vlotage across both 2 ohm and 3 ohm would be 2 so i did 3/5 multiplied by 2 =1.2 so then 1-1.2 =0.2 so like is my concept of s to q correct?
thanks alot, jazakAllah khair

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

kinglynx · Jun 7, 2012

geek101 said:
kinglynx
here I is directly proportional to 1/d^2 and the graph which shows this is D. and it can not be C cuz when ^2 is zero there is gonna be zero intensity as well, but in C when d^2 is zero...intensity is max, thats wrong.

second pic: http://www.animations.physics.unsw.edu.au/jw/SHM.htm see this...

3rd: ok the strain energy = work done
take the first part as a triangle and the second as a trapezium...
so the area = 0.5 x 500 x 10 x 10^-3 = 2.5
and 0.5 (550 + 500) x 2 x 10^-3 = 1.05
so the wd = 2.5 + 1.05 = 3.55

Why didnt you assume the whole thing as a triangle?
and also why didnt you use 0.5 * k * x^2

geek101 · Jun 7, 2012

kinglynx said:
Why didnt you assume the whole thing as a triangle?
and also why didnt you use 0.5 * k * x^2

you can do that as well, but then you have to take care that the value you get is not accurate, and i think you get 3.3 you have to add the extra bit of the curve shape, so we might get confused whether it should be 3.55 or 3.6...now its up to ya!

kinglynx · Jun 7, 2012

geek101 said:
you can do that as well, but then you have to take care that the value you get is not accurate, and i think you get 3.3 you have to add the extra bit of the curve shape, so we might get confused whether it should be 3.55 or 3.6...now its up to ya!

I DID GET 3.3!!!! but i dont know why. should i always just work out the area to be on the safe side?

geek101 · Jun 7, 2012

kinglynx said:
I DID GET 3.3!!!! but i dont know why. should i always just work out the area to be on the safe side?

if its a straight line then you can use the formula, definitely! but if its a curve or something, the examiners intent is to make us use the area (saw in the examiners report

)

Silent Hunter · Jun 7, 2012

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf

Q7,8,16,22,25,27,20 thanks alot

confused123 · Jun 8, 2012

Knight said:
Q6; P=I^2/R
%age uncertanity in P = 2*(%age uncertanity in I) + %age uncertanity in R)
= 2*((0.05/2.50)*100) + 2
=6% ( C )

Q17; K.E=0.5*m*v^2
Let
K.E of Y=0.5mv^2
for Y having mass=m speed=v
Then
K.E of X=0.5*(2*m)*(v/2)^2
=0.5*2m*(v/4)
=0.5*m*v/2
=0.25*m*v
Therefore
K.E of Y = 0.5 the K.E of X ( A )

Q26; Intensity=E
E is directly proportional to A^2
so E will increase 4 times when amplitude A is increased to 2A but the surface area is halved so E will become 2E when amplitude is increased to 2A and surface area is decreased to 0.5S ( B )

Q31; I=nAve
as the wire used are made from the same material so n,v,e will be canceled in the ratio
(I in P)/(I in Q)=(area of P)/(area of Q)
=4 ( D )

Hope you understand it...............

Hi, Thanks for the answers. But i didn't got this formula I = nAve. what's nve here, how is this formula derived..

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AS Physics P1 MCQs Preparation Thread.

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