AS Physics P1 MCQs Preparation Thread.

[A.ELWY 7]

Can someone explain me June 2008 Q.24 I am getting answer 2E.??

Thank you

please next time post the link to the question paper...the answer is simply C as no calculations is needed because he said that in the other experiment the wire was of the SAME TYPE OF STEEL as in the first experiment and so same spring constant and so the Same young modulus !!

 

[anythin576]

Hey guys. For O/N/05 question number 34, why is the answer C not B? Since B has the highest gradient in which reciprocal of the gradient represents resistance? Thx in advance

NO RESISTANCE IS NOT THE RECIPROCAL OF THE GRAPH!!!! This is a very common misconception. This has been mentioned in the examiner report too( if i'm not wrong the examiner paper of the same paper). Resistance is the ratio of the V to I which means you read off the values from the graph at a particular point and divide V by I. This question can be done by measuring the x and y distances of the points and dividing. C will have the lowest value for V/R (which ofcourse is resistance) hence the correct option is C. hope made it clear.

 

[Unicorn]

In A,B and D, work is done because energy has been converted from one form to another. In C, energy isn't transformed so no work is done.

oh ok thanks

 

[anythin576]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf
question 37 was quite difficult.....anyone who know's why c is the correct answer???? Will greatly appreciate any help

 

[master_11]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
Questions 7, 8, 16, 17 and 31. please explain these to me. thank you.

For Q7. Maximum displacement occurs when velocity=0 and and Q, the point already is at is maximum velocity, the acceleration must be zero.
For Q8. you know that gradient of the displacement-time graph is velocity. In this situation, initially, velocity will increase as air resistance acting on it is lesser than acceleration due to gravity and after a while, velocity becomes constant (air resistance=acceleration due to gravity). Bearing this in mind, as initially velocity increases, gradient MUST increase and then when velocity gets constant, gradient becomes constant by implied by the straight line.
For Q16. By taking moments, 0.5l x 10= 2l * Fcos60
For 17. First Using triangle XYZ, find out the angle YX and mid point of YZ and by using that angle find 4cos(angle0 and multiply it by 2 the force is being exerted by both the strings and it is the horizontal component of the tension which is causing acceleration.
for 31. answer is B because by moving the positive charge in that direction, repulsion increases as the positive charges repel but D is the distractor as is not known that what is the distance between the charge and the positive plate.

 

[Henry930821]

NO RESISTANCE IS NOT THE RECIPROCAL OF THE GRAPH!!!! This is a very common misconception. This has been mentioned in the examiner report too( if i'm not wrong the examiner paper of the same paper). Resistance is the ratio of the V to I which means you read off the values from the graph at a particular point and divide V by I. This question can be done by measuring the x and y distances of the points and dividing. C will have the lowest value for V/R (which ofcourse is resistance) hence the correct option is C. hope made it clear.

Okay okay wow thanks! I carelessly forgotten about taking the ratio at different points to obtain the resistance. Thanks again!

 

[Oliveme]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
qs 11 plz!!

Using equation - m1*u1 + m2*u2 = m1*v1 + m2*v2 where m1 = 20 kg, m2 = 12 kg, u1 = 6m/s, u2 = -15 m/s as it's traveling in the OPPOSITE direction. You MUST NOT ignore directions! so the equation becomes m1u1 + m2u2 = v*(m1 + m2) as they now stick together and have the same final velocity.
(20*6) + (12*-15) =v*(20+12) -------> -60 = -32v as they are traveling towards the bigger speed (-15) and we have taken this direction as negative (-).
answer is v = -60/-32 = 1.875 rounding off 1.9 m/s. answer is A.

 

[master_11]

please next time post the link to the question paper...the answer is simply C as no calculations is needed because he said that in the other experiment the wire was of the SAME TYPE OF STEEL as in the first experiment and so same spring constant and so the Same young modulus !!

Thank you so much.

 

[A.ELWY 7]

Thank you so much.

welcome

 

[A.ELWY 7]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
question 37 was quite difficult.....anyone who know's why c is the correct answer???? Will greatly appreciate any help

Because both are made of the same material and so bith have the same resistivity...R=pl/A....and he also said in the question that P nad Q have the same length..so A is left...but at the beggining of the question he said both P and Q have same volume and volume is height x Area...and so using the equation...resistivity, length...and area are the same,and so both have same resistance

 

[Oliveme]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
Questions 7, 8, 16, 17 and 31. please explain these to me. thank you.

please answer questions 16, 17 and 31 above. thank you.

