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As physics p1 MCQS YEARLY ONLY.

dumb human

dumb human

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A.ELWY 7 said:
Nov 2008:
1: C...v=f x (lambda)..he said expressed as the number of the waves he meant the frequency..so speed of light is ( 3 x 10^8) so 3 x 10^8 / 600 x 10^-9 = 5 x 10^14
2: C...i dont know why they put a question of a specific heat capacity in AS level..but anyway u must know the equation of the specific heat capacity then derive it in the equation
3: A...use pythagoras theorm and try it for every choice..A will give the greater one
4: D..he said there is a Systematic uncertanity of 1% and the numbers show that there is another1% as it fluctuates between 1.98 and 2.02 so it is 2 +/- 0.02 so this is the second 1%...1% + 1% = 2% so answer is D
5: D...when the true current is 0.2 the false current > 0.2..when the true one is 0.4 the false one is >0.4...when the true current is 0.6 the false current is > 0.6..but at 0.8A ...they both meet at that point so D
6: D...just by using the area under the graph..there is area for triangle and area for rectangle
7: A...basic stuff
8: A ...also basic stuff..uniform increase in velocity means constant acceleration
9: B...the force of the ball is opposite and equal in magnitude to the ground force...not the weight
10: D..this is the usual equation used in a perfectley elastic collision
11: A... for the 2Kg box...because it is accelerated downward so the weight is greater than the tension, so W-T=ma...so ( 2 x 9.81) - T = 2a , then for the 8 Kg box...the tension is grater than the friction because the 2 kg box is pulling it downwards so, T-F=ma, so T - 6= 8a...add both equations..so [((2 x 9.81)-T= 2a) + (T - 6)= 8a] so T will cancel each other and it will be like this: (2 x 9.81) - 6= 10a so a = 1.4
12: D...friction is up the plane as it opposes the block motio..resultant force is zero because he said it is falling with uniform velocity so no acceleration so no resultant force
13: A..this is easy and doesn't need explanation
14: A...torque of couple is when 2 forces of equal magnitude but act on an object in the opposite directions of each other..so A
15: A...work is force in the direction of the motion x ditance...so F is the same direction as X so Fx...and PE is mgh so Wh
16: A...PE increases with h so B is wrong..D is wrong because the question said it is thrown upwards not downwards..C is wrong because he didnt say it fell again to the ground
17: C...mgh = 0.5 x m x v^2
18: D...by trial using the equation P=Fv
19: D...ductile because it can be drawn to a wire and plastic because it wont return to its original length and shape
20: C.... basic IG stuff
21: C...F is proportional to x...in A it is F/2 so x is x/2..in B it is F/4 so x/4 in D it is 4F/3 so 4x/3 means 1.33x but in C it is 3F/2 so 3x/2 means 1.5 so C is the answer
22: C...basic stuff
23: C...again basic stuff
24: D...wave y has half the amplitude but 3 times the frequency as the diagram shows so D
25: B.. sound can never be polarised but can have interference and reflection
26: A...I = a^2 so 3^2/1^2 so 9/1
27: C...v of light is 3 x 10^8 ....distance between two adjacent nodes or antinodes is lambda/2...so one lambda is 15mm x 2 then f= V/ lambda = ( 3 x 10^8) / (30 x 10^-3)= 1 x 10^10
28: D...A&B will increase the seperation and C will affect the intensity..but increasin the frequency decreases the wavelength so decreasing the seperation
29: B..obvious and mentally solved...
30: A...F=W so W=EQ and E = v/d so W = v/d x Q subtitute so the answer is A
31: D...I= V/R..and R = pl/a and A= pie r^2 so for P: I= V/ pl/(pie) x 1^2 and for Q: I=V/pl/(pie) x 0.5^2 then divide the equation of P by Q and it will be 4/1
32: B....the copper wires are in parrallel so total resistance in parrallel = 1/Rt = 1/R1 + 1/R2 etc... so 1/Rt= (1/10) x 6 = 3/5 so 1/3/5= 1.67 ...then also the steel core is parralell so 1/Rt= 1/1.67 + 1/100= 0.608...1/0.608 so 1.6 is the answer
33: B...basic stuff
34: D...i didnt understand it but what happened is that they divided 100KC/ (200 x 2) = 250
35: D...get the total resistance using R=V/I it will be 3.75 then use the equation of the resistors in parrallel..so 3.75= 1/10 + 1/10 + 1/x...so by trial, D is correct
36: B...7.5/15 = 0.50
37: B...we need the length and the EMF 1 to calculate the pd
38: D...basic stuff
39: A...same number of protons but different number of nutrons so different nucleon number so A
40: A...because as we all know alpha particle have proton number of 2 but beta particle have -1 and nucleon number of 0 while aplha have 4 so A
Click to expand...

2 is B iN MARK SCHEME o_O
 
H

Hanona

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Thought blocker

Thought blocker

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Adira Luis said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
can someone please explain how Q 13 and 14 are solved :/
Click to expand...
13)
Anticlockwise moments= Clockwise moments
20 x 9.81 x60= 100x 10x 9.81 + 50x9.81 x
11772-9810=50x9.81x
x=4 cm clockwise
40+4=44 cm

14)
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2
 
Adira Luis

Adira Luis

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Thought blocker said:
13)
Anticlockwise moments= Clockwise moments
20 x 9.81 x60= 100x 10x 9.81 + 50x9.81 x
11772-9810=50x9.81x
x=4 cm clockwise
40+4=44 cm

14)
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2
Click to expand...
Thought blocker said:
13)
Anticlockwise moments= Clockwise moments
20 x 9.81 x60= 100x 10x 9.81 + 50x9.81 x
11772-9810=50x9.81x
x=4 cm clockwise
40+4=44 cm

14)
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2
Click to expand...

hey thanks :) i might bring up other doubts too!
 
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Hanona

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MaaAsif said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf
Can somebody please help me with Q19 and Q35?
Click to expand...


for Q no. 19 you need to find the power input... power = work done / time
work done = F x D or rather F here as Weight =mg = to find mass you use the formula density = mass/volume= 1000 x 340 = 340000... now converting it to weight x 9.8 = 3332000 ... Work done = 3332000 x distance (30 ) = 99960000
for power it has to be per second and youre given in the question per minute so you divide by 60s = 99960000/60 = 1666000 mnow thats the power in to find the power output you multiply bv the efficiency of the turbine generator = 1499400 = 1.5 MW

q 35 resistivity is doubled so R is doubled and diameter is doubled so R reduced by the factor 4 so the final R = 2/4 = 1/2
 
M

MJAX05

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leosco1995 said:
OK here's all of June 2002.

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. :p

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. :p

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.
Click to expand...

God bless you, you're a life saver!!!!!
 
Adira Luis

Adira Luis

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Thought blocker said:
13)
Anticlockwise moments= Clockwise moments
20 x 9.81 x60= 100x 10x 9.81 + 50x9.81 x
11772-9810=50x9.81x
x=4 cm clockwise
40+4=44 cm

14)
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2
Click to expand...
hey can yu elaborate how yu solved question 13) why did you take distance frm 100g = 10 ,why not 20. i dont get it :(
 
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