As physics p1 MCQS YEARLY ONLY.

[A.ELWY 7]

10 has 2sf
0.01 has 3sf

Im really sry...but u r Wrong!!!!
10 is 1 SF and 0.01 is also 1 SF

 

[Haniya Khan]

november 2003???????

 

[biba]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
Q ^26 plzzzzzz help me out!

 

[Haniya Khan]

OCT 2003 MCQ 30 PLSSSS HELPPPPP

 

[Thampi4]

June 2007

1: C...basic stuff
2: D..also basic stuff
3: im to0 sry i couldnt do this one and i would appreciate if some 1 can help me with it
4: C...one wavelength is 4 squares..so 4 x 2.5 = 10...amplitude is about 3.5 so 3.5 x 5= about 17
5: A...V=IR...1.2V / 0.48A= 2.5
6: D...again basic stuff
7: C...v is the vector sum of u and x and v=u +at so X is at so C
8: A...s= ut + 0.5at^2...get the time when the distance is 40m ..then when the distance is 30 m...then T1 - T2 = 0.38s
9: B..basic stuff
10: D...F=( mv - mu)/t ...so substituting into the equation it is ( p1 - p2)/ ( t2 - t1)
11: B... (20000 x 20) - (900 x 30) = 373000Ns
12: B...1st get the final V by: m1u1 + m2u2 =(m1 + m2)v so (2 x 4) + (4 x 1)= 6V..V = 2 then he said it stuck together so 0.5 x m x v^2 so 0.5 x 6 x 2^2= 12J
13: B...torque of a couple is defined as the 1 force x perpendicular distance between the 2 forces..so geth the perpendicular force by drawing a straight line perpendicular to the force..then use adj/hyp = cos( theta) so adj= 0.6 sin (30) so 0.52...0.52 x 8 = 4.2
14: C...basic stuff
15: A...P=pgh...so 100000/ (13.6 x 10^3) x 9.81 = h... so h = 0.75m
16: C... A is wrong because liquids doesnt have a very big seperation..B is wrong because it describes a gas..D is wrong because liquid atoms are not in fixed positions
17: C...copper is not brittle..and when drawn to a wire it will not return to its original shape again so plastic only
18: B... the whole area under the graph is the strain energy...and the unhighlited part which is between the 2 drawn lines is the heat enerfy..so the shaded are is B
19: A...1/2 F X so 0.5 x 2 x (0.90 - 0.50)= 0.40J
20: B...X is in tension as it is opposing the weight of the horizontal bar and the foce W...Y is aso in tension as it is opposing W..Z is in compression as it is between a force from X and a force of Y and W
21: C...A & B are wrong because no longitudnal waves are polarises..D is wrong because sound is longitudnal not transverse so C
22: A...I is proportional to a^2..try ti solve tis example by giving the intensity I of y a magnitude of 3600 and the intensity of X a magnitude greater than y by 10^12 then find the amplitude for both..it will be A the answer
23: D...im weak inthis kind of a question but i know that at Q it is not always zero at and R it is not entirley kinetic and at p the speed is not at maximum..so D
24: D..you must memorize this
25: C...basic stuff
26: D...A is wrong because decreasing the distance between the slits and the screen will decrease the fring seperation, B is wrong because this will also decrease the seperation..C is wrong because this will affect the intensity not the seperation..so increasing the frequency decreases the wavelength so seperation increase so D
27: A..stationary wave is when 2 owaves travelling opposite to each other ( so B and C is wrong) then he said 2 nodes and antinodes so a have 2 points shown so 2 nodes and anti nodes
28: D..basic stuff
29: A...again basic stuff to ne known..electron is attracted to +ve plate
30: B...E= v/d...so 4v / 2d so 2E
31: D..just memorize ohm's law..and becareful it is not A because he said they are equal not proportional
32: A...Q=It...(8 x 10^-3) x 0.020 = 0.16 mC
33: C....V1= 5000/ (5000+5000) x 2 = 1V then V2= 3000/ (3000 + 2000) x 2 = 1.2 so V1 - V2 = -0.2 so C
34: A...V was 4...then light increases so R decreases so V decreases and the only number less than 4 in the answers is 3V so A
35: C...W= QV so V = W/Q and I = Q/t
36: C... V= IR, i= v / r...3/ 2 + 4= 0.5..then pd= 0.5 x 4=2 then I= E/ (R+r) sp P = RI^2 so P= (E/R+r)^2 x R so (3/ ( 4+ 2)^2 x 4 = 1 so C
37: C...R = pl/A he siad they have same length..and they are of the same material so same resistivity..then he said of same volume and volume = A x h so same area so same Resistance so C
38: A...basic stuff
39: D..again B particle is 0 nucleon number and -1 is the proton number so D
40: C .... divide the proton number by the nucleon numbre for each element lithium will give the smaleest number so c

