As physics p1 MCQS YEARLY ONLY.

[kitkat <3 :P]

Take centre of moment 2 b on pivot n resolve
Clockwise moments = Anticlockwise moments
(10 × 100) + (50 × D) = (20 × 60)
1000 + 50D = 1200
50D = 1200 - 1000
D = 200/50
[ D = 4 cm ]

So distance from pivot = 4 cm
Distance from 0 mark = 40 + 4 = 44 cm
kitkat <3 :P

why is it 60

 

[Hadi Murtaza]

why is it 60

60cm is da distance of da pivot from da string which has da 20g weight

 

[kitkat <3 :P]

60cm is da distance of da pivot from da string which has da 20g weight

ohhh got it thnk u

 

[Hadi Murtaza]

ohhh got it thnk u

most welkum

 

[Thought blocker]

Thought blocker

Yes ._. ?

 

[kitkat <3 :P]

Yes ._. ?

kuch nai wapic chaly jao doubt solve hogya

 

[Thought blocker]

kuch nai wapic chaly jao doubt solve hogya

Kaunsa, wo Equi wala ._. ?
Wo nay aya ta kya teko ?

 

[kitkat <3 :P]

Kaunsa, wo Equi wala ._. ?
Wo nay aya ta kya teko ?

nai but aab aagya

 

[forever_chocoholic]

Take centre of moment 2 b on pivot n resolve
Clockwise moments = Anticlockwise moments
(10 × 100) + (50 × D) = (20 × 60)
1000 + 50D = 1200
50D = 1200 - 1000
D = 200/50
[ D = 4 cm ]

So distance from pivot = 4 cm
Distance from 0 mark = 40 + 4 = 44 cm
kitkat <3 :P

i get it.. that was simple... im so stupid!
thank you!!!

 

[Hadi Murtaza]

i get it.. that was simple... im so stupid!
thank you!!!

Dont call ur self stupid
Ur most welkum

 

[forever_chocoholic]

Dont call ur self stupid
Ur most welkum

heehe okay, =)

 

[Zepudee]

Q4,5,28,29,34,37,40!! HEHEHEE boohoo, someone please help me

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf

 

[forever_chocoholic]

Q4,5,28,29,34,37,40!! HEHEHEE boohoo, someone please help me

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

Q.4 A
power=energy transformed(or workdone)/time taken
since they said between two times..its the change in energy which is kinetic energy here

Q5 B
check the first page of this thread...this question has been repeated

Q29 C
wheteher its stationary or moving,in both cases u have forces acting on it
and the forces are parallel to the plates.. towards positive plate if its negatively charged n towards negative plate if it is positively charged so itsparallel to the electric field

Q37 A
when a beta particleis emittedthe proton number increases by 1 but the nucleon number remains the sam
so now u have 4 protons (3+1 = 4)
and 4 neutrons (A-Z = 8-4 =4)

Q40 C
in nuclear processes the following things are conserved:
A number (nucleon number)
Z number ( proton number)
mass-energy
electric charge

you dont have momentum here so the answer is neutron number

I skipped some... sorry i know the answers but im not sure if my explanation is rite.. dont want to confuse u
hope that helps!

 

[IGCSE O/L student]

Q4,5,28,29,34,37,40!! HEHEHEE boohoo, someone please help me

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

Q 28 the charged particle moves with a displacement perpendicular to the direction of the force
when the force and the displacement are perpendicular to each other, the workdone is 0
F cos (90) x distance = 0

Q 34 All the power in the battery is dissipated in the resistors respectively.
12 = I^2R + (I/2)^2R + (I/2)^2R <the current in the parallel resistors is halved as they are identical)
12 = I^2R + (I^2/4)R + (I^2/4)R
12 = 1/2(2I^2R + I^2R)
24 = 3 I^2R
8 = I^2R

the power dissipated in R is (I^2/4)R
1/4 of 8 is 2

 

[DeViL gURl B)]

Hi all, i've made this thread only for yearly papers starting from june 2002. We'll do 2-3 years each day. And anyone who wishes to help is most welcome. But please once again, this is only for YEARLY papers.

MOD EDIT

AsSalamoAlaikum Wr Wb!


