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haha!
u didn't sleep yet??! (just kidding).... i just solved nov 10 varient 3.......only june & nov 11 papers remaining....INSHA ALLAH i try to solve all of them tomorrow ...and now me off to bed ....its 2 in the midnight....bye for now 
yup i didnt sleep i hav sum unfinished business 2 attend 2haha!
THANKS TO U TOO FOR EVERYTHING
BEST OF LUCK FOR UR EXAMS
hmm ok! enjoy attending ityup i didnt sleep i hav sum unfinished business 2 attend 2
clockwise moment=20*3=60Nm
OK here's all of June 2002.
R M I V U X Y
can you please explain what each alphabet stands for?
June 2002
==========
1. B
Fact. A would be right if K was given instead of °C.
2. B
You go FORWARD in the direction of X and BACKWARD in the direction of Y.
3. A
The units of speed (msˉ¹)s are equal on both sides.
4. B
You can eliminate A, C and D because they are all nonsense (in my opinion). B is correct because if the timer was started and THEN the ball thrown, you would get 0
height for some time 't' on the x-axis.
5. C
Uncertainity = 2(0.03) + 0.02 = 0.08 = 7%. You multiply the uncertainty of V twice because of the square.
6. D
Air resistance isn't negligible because the speed eventually becomes constant (terminal velocity), so rule out A and C. The Y-axis can't be distance because it
eventually becomes constant, the distance can't become constant during the fall of a body.
7. A
Acceleration is constant with uniformly increasing speed. B is increasing rate of speed, C is constant speed and D is at rest.
8. D
S is the distance from cliff to highest point. R is the distance from highest point to sea-level. We are looking for the distance of sea-level to cliff, which is R -
S.
9. B
K.E is ALWAYS conserved in elastic collisions, so K.E before impact is 0.5mv² + 0.5mv² = mv². That means after impact, the K.E should also be mv².
10. B
Fnet = ma.
12 - x = 4 * 0.6
x = 9.6 N.
11. B
Momentum is always conserved so forming an equation,
0 = M1V1 + M2(-V2)
M1V1 = M2V2
V1/V2 = M2/M1
12. D
Fact. Upthrust is very small compared to weight. Also, drag is almost as large as weight not they are not the same (weight is a bit larger) so A and B are wrong.
13. C
Torque = 2 * PD
To find the perpendicular distance, find the perpendicular distance from the force to the pivot and then multiply that by 2. To find the distance from the force to
pivot, construct a triangle and obtain the equation x = 0.15 sin 30.
14. C
Upthrust is the pressure of the block (Pb - Pt) * area, I think.
15. D
Resolve the horizontal 3N and vertical 4N force to get a sideways 5N force which is in the same line as the diagonal 4N force. The resultant force has a magnitude of
1N and the direction is towards the upper-right.
16. D
Efficiency = (useful output)/(total input)
17. C
The only work done is reducing the volume, and since the pressure is constant, work done is p(V1 - V2).
18. B
At Q, the potential energy is 50kJ less than P. This means that the 50kJ must have been converted to kinetic energy. So, K.E (Q) = K.E (P) + 50kJ = 55kJ. And 10 was
lost in friction, leaving us with 45 kJ.
19. D
Simple Power = Force * Velocity.
24 * 10³= 600 * V
20. B
Evaporation occurs over a range of temperatures, while the rest only occur at a fixed temperature.
21. A
Total density = total mass / total volume.
The total mass is m1 + m2 = 2m (since they are equal).
The total volume is MD1/MD2. D1 is ρ and D2 is 2ρ, and that gives us the total volume has 3M/2ρ. Then just use the total density formula I wrote above.
22. C.
Stress = F/A
Strain = extension/length
YM = stress/strain.
23. B
Simple ratio stuff with the YM formula FL/Ax.
24. B
Area below a force/extension graph is energy (i.e. work done). In reducing l2 to l1, this part of the graph is MNQP.
25. C
Speed of electromagnetic waves is always constant, but the frequency will decrease because the wavelength increases.
R M I V U X Y
--> increasing frequency
<-- increasing wavelength
26. B
λ = 4cm.
Time period for one wave is therefore 4 * 0.002s = 0.008s.
F = 1/T = 125 Hz.
27. B
Fact. I α a² and I α 1/r².
28. C
Use the path difference formula (distance from one source - distance from other source) to create an equation involving X. S2x - S1X = λ/2. λ/2 because X is a minimum point.
29. D
λ = 2(1.5) = 3m. 2 nodes are 0.5λ apart.
v = fλ
v = 300(3) = 900.
30. B
Graph X - Diode
Graph Y - Ohmic conductor/metal wire
Graph Z - Lamp
You have to learn these I/V graphs. Also note that a thermistor/semi-conductor has a graph which is like the lamp but has an increasing gradient instead.
31. C
Basic formula recall needed here.
32. A
R = V²/R = 240²/100 = 576. However, this is when the filament is heated so it has increased by 16 times. The normal room temperature would therefore be 576/16 = 36.
33. C
Basic Kirchoff's first law.
34. C
Diagram 2 has the same setup as diagram 1 (2 lamps connected in each parallel setup). If you are confused about diagram 2, then just rotate it. Since they have the same setup, the brightness would obviously be the same.
35. C
The way I solved this was by trying each option until I calculated a 2V drop at R1 and 1V drop at R2. Find the total current using V = IR (where V is 5 and the total resistance is the sum of R1, R2 and R3) and then use the formula again at each resistor to find the voltage drop (5 - V).
36. A
The electron will accelerate towards the +ve plate so A. A is +ve because the field lines are directed from +ve to -ve.
37. A
E = V/d
Increasing the value of 'd' will decrease the value of 'E'. Therefore A is correct.
38. C
Fact.
39. A
Basic stuff.
40. C
Work backwards to figure this one out, and see what is happening to the nucleon number and proton number individually. Then # of neutrons is nucleon number - proton number.
November 2002 will probably come next, and then I'll do 2011 backwards.
q9 for these type of calculation u r always suppose 2 use a straight so u make a straight line from 4m/s 0sec till 8m/s 3sec and find acceleration u get 1.33m/s^2
Im really sry...i reached my limit..now i need sometime for me to studyi feel piety for u coz all the time i have doubts, u just pop up in my mind. So i try to ask my doubts from others coz u already do so much stuff.....anyways thanks for everything and if its possible for u AND u have time, just post june 11 varient 1 & 2
BEST OF LUCK FOR UR EXAMS!!!
And thnx very much for you feelings and your manners.....rare ppl are like this nowi feel piety for u coz all the time i have doubts, u just pop up in my mind. So i try to ask my doubts from others coz u already do so much stuff.....anyways thanks for everything and if its possible for u AND u have time, just post june 11 varient 1 & 2
BEST OF LUCK FOR UR EXAMS!!!
u welcum
no problem if i have doubts in these two paper, i'll try to clear them someone else if possible and yeah u too need to study. I cant post papers coz i often hav doubts and fearing that i'll post sumthing wrong...
answers id because Imagine a velocity-time graph, the graph will slope downwards until from when the ball is released until it hits his down again, thus a negative acceleration.thanks a lot
hmm i am a little confused wid nov 10 question 7); u said: acceleration remains constant then when the ball bounces acceleration decreases and then wen it cums bac it is da same again......thats perfectly fine i get it but why cant the answer be B.....why is it below the x-axis??!
done...i posted a note in the thread u made that'll help
aha! now i get it
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