For no 3, its dimensional analysis(Finding units):
ur looking for v which is ms^-1
g=ms^-2
h=m
lambda=m

its A, because

Square root((ms^-2)*m)=ms^-1, which is the unit for speed

 

[A.ELWY 7]

For no 3, its dimensional analysis(Finding units):
ur looking for v which is ms^-1
g=ms^-2
h=m
lambda=m

its A, because

Square root((ms^-2)*m)=ms^-1, which is the unit for speed

Thnx....Thnx very much

 

[Capricedcapri]

DO ANY1 KNOE WHERE I CAN FIND PAPER 1 OF OCT NOV 2011?.........

 

[Thampi4]

Thnx....Thnx very much

no probs and thanks for explaining some of the answers, it really helped

 

[Sanis]

am sry...but i think i wrote that im the one who needs help with it..so how can i explain it to u??

v is in ms^-1,, u need 2 check which choices give the ratio of m:s 1:-1 .... it is A because g is ms^-2 * lamda in metres,, so the unit will be m^2s^-2 ,,, the square root of exponents, as in radical uncertainities is multiplied or divided,, if answer power 2 is given, the answer is multiplied by 2, if the answer is square rooted, then the answer is divided by 2, so square root of m^2s^-2, will be ms^-1

 

[Sanis]

guyz plz help in november 2009, it is the hardest paper ... if u have doubts in some questions leave them, other ppl will be there 2 make them clear

 

[XPFMember]

DO ANY1 KNOE WHERE I CAN FIND PAPER 1 OF OCT NOV 2011?.........

AsSalamoAlaikum Wr Wb!

They're available in the download section..
anyway here are the direct links...these are for all the variants...u just need to do variant 2 and 3.....1 is similar to one of the papers..


9702_w11_ms_11.pdf
9702_w11_ms_12.pdf
9702_w11_ms_13.pdf

9702_w11_qp_11.pdf
9702_w11_qp_12.pdf
9702_w11_qp_13.pdf

 

[Arpit17]

بسم الله الرحمن الرحيم
Hope I complete the post before I fall asleep.
May/June 2006
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf

1) B; force has a direction and energy does not.

2) D; Wavelength of visible light ranges from 4 x 10^-7 m to 7 x 10^-7m, option D lies in the range as 500 nm = 5 x 10^-7m.

3) D; voltmeter reading, temperature and charge have no direction, only displacement has.

4) D; I is proportional to the inverse of d^2. If 1/d^2 is taken on an axis, the graph formed is a straight line as that for a directly proportional graph.

5) C; simple unitary method, one division= 10ms. so four divisions as occupied by one wave = 4 x 10 = 40ms.

6) C; Uncertainties are added simply. 3+2 =5%

7) A; We need to find acc. (a). we have initial speed (u) = 0. Check for the three equations of motion. If we get the quantities mentioned in A,
s = 1/2 at^2 + ut can be applied to find a.

8) C; The area under the graph is the distance. 1/2 x 5 x 20 = 50m

9) D; mass is moving up and down. It has a zero velocity at two positions, once at its highest point and the second at lowest. B is that when it reaches the highest point. therefore at lowest D is the velocity.

10) B; Newton's third law states that action and reaction are always equal but opposite in direction. so the reaction of force exerted ON road will be the force exerted BY the road. so, push of the road on the back wheel is equal and opposite to that stated in question.

11) B; a fact.

12) A; m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
initial momentum is (m)(-2v) + (3m)(v) as they;re travelling in opposite directions. total initial momentum = 3mv-2mv = mv
after collision, both stick hence the masses added, momentum after collision = (4m)(V)
initial = final
mv = 4mV
V = mv/4m both the *m* cancelled out, so V = v/4

13) D; For a body in equilibrium, all arrows are joined head to tail.

