As physics p1 MCQS YEARLY ONLY.

[arlery]

MashAllah this is great. Keep up the awesome work!

(I was about to start with this after about an hour or two, but it's good you did and posted it first. )

Thanks!
Haha yeah
You can solve another paper now.

 

[leosco1995]

Thanks!
Haha yeah
You can solve another paper now.

I would, but I'm too tired at this time. Will definitely do one or two tomorrow though.

 

[A.ELWY 7]

2 say the truth this thread had motivated me to solve atleast an exam paper a day..i wanted to take atleast 1 week free after the practical but now i know i was wrong...Thnx for every 1 who participated in this thread as each one of us motivates the other...keep up the good work..i just finished Nov 2008 and ill post it now.

 

[MA1234]

you didnt post nov 08 plzz do it asap im waiting

 

[MA1234]

thank u very mch. Posting the solutions helped alot to understand the ques in a better way!!!

 

[A.ELWY 7]

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

 

[A.ELWY 7]

you didnt post nov 08 plzz do it asap im waiting

here it is...sry to be late

 

[MA1234]

here it is...sry to be late

thank uuuuu soooooooooo mch...

 

[xyz!]

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.

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. D

Electric field strength is always a constant value for a field.

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.

hey can u explain the Q.15 again pls?
n for Q.14...y dont v consider W? usually ven solving such questns v consider all the possible forces acting on the object ryt?

 

[A.ELWY 7]

thank uuuuu soooooooooo mch...

welcom

 

[Gémeaux]

May/June 2008

1) C; kilo(k): 10^3
milli(m): 10^-3
Mega(M): 10^6
nano ( n):10^-9

2) D; Force = mass x acc. = kgms-²
work done = force x displacement = kgm²s-²
e.m.f. = work done/ charge = kgm²s-²/As

3) C; speed= distance/time = 100/10 = 10m/s
the approximate mass of an athlete can be estimated between 60 to 85 kg, if 80 kg is taken, by K.E.= 1/2 m V ² = 1/2 x 80 x 10 ² = 4000J

4) C; Here, R= V/I = 1/0.5 = 2 ohms.
ΔR/R = ΔV/V + ΔI/I
ΔR/2 = 0.05/1 + 0.o1/0.5
ΔR = ± 0.14
%age uncertainty = (0.14/2)x 100 = 7.0%

5) D; precise and not accurate result would be very different from 9.81 but within close range of one another.

6) C The case is similar to that when an object moves in a circular path. The speed remains same, but the direction keeps changing (direction at each instant is tangent to the circle). The acc. is because the velocity changes, by the changing direction.

7) B; Acc. of free fall on this planet would be, 9.81/6 = 1.635 m/s^2
W=mg
= 30 x 1.635
= 49.05N

8) C; At the top when it is released, it has a maximum acc. of 9.81 m/s^2. After that it keeps decreasing and finally when terminal velocity is reached, acc. becomes zero. The graph is often confused with that of D, but remember that shape is for a V-t graph.

9) D; The principle states that for a closed system (in which no external forces act) the momentum is always conserved.

10) D; Relative speed of approach = Relative speed of separation. Also, in an elastic collision, total kinetic energy is conserved.

11) A; F=ma
2000-R = 750 x 2
R = 500N

12) A;

The drag is a resisting force in fluids ( like air resistance in air). This is similar to friction.
Upthrust is only because of the pressure difference on top and bottom of an object when in a fluid.
Weight is the greatest of all.

13) A; Forces in equilibrium form a head to tail vector triangle (clockwise or anti-clockwise).

14) A; moment = force x distance
clockwise moment = anticlockwise moment
W x a + F x h = W x 2a

15) D; fact. the concept of ρgh.

16) A; for potential energy, vertical height is always taken, so distance would be s.
The field is towards the right side, so left is positive and the right negative, which means that the potential energy will decrease. energy = workdone, therefore, w.d. = F x s

17) C; linear momentum is Always conserved. B and D rejected.
The collision here is inelastic, so K.E. is not conserved. However, the Total energy is still conserved because of the law of conservation of energy.

18) B; The speed is constant, therefore the Kinetic energy is constt, which would give a straight line graph.
The ball is falling in a fluid of constt density. The potential energy varies linearly with the height. Ball is falling, h is decreasing so P.E. decreasing at a constt rate.

19) C; efficiency = (*useful* output/ total input) x 100

20) B; same old repeated question. fact.
A ignored cuz container's pressure is nowhere discussed.
C rejected bcuz the collisions are elastic.
D put down cuz the weight is downwards, and the statement is just wrong.

21) A; In the derivation of this formula, these formulae/principles are used, P = F/A, F=W, W=mg, m= ρv and v=area x height. Out of these, only mass= density x volume is given.



22) B; The metal stretches and then returns to its original shape. there is an extension, therefore, elastic behaviour. It follows the *same* curve when it contracts, therefore no plastic deformation.


23) A; The spring is being compressed, therefore there is a decrease in its length. The original length was 100mm. When the l is 70mm, the *change* in length = 30mm
so, the energy stored is area under the graph between 70 and 100 mm.
i.e. 1/2 x 30x10³ x 6 = 0.09J

24) C; the young modulus is same for a material.

25) B; p is not the height of the highest point on the wave so is not the amplitude. on x-axis is time, therefore q is time taken by one complete wave to pass through a point, time period.

