AS Physics P1 MCQs Preparation Thread.

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help me in Q NO 16...PLEASE.
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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. :)
 
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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
hey! thank you so much. just one question, how did you get [ 4 /(4+1)] x 25 = 20V? is it the equation V1/V2 = R1/R2? thanks once again. :) and yes, I'll try to get the explanation for ques 10 and will tell you when I do. :D
 
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plz give me an explanation :(
I'll try to help but I can help only so much because This is a tricky one.
first of all, the answer is B, not C.
so, since we're given V, d, l and I, we can use the equation R=*rho*L/A where rho is the resistivity of the wire and so is a constant so we can eliminate it. we don't have R and R=V/I
thus, V/I=L/A --------> V/I = L/(d/2)^2 since A=pi x r^2
so making I the subject, I = (V x (d/2)^2)/L
when V becomes 2V, L = 2L and d=2d we have I = (2V x (2d/2)^2)/2L. now from this you can see that (d/2)^2 is proportional to I, so if d is doubled the I is also doubled regardless of V and L. now d is also squared so you may ask why is it not 4I then? because we have d/2 to get radius. so when I = (d/2)^2, then 2I = (2d/2)^2 -----> this can written as 2I = d^2.
hope this leads somewhere. :)
 
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hey! thank you so much. just one question, how did you get [ 4 /(4+1)] x 25 = 20V? is it the equation V1/V2 = R1/R2? thanks once again. :) and yes, I'll try to get the explanation for ques 10 and will tell you when I do. :D
4 was the variable resistance, the other fixed resistor had 1kΩ. Therefore the *total* resistance of the circuit was the sum of both resistors, i.e 4+1 = 5 kΩ
P.S. The ratio of p.d. is always same as the ratio of resistances in a series circuit.
 
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I'll try to help but I can help only so much because This is a tricky one.
first of all, the answer is B, not C.
so, since we're given V, d, l and I, we can use the equation R=*rho*L/A where rho is the resistivity of the wire and so is a constant so we can eliminate it. we don't have R and R=V/I
thus, V/I=L/A --------> V/I = L/(d/2)^2 since A=pi x r^2
so making I the subject, I = (V x (d/2)^2)/L
when V becomes 2V, L = 2L and d=2d we have I = (2V x (2d/2)^2)/2L. now from this you can see that (d/2)^2 is proportional to I, so if d is doubled the I is also doubled regardless of V and L. now d is also squared so you may ask why is it not 4I then? because we have d/2 to get radius. so when I = (d/2)^2, then 2I = (2d/2)^2 -----> this can written as 2I = d^2.
hope this leads somewhere. :)
the marking scheme says its C :)
 
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27)
For a charged particle, the force on it remains the same in the entire electric field. We can prove this mathematically:

E = F/Q

Since E is always constant and Q is always constant for that particle, F will be the same. Hence D.

28) When a charge is placed in an electric field the force acting upon the charge will be parallel to the electric field (directions change if the particle is negative/positive but that's irrelevant). That rules out B and D. And a force can be exerted on both stationary and moving charges, it's sort of a fact. You can have e.g. a stationary particle held between the plates due to both weight and the force, and e.g. a moving particle be accelerated because of the force. So a force can act on both stationary and moving particles.

29) Since point P and Q are both in the same electric field, they experience the same electric field strength 'E', so both have the same potential. This means no work will be done in moving the charge from P to Q.

35) Q and R are in the parallel, so they get 1/3 V of the total power supply. The series one gets 2/3 V but that's irrelevant. Also, in parallel the current is divided, so they both get 1/2 A. The power loss would be I * V = 1/3 * 1/2 = 1/6 of the total power supply, so 2W is correct.

Another way of solving this question would be to assume some fake value for the resistance of the resistors (since they are all equal), find the total resistance and therefore current of the circuit using P = I²R. You would find the current going into any of the parallel resistors and use I²R again to find the total power dissipated in the resistor. This method might be slower because you have to do a bit of working, though.
 
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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.
 

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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 Q33 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf
MAY/JUNE 2004: Q21 ans C Q25 ans A Q26 ans B Q39 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
OCT/NOV 2004: Q20 A ans Q31 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
MAY/JUNE 2006: Q9 ans D Q36 ans C Q38 ans A
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
OCT/NOV 2006: Q3 ans A Q31 ans A
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
MAY/JUNE 2007: Q20 ans B Q23 ans D Q 10 ans B Q40 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf
OCT/NOV 2007: Q7 ans C Q 16 ans C Q37 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf
MAY/JUNE 2009: Q15 ans B
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf
OCT/NOV 2009: Q9 ans A Q14 ans A Q27 ans D Q30 ans C
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
MAY/JUNE 2010 paper11: Q 33 ans A Q 34 ans A
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf
Oct/NOV 2010 paper11: Q5 ans B
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf

THIS IS IT FOR NOW, please help me as im really frustrated, it took me a while to make this thread please do reply, thankyou so much, i will remember All of you in my prayers. Thanks a million :)
 
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Does anyone have a sheet with approximates like mass of apple, volume of a man's head etc? Any reliable source for such approximates? And ALSO, the electromagnetic spectrum wavelengths and frequencies?
 
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question23.png

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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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 Q33 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf
MAY/JUNE 2004: Q21 ans C Q25 ans A Q26 ans B Q39 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
OCT/NOV 2004: Q20 A ans Q31 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
MAY/JUNE 2006: Q9 ans D Q36 ans C Q38 ans A
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
OCT/NOV 2006: Q3 ans A Q31 ans A
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
MAY/JUNE 2007: Q20 ans B Q23 ans D Q 10 ans B Q40 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf
OCT/NOV 2007: Q7 ans C Q 16 ans C Q37 ans C
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf
MAY/JUNE 2009: Q15 ans B
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf
OCT/NOV 2009: Q9 ans A Q14 ans A Q27 ans D Q30 ans C
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
MAY/JUNE 2010 paper11: Q 33 ans A Q 34 ans A
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf
Oct/NOV 2010 paper11: Q5 ans B
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf

THIS IS IT FOR NOW, please help me as im really frustrated, it took me a while to make this thread please do reply, thankyou so much, i will remember All of you in my prayers. Thanks a million :)
o/n 03- 33- A n B are out coz they dnt hav a variabl output tat can divide da pd..D is out cz it is nt conncted to da lamp..
o/n/06-3- its coz..elcctric fld strngth =F/Q
so unit of F u find it using F=ma which is Kgms^-2 (since mass is Kg and acclrtion is ms^-2)
now use ths in da equation E=F/Q
=Kgms^-2/C (since chrge is Coloumb)
thrfr wen u brng C to da numerator u g t= Kgms^-2C^-1..whch is A
m/j/07-10.. the equation Force=Chamge in momentum/change in time must be applied..so change only B is posibl
0/n/07- 7..since w=mg..g will b equal to w/m.. ans C
A B n D r wrong as da statements r incorrect
- 16- Kintc energy is alwys due an objects motiom..at position 3 since the prson dsnt move KE is min..posibl ans are C n D
potntial energy of an object is due to its postion..elstic PE is due to sumthng tat is strchd n has a PE..so at postion 3 the rope strches to its max...so his elastic PE ismax at postn 3..so ans C
m/j/09- 15- Its B coz..initally in x the PE is mgh..k.so once the tap is open half of y is filled..now in x da mass decreases by 1/2 and so is the height..
thrfr da new chng in PE is 1/2m*g* 1/2h = mgh/4..

ths is not mch..hped it has hlped u in sumwy or the othr :) ..sry with da othrs!!!
 
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