14) B; Greater the force F, greater moment produced. More the distance from pivot (d) of F, more the turning effect. Smaller the θ, greater would its cosθ be therefore greater vertical component it shall have.


15) A; midpoint of the bar is at 1.2m, distance of 300N from pivot = 1.2 - 0.8= 0.4m
clockwise moment = 0.4 x 300 = 120N
anti-clockwise moment = 0.8 x 200 = 160N
According to the principle, both should be equal, therefore 160-120 = 40N is to be applied to the bar in clockwise direction to maintain equilibrium.

16) D; definition of internal energy.

17) D; For the first situation, u=10m/s , v=0m/s , s=10m
2as= v^2 - u^2
2 x a x 10 = 0^2 - 10^2
a= 100/20
a= -5 m/s²
so, for second situation a remains same.
u = 30m/s v=0 m/s and a = -5m/s²
same equation again, gives us 90m braking distance.

18) A; potential energy= mgh
mg means weight, here already given i.e. 4.0N
P.E is weight x vertical height
therefore, 4 x 30 = 120J

19) C; A is wrong cuz not all molecules are at same speed
B is wrong cuz there are forces of attraction between molecules.
D is wrong cuz fastest molecules leave the surface by evaporation.

20) A; seriously, you need an explanation for THIS???

21) C; F(load) is directly proportional to x, so F=kx. k is the constant n can be found by F/x where F is the load and x the extension of the spring.

22) B; area under graph., i.e area of triangle + area of trapezoid/trapezium
(1/2 x 500 x 10x10^-3) + [ (1/2) x (2x10^-3) x (500+550)]
2.5+1.05 = 3.55 J

23) A; fact

24) B; intensity α (amplitude)^2, so if intensity is doubled, amplitude becomes √2 that is almost 1.4 something. so options A and C out. the question than says that the frequency is halved, which means that the wavelength increases. D ignored cuz there wavelength decreased.

25) B; f=500Hz V=340m/s
V=f λ
so, lambda comes 0.68m
one wavelength means 360 ⁰ i.e, 2π
unitary method,
0.68 : 2π
0.17 : X
so, X = π/2

26) C; nodes are the only points stationary.

27) B; apply dsinθ = n λ
make d the subject of the formula, which shows that d is inversely proportional to angle, and directly proportional to n( no. of orders).
More lines per metre mean lesser distance between them, i.e smaller d, hence n decreases and angle increases.

28) D; λ= ax/d which can be re-arranged to give, x= λd/a
d is doubled, x increases. a decreases, x again doubled. i.e. it is doubled twice, 4x2x2 =16mm

29) B; More positive to less positive. between the plates field is uniform and circular on sides as in B and D.

30) A; E = V/d also, E= F/Q
here Q=e
V/d = F/e
so eV/d =F

31)C; I=Qt
time is taken as 1 second, so Q= 4.8C
also, Q=ne where e is elementary charge.
Q/e =n
n= 4.8/(1.6x10^-19) = 3 x 10^19 per sec
the direction of flow of electrons is asked, which is from negative to positive, therefore Y to X.

32) B; V=IR defines resistance... always.

34) A; V= I (r+ R)
r being the internal resistance.
so, 12 = I (1+3)
12/4 = I = 3A
rate of energy supplied to the *heater* is required,
the p.d across heater n voltage source is in the ratio as that of their resistances.
so p.d across heater is 3/4 of 12V = 9V
by P=VI
3 x 9 = 27W

35) B; there are two loops, one with a single resistor and another with three resistors. each loop's resistance is 10 and 30 Ω respectively. In parallel the effective resistance of circuit is even lower than the lowest resistance of any loop, so it wud be between 1 and 10.

36) C; Brightness depends upon the current. In each diagram, the current is divided among two loops each with two resistors so curent reaching each bulb stays the same, so does its brightness.

37) B; Resistance is directly proportional to the voltage/p.d. More resistance more p.d. across thermistor. Less resistance of LDR, less p.d across it, also contributing to more p.d. across thermistor.
38) A; has to be learned... again a fact.

39) D; there is no change on the proton or nucleon nmbr by emission of gamma rays. By beta emission, however, only the proton nmbr increases by one. it has no chnge on the nucleon nmbr.

40) B; neutron nmbr = nucleon nmbr - proton nmbr
220-86 = 134
216-84 = 132

hey! sorry i'm late,but wat about question 33?! :S

 

[Capricedcapri]

AsSalamoAlaikum Wr Wb!

