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AS Physics P1 MCQs Preparation Thread.

Pals_1010

Pals_1010

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xxfarhaxx

xxfarhaxx

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Pals_1010 said:
J06 No 23 and No 27:

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf

J07 No 23 :

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

N07 No 23:

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf

J08 No 28:

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf


I would really appreciate if someone could help me with these MCQs. Thank You :)
Click to expand...
j/o6
23) usualy it is only transvers waves that can be polriszed...option B n C n can b done using both kind of waves..
26)its between 2 nodes that u will find an antinode and all the other particles..thrf they vibrate. ans C..the ans cant b D coz..btween 2 antinodes u will find a node which is stationary.. so D is wrong

j/08
28) use da equations d sin O= n lamda...( substitue 1/N for d , 3 for n)
make sin O the subject = sin O= 3*lamda / 1/N
sin o= 3*lamda*N---------ans B

did u undrstand????
 
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Ahmed Ali Akbar

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Mubarka said:
View attachment 12195
There are three forces in total acting upon it to keep it balanced. Therefore, the resultant must be zero.
Consider the vertical components of all the forces.
Upward force must be equal to the sum of the downward forces. For this to be correct, T must be greatest whose vertical component equals the downward forces. The only option with the last force T is C.
Hope it helps. :)
Click to expand...
thanks...
 
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Ahmed Ali Akbar

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SalmanslK said:
View attachment 12217

Area under curve can be regarded as energy stored or lost.But at the same time can also be regarded as WORKDONE.
the answer is B.I know what the reason for answer to B is.But Why not D ?
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work is said to be done when we apply force and object/mterial movess...here we are removing force nt adding...rubber is returning to its original position on its own...as force is not applied to get it to original position...D is wrong..no work is done..its zero....
 
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MA1234

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A spindle is attached at one end to the centre of a lever 1.20 m long and at its other end to the
centre of a disc of radius 0.20 m. A cord is wrapped round the disc, passes over a pulley and is
attached to a 900 N weight.What is the minimum force F, applied to each end of the lever, that could lift the weight? PLZZZZZZZZZZZ HELP JUNE 09 QP 1
 
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leosco1995

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Pals_1010 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

No 25 ? Anyone Please?
Click to expand...
The wave is undergoing simple harmonic motion.

Think of the displacement-time graph as that of a pendulum bob. At the equilibrium position, the particle has maximum velocity and at the maximum displacement position, the particle comes to rest momentarily before changing direction. Since point Q is at a maximum displacement, it must be stationary. Therefore options B and C are wrong.

As for P, it could either be going up or down. But since the area below the graph before point P is all negative it means P was going downwards. This also means that the next point which cuts the x-axis would be going upwards, one after that downwards and so on.
USMAN Sheikh said:
ANY ONE PLS RESPOND TO MY QUESTIONS AS WELL PLS :( :'( :cry:
Click to expand...
Bro don't worry I will try to answer your questions eventually if nobody else does. :)
 
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MA1234

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A spindle is attached at one end to the centre of a lever 1.20 m long and at its other end to the
centre of a disc of radius 0.20 m. A cord is wrapped round the disc, passes over a pulley and is
attached to a 900 N weight.What is the minimum force F, applied to each end of the lever, that could lift the weight?PLZZZZZZZZZZZZZZZZZ HEEEELLPP!!!!! JO9 QP1
 
xxfarhaxx

xxfarhaxx

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MA1234 said:
A spindle is attached at one end to the centre of a lever 1.20 m long and at its other end to the
centre of a disc of radius 0.20 m. A cord is wrapped round the disc, passes over a pulley and is
attached to a 900 N weight.What is the minimum force F, applied to each end of the lever, that could lift the weight?PLZZZZZZZZZZZZZZZZZ HEEEELLPP!!!!! JO9 QP1
Click to expand...
ans B...u mst tke the torque... F*1.20=900*0.20..wen u find F u will gt it as 150 N..
 
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Oliveme

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leosco1995 said:
Q14)
Forces in equillibrum must follow the closed triangle rule, there can't be head to head in any of the vectors, making B and C invalid. D is wrong because weight acts downwards so A is right.

Q15) See the attached diagram.

torque = one of the forces * p.d. between the forces

The ruler is 0.30m so one side of it to the pivot will be 0.15m. Then we can find 'x', half of the p.d. between the forces. The p.d. between the forces will be '2x'. Total torque can then be calculated by 2 * whatever the p.d. is.

Q17) Initially K.E = KE of X + KE of Y

0.5mv² + 0.5mv² = mv²

After collision, both 'stick together' as stated in the question so their masses add up. When this happens, the speed has to be halved because momentum will be conserved. Look:

mv + mv = 2mx ('x' being their speed after sticking together)
2mv = 2mx
x = v/2.

So K.E after is 0.5 * 2m * (v/2)² = mv²/2

K.E lost is Initial K.E / 2, so C.

21. Young Modulus is the same for the same material, regardless of its dimensions. So the answer is C and no calculations have to be done.

27. When one of the slits is covered, the amplitude gets halved (fact), so:

I1/I2 = (A1/A2)²

Plugging in the values,

I1/I2 = (A/0.5A)²

I2 = 1/4 I1.

So D.

40. Use the potential divider formula for this one.

Minimum is when the resistance is 0, so voltage will also be 0.

Maximum is given by:

9 * (5000/1000) = 4.5 V

Let me know if this needs more explanation.
Click to expand...
Hey there. Thank you so much. Mashallah, you're like a physics genius. :) Honestly, we had a very bad teacher for our AS level, so yeah I pretty much suck at physics. Thanks again! :)
 
USMAN Sheikh

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leosco1995 said:
The wave is undergoing simple harmonic motion.

Think of the displacement-time graph as that of a pendulum bob. At the equilibrium position, the particle has maximum velocity and at the maximum displacement position, the particle comes to rest momentarily before changing direction. Since point Q is at a maximum displacement, it must be stationary. Therefore options B and C are wrong.

As for P, it could either be going up or down. But since the area below the graph before point P is all negative it means P was going downwards. This also means that the next point which cuts the x-axis would be going upwards, one after that downwards and so on.

Bro don't worry I will try to answer your questions eventually if nobody else does. :)
Click to expand...
pls i need ur help buddy :)
 
hussamh10

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USMAN Sheikh said:
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
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In ques 24 the steel is of the same material so young modulus will be the same E is the ans... in 30 to get the final velocity u will have to do ucos theta=v this is the only valid option which is A.39 only requires book reading the ans is C
 
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