26) D; I is inversely proportional to x² so doubling x means I would be divided by 4. I is directly proportional to A², so √(8²/4) = √16 = 4.0 μm
This cal also be simplified to this,
I1/ I2 = A1² / A2²
1/(1/4) = 8² /x²
4/8² = 1/x²
x²= 16
x= 4.0 μm

27) D; V= f λ where speed is c and wavelength is x+x+x+x =4x (as x is the distance between an antinode and a node, and one wave has three nodes and two antinodes).

28) B; d= 1/N
dsinθ = nλ
(1/N)sinθ = 3λ

29) A; λ = ax/D
re-arrange it to give, x = λD/a
this clearly shows that fringe spacing "x" is directly proportional to wavelength and inversely proportional to "a" (slit separation).
when λ is halved, x decreases, when a is doubled, x again is halved, so its like (3/2) /2 = 3/4 = 0.75 mm

30) A; the horizontal component of v is vcosθ
and u is in the same direction as vcosθ
re-arranging this gives u/cosθ = v

31) C;
E = V/d = F/Q
5000/0.8x10-² = E = 625000 N/C
now, Q=ne (where e is elementary charge)
625000 = F/5(e)
625000 x 5 x elementary charge = 5 x 10-¹³N
the drop has *gained* five electrons so is now negatively charged and will be attracted to the positive plate i.e. will experience an upward force.

32) B; P = I² x R
[(1/2)² x 2] / [1 x 1]
= 1/2

33) C; P= work done/ time
12 = w.d. / 50
w.d. = 600 J
v = work done/ charge
V= 600/100
V= 6.0V

34) C; R= ρL/A
which can also be written as R = 4ρL/πd²
the volume stays the same. L is increased four times, so the diameter is halved.
X has R1 = 4ρL/πd²
Y has R2 = 4ρ4L/π(d/2)² = 16ρL/(πd²/ 4)
ratio comes, (ignoring the constants) (16 x 4)/4 = 16.

35) C; e.m.f. = w.d./charge
w.d. = 12 x 4 = 48 J
P = w.d. /t
t = w.d. /P
t = 48/24 = 2 sec.

36) D;
basic concept of resistance of thermistor. More the resistance of a component, more will the voltage across it be. No other changes are made therefore those with LDRs rejected. when the temp. is reduced, resistance of thermistor increases and so does the p.d.

37) A; net resistance of R2 and voltmeter = 50kΩ
effecctive R of circuit = 150 kΩ
Total voltage across R2 and voltmeter is 1/3 rd of 6 V (same as ratio of resistances i.e. 50/150) = 2 V
V= IR
2= I x 100 x10³
I = 2 x 10-⁵ A

38) B; simple set up for a potentiometer.

39) C; estimations we're supposed to learn.

40) B; By beta decay, the proton nmbr increases by one and there is no change on the nucleon nmbr. After alpha emission, N decreases by 4 and Z decreases by 2.
On these graphs, x-axis is labelled Z and y-axis N.

 

[geek101]

THIS IS GREAT! THANK YOU! you guys can also make videos to save all the typing

 

[arlery]

Has November 2005 been posted? What about November 2003 and 2002?

 

[Soldier313]

THIS IS GREAT! THANK YOU! you guys can also make videos to save all the typing

videos would be inconvenient i feel, just an opinion i would suggest pdf files, would make things a lot easier)

 

[XPFMember]

Has November 2005 been posted? What about November 2003 and 2002?

AsSalamoAlaikum Wr Wb!

nopes....they haven't ....
I've updated the first post..sorry for the delay..
@all those who posted those answers...JazakAllahu khairen kaseeran..may Allah reward you loads for that...Aameen!

 

[XPFMember]

MOD EDIT
AsSalamoAlaikum Wr Wb!

Here are few solved explanations for 9702 Physics Paper:1

June:2002

Nov:2002

June:2003

Nov:2003

June:2004

Nov:2004

June:2005

Nov:2005

June:2006

Nov:2006

June:2007

Nov:2007

June:2008

Nov:2008

^AsSalamoAlaikum Wr Wb!

First post updated with all the links to the solutions posted in this thread....!

 

[abcde]

OK here's all of June 2002.

June 2002
==========

37. D

Electric field strength is always a constant value for a field.

Hi! I was just going through this when I stumbled upon this mistake. For 37, the correct answer is A. This is because the plate separation is increasing so the field strength would decrease. I'm sure it was just a slip. The rest is all correct, Masha'Allah.

 

[leosco1995]

Hi! I was just going through this when I stumbled upon this mistake. For 37, the correct answer is A. This is because the plate separation is increasing so the field strength would decrease. I'm sure it was just a slip. The rest is all correct, Masha'Allah.

Hmm, yeah you're right. I got this wrong also the first time I did this MCQ because I didn't read the question properly. Thanks for letting me know.

 

[A.ELWY 7]

^AsSalamoAlaikum Wr Wb!

First post updated with all the links to the solutions posted in this thread....!

AsSalamoAlaikum Wr Wb!
im sry but i didnt understand what is this exactly !!!

 

[A.ELWY 7]

^AsSalamoAlaikum Wr Wb!

First post updated with all the links to the solutions posted in this thread....!

oooooohhhhh ok im sry now i understood...thnk u very much..