They're available in the download section..
anyway here are the direct links...these are for all the variants...u just need to do variant 2 and 3.....1 is similar to one of the papers..


9702_w11_ms_11.pdf
9702_w11_ms_12.pdf
9702_w11_ms_13.pdf

9702_w11_qp_11.pdf
9702_w11_qp_12.pdf
9702_w11_qp_13.pdf

thank you soo much........actually my net used to get stuck there............

 

[Gémeaux]

hey! sorry i'm late,but wat about question 33?! :S

uh oh, somehow missed the question. Here is its answer.

 

[A.ELWY 7]

v is in ms^-1,, u need 2 check which choices give the ratio of m:s 1:-1 .... it is A because g is ms^-2 * lamda in metres,, so the unit will be m^2s^-2 ,,, the square root of exponents, as in radical uncertainities is multiplied or divided,, if answer power 2 is given, the answer is multiplied by 2, if the answer is square rooted, then the answer is divided by 2, so square root of m^2s^-2, will be ms^-1

Thnx

 

[Arpit17]

uh oh, somehow missed the question. Here is its answer.
View attachment 12659

owkay,got it! thanx a lot! and goodluck with ur paper! (if u're doing it)

 

[Tkp]

Im really sry...but u r Wrong!!!!
10 is 1 SF and 0.01 is also 1 SF

bro can u please give me the links of o/n 01 qp only and mj01.pls

 

[Arjun Dhanak]

Hey I have a doubt in qs 8 paper 9702/11/O/N/11..
Pls help me out...
thx in advance
n gud luk to all...

 

[Tkp]

15 is a becaus

November 2005
=============

Q1. B

Fact

Q2. C

Fact

Q3. D

volt = work/charge
kgm^2s^-2 / As

= kg ms^-2 s^-3 A^-1

Q4. B

Results are said to be precise if the values are within 1 mm of their mean. Accurate is how close the values obtained are to the true value. Here, none of them are within 1 mm to 895 mm.

Q5. C

Vo = (5*2*1) = 10 (principle volume value)
Vu = (0.01/5) + (0.01/2) + (0.01/1) * 10 = 0.17

So volume = 10 +/- 0.17
And mass = 25 +/- 0.1

Uncertainty in density = (0.17/10) + (0.1/25) * 2.5 = 0.05

Q6. C

The acceleration will decrease until it reaches 0

Q7. From 0 to x,

s = 0 + 0.5a * t1^2
s = 0.5a * t1^2

From h to x,

s = 0 + 0.5a * t2^2

For h - x,

h = 0.5a * t2^2 - 0.5a * t1^2
h = 0.5a (t2^2 - t1^2)
a = 2h / (t2^2 - t1^2)

Q8. A

Initially, as the force is 0, acceleration is 0 (F = ma)

Therefore the speed will initially be 0, as in all graphs

Once the force becomes a constant value, the acceleration is constant but non-zero, so the velocity increases linearly

9. D

Fact

10. A

Mass is always constant, so C and D are wrong

gravitation field on P = W/M (since mg = W)
= 1/1 = 1

on Q, it is one-tenth so 1/10 * 1 = 0.1

Weight of mass on Q = 1 * 0.1 = 0.1 N

Q11. A

Only acceleration will act and that too in direction XY only since its part of the vertical component

Q12. D

Clockwise = 20 * 0.4 = 8 Nm
Anti-clockwise = 10 * 0.6 + 100 * 0.1 = 16 Nm (don't forget the weight of the beam!)

Therefore we need a clockwise moment of 8 more Nm

(20 * x) = 8, x = 0.4m from the pivot, so D

13. A

Resultant torque = 45 N and resultant force = 60 N to the right

14. C

0.5 * 1400 * 30^2 = 630 kJ

15. B

Ep decreases linearly with height above the ground.

EP = mgh

If h is on the x-axis and EP on the y-axis, then the gradient mg would be a constant

Q16. C

Tension = mg sin θ = 10^3 sin 30 = 500 N (note: weight was already given, so need to multiply by 9.81)

Work = force * distance moved in direction of force = 500 * 5 = 2500 J

Q17. C

Fact. Heating a gas gives more K.E to the gas molecules so they hit the wall containers more often. Some statements, e.g. B are correct but aren't relevant to the question, so it's important

to read these type of questions carefully.

