As physics p1 MCQS YEARLY ONLY.

[Albert Einstein]

Nov 2010 paper 12. soln missing plzz upload

 

[bamteck]

Nov 2010 paper 12. soln missing plzz upload

June 2010 & Nov 2010 with all variants missing !
Please upload. It helps tremendously !

 

[leosco1995]

For the time being, I could answer specific problems to those papers. I am not sure if anyone would be willing to make solutions for entire papers again though. It takes a lot of time, especially on a computer. But hopefully someone will post solutions to all papers in October when the exam is around the corner.

 

[bamteck]

For the time being, I could answer specific problems to those papers. I am not sure if anyone would be willing to make solutions for entire papers again though. It takes a lot of time, especially on a computer. But hopefully someone will post solutions to all papers in October when the exam is around the corner.

Why can't you upload them ?

 

[leosco1995]

Why can't you upload them ?

Upload what? Detailed solutions? I could but I'm really busy at the moment (SAT, school, university, etc). It takes a lot of time to make detailed solutions for a single paper.

 

[bamteck]

Upload what? Detailed solutions? I could but I'm really busy at the moment (SAT, school, university, etc). It takes a lot of time to make detailed solutions for a single paper.

Yeah the detailed