Nov:2001 Answer Key

Here are few solved explanations for 9702 Physics Paper:1

June:2002

Nov:2002

June:2003

Nov:2003

June:2004

Nov:2004

June:2005

Nov:2005

June:2006

Nov:2006

June:2007

Nov:2007

June:2008

Nov:2008

June:2009

Nov:2011 # 1

Nov:2011 # 2

Dude where is Nov 2003? :$
Thank u

 

[forever_chocoholic]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf

can someone please explain Q.12 and Q.31... pleaseeeeeeeeeeee!!!!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf

can someone please explain Q.12 and Q.31... pleaseeeeeeeeeeee!!!!

12)
It's easy
Hope this helps you --> Click here
31)
Its easy too
Total resistance in 6 and 3 ohm is 2 ohm, so the voltage will be 6v in d 2 ohm resistor. n 6v across d 6 n 3 ohm resistors respectively. we have the voltage n resistance. so v=ir, 6=6i. so current is 1A

 

[forever_chocoholic]

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

Q9, Q12, Q17, Q18, Q25, Q31, Q33

can anyone explain these questions pleaseee!

 

[forever_chocoholic]

12)
It's easy
Hope this helps you --> Click here
31)
Its easy too
Total resistance in 6 and 3 ohm is 2 ohm, so the voltage will be 6v in d 2 ohm resistor. n 6v across d 6 n 3 ohm resistors respectively. we have the voltage n resistance. so v=ir, 6=6i. so current is 1A

Thanks for the answers!!!

 

[Thought blocker]

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

Q9, Q12, Q17, Q18, Q25, Q31, Q33

can anyone explain these questions pleaseee!

9) Get the time from eqn s = ut + 0.5at^(2)
S= 1.25, u = 0, a = 9.8, t = ?
t = 0.5
Now as v = d/t
V = 10/0.5 = 20 m/s

12)upthrust is always due to a difference in PRESSURE

17)First take the any random value of X that is half than Y, Say X = 2 and Y = 4 hence they are the respective velocities of X and Y
Now same way, for MASS Take Y as 2 and X as 4
Now for X, m = 4 and V = 2
---> 0.5*4*4 = 8
Now for Y, m =2 and V = 4
---> 0.5*2*16 = 16
Hence it is shown, that,
Car X has half the kinetic energy of car Y.

18)W.D = F*d Where F = force and d = perpendicular distance
so force is 50 times 9.8 that is 490 and d is 1.6
W.D = 490*1.6 = 784 that is near to 780

25) The image of a wave has been given, and so has the direction it is traveling in.

Note that since this is a transverse wave and the disturbances are traveling from left to right, the particles at any point are vibrating in a direction perpendicular to that propagation.
In other words, if you focused on any one segment of the rope and that one segment alone (let's say by painting it a different color from that of the rope), then you would find it does not experienceany sideways displacement; it only moves up, then down, then up, then down, etc.

At the same time, the waveform shifts towards the right, since that is the direction in which it is traveling.
So just imagine the waveform displayed in the question shift to the right; imagine each part of the waveform progress, and you would then see that P is moving downwards; since the trough just before P has to travel to the right, P has to "fall" into that trough, and if it "falls" into that trough, it is moving downwards.

Of course, this eliminates all the options in the question, leaving behind only A, the correct answer, but it is worth discussing the motion of Q:

At the instant of the displayed image, Q is a segment at the crest of the waveform; it is debatable that after some time, Q has to fall towards the equilibrium position, but taking a closer look at the wave motion (and seeing that we can estimate the motion of any one segment as Simple Harmonic Motion, SHM) it turns out that Q would be stationary.

It's just like the motion of a pendulum; at any extreme, the rate of change of it's position - it's velocity - is zero, but due to the forces acting on the pendulum (in THIS case, the force is the tension in the string) this changes slowly to make it change position. So that's most likely why the velocity of Q is zero at the instant shown.

31)I=V/R AND R=pL/A So you can write it as I= V/pL/A ; A is pi*r^(2)
I = [V/pL/pie *(1 * 10^-3)^2 / V/pL/pie * (0.5 * 10^-3)^2]
substitue u will reach to a point where it will be 1/0.25=4