 

[Oliveme]

For Q7. Maximum displacement occurs when velocity=0 and and Q, the point already is at is maximum velocity, the acceleration must be zero.
For Q8. you know that gradient of the displacement-time graph is velocity. In this situation, initially, velocity will increase as air resistance acting on it is lesser than acceleration due to gravity and after a while, velocity becomes constant (air resistance=acceleration due to gravity). Bearing this in mind, as initially velocity increases, gradient MUST increase and then when velocity gets constant, gradient becomes constant by implied by the straight line.
For Q16. By taking moments, 0.5l x 10= 2l * Fcos60
For 17. First Using triangle XYZ, find out the angle YX and mid point of YZ and by using that angle find 4cos(angle0 and multiply it by 2 the force is being exerted by both the strings and it is the horizontal component of the tension which is causing acceleration.
for 31. answer is B because by moving the positive charge in that direction, repulsion increases as the positive charges repel but D is the distractor as is not known that what is the distance between the charge and the positive plate.

thank you. that cleared some confusions.

 

[DARK DRAGON]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf
Q28 answer is A. plz explain

 

[master_11]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf
Q28 answer is A. plz explain

Beacause the movement from P to Q is at a 90 degree angle and in the case of electric field, forces act on parallel to the field NOT perpendicular to it so work done in moving from P to Q is 0.

 

[Jaf]

Thanks alot
explanation needed for this one

3. Intensity = energy/(area * time) (also given in the question)
37. The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).

In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf

First find the combined resistances of each of the two sets of parallel resistors.
It's 2Ω and 1Ω respectively.

Since the combined resistance of the first set of parallel resistors is more, we know it's share of voltage will be more. So V₁ > V₂.
For ease of calculation, suppose the battery has an emf of 3V. The voltage across the first set of resistors would be 2V and the voltage across the second set of resistors would be 1V. I₁ would then be 2/3 [V/R = 2/3] and I₂ would be 1/2 [V/R = 1/2]. Therefore, I₁ > I₂.

 

[leosco1995]

Can someone explain me June 2008 Q.24 I am getting answer 2E.??

Thank you

The Young Modulus is a characteristic of a material, not its dimensions. Therefore the Young Modulus will still be E. Even if you triple the length and do whatever with the area, it will still be E.

No calculations were needed to solve this question.

 

[Sanis]

November 2010 V2
*7
*8
*9
*15
*25
*26
*32

 

[Oliveme]

can someone please answer these questions? questions 16, 17 and 31 above. thank you. I'll be really grateful.
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf

 

[leosco1995]

can someone please answer these questions? questions 16, 17 and 31 above. thank you. I'll be really grateful.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf

Q16)

Take moments about the point X on the wall. You know that the weight lies in the middle, so if you assume the distance XY to be 1m (or 'x' m or whatever you want), the weight would be 0.5m from X. And the anti-clockwise moment, F is given by F sin 30 and it is 1m from X. So form an equation and solve:

F sin 30 * 1 = 0.5 * 10
F = 10.0 N

Q17) ..This one is kind of tricky, I think. I'm not sure if my method is correct. Check out the attachment. I basically found the angle between the strings Y and Z and then resolved forces to find the component 'x'. I did 4 cos 53.1 and then multiplied by 2 because there is a force acting both upwards and downwards. It turns out I got the right answer, but again, don't know if this is the proper method.

Q31) It can't be A because a +ve particle is going away from the +ve charge in the centre, so the repulsive force is decreasing.
It can definitely be B because that's basically the opposite of what I said above.
It can't be C or D because the force in an electric field is the same. In D, the +ve particle will experience the same force at any point. So the answer should be B.

 

[Parfals]

May/june 2002
Question:
3,14,21,26,28,29,35
em weak in physics
Thnx in advance

http://www.freeexampapers.com/get_p... Level/Physics/CIE/2002 Jun/9702_s02_qp_1.pdf
plz helpp