Q18. C

P(X) = P(Y)

ρgh = pgh

800 * g * h1 = 1200 * g * h2
800h1 = 1200 h2

C is the only answer which is correct for this equation

Q19. B

White sugar granules appear as white small crystals, obviously so it's crystalline. When something is melted quickly (i.e. supercooled) and appears to be sort of brittle, then it becomes

amorphous. So B. You need to learn the properties of crystalline, amorphous and polymeric solids well.

Q20. B

B is the net work done stretching the sample

Q21. C

E = FL/Ax (where x = extension and E = Young Modulus), rearranging to give 'x' as the subject gives us:

x = FL/AE (E is a constant which MUST remain the same because its the same material)

Half diameter = 1/4th of the area and quarter length = 1/4th of length

ratio of new x = (F * 0.25L) / (0.25A * E)
= 1

Therefore the extension remains the same, 8 mm.

Alternatively, you can use the spring constant to solve this:

60 = k * (8/1000)
k = 7500

Since the forces are the same,

F1 = F2
ke = kz (where z is the new extension)

7500 * (8/1000) = 7500 * z
z = 8 mm

Q22. D

If a wave is to be polarized it must be transverse

Q23. B

In A and C, the amplitude is marked incorrectly. In D, λ is actually the time period.

Q24. B

I α a^2 and I α f^2.

Rather than doing all the math to do this, compare the amplitude and frequency of the waves and use the formula to figure out this stuff:

If Q's amplitude is twice as much, the intensity will be four times as much.
If Q's frequency is half that of P, the intensity will be one-fourth.

Net change = 0, so the intensity remains the same.

Q25. C

λ in water = 1500/150 = 10m and λ in air = 300/150 = 2m

Q26. D

X and Y are adjacent anti-nodes, and they (as well as adjacent nodes) are 0.5λ apart.

Q27. C

Fact, from the definition of diffraction. Light bends when it passes through an aperture or narrow slit

Q28. A

x = λd/a

Halving λ also halves x so 0.75 mm.

Q29. C

Since the adjacent 1st orders are 60 apart, that means one 1st order from the undeviated beam is 30 apart.

1.15 * 10^-6 sin 30 = 1 * λ
λ = 575 nm

Q30. A

Electric field lines go from + to -, so the +ve particle will move down towards the -ve side.

Q31. A

Fd = VQ
F = (200 * 0.005) / 1.6 * 10^-19
F = 6.4 * 10^-15 N

Q32. B

Graph X = diode
Graphy Y = ohmic
Graph Z = lamp

Q33. D

Increasing the strain (i.e. extension) increases the length and reduces the cross-sectional area, and since R = ρL/A, this means R increases.

Q34. C

One way of doing this accurately is drawing a straight line from the origin and seeing which option is the closest, since R is the ratio of V:I. Or if you are too unsure about that, make an accurate scale with your ruler and calculate the values. That's a waste of time though.

Q35. B

Variable resistor = box with diagonal arrow through it
Fuse = box with straight line through it
LDR = box with 2 arrows "shining" onto it
Thermistor = box with a diagonal line through it, with an added small straight, horizontal line at the bottom

Q36. B

Total I = V/R = 6/450 = 0.0133... A

V through 180 resistor = 0.0133... * 180 = 2.4 V

Q37. C

Readings on both VT and VL are high, so the voltmeter readings must be high. Since the voltmeter is connected directly to the LDR, a high resistance indicates a high voltmeter reading (because V = IR). An LDR gives high resistance in the dark, so the light level should be low.

For VT, the thermistor is NOT connected directly to the voltmeter, but the fixed resistor is, so V = IR doesn't apply for the thermistor but for the fixed resistor instead. Decreasing the resistance on the thermistor will give a high VT reading (you can prove this using the potential divider formula). And a low resistance on the thermistor means a high temperature.

Q38. D

Emission of a β particles increase the proton number by 1 but doesn't affect the nucleon number.

Q39. B

Both particles will be deviated upwards, but the one closer will deviate more because it's closer.

Q40. D

To balance the equations, the nucleon and proton number of X must be 1 and 1 respectively. This is a proton.

15 is a because they are thrwing the ball upwards so ep increases with height

 

[A.ELWY 7]

bro can u please give me the links of o/n 01 qp only and mj01.pls

sry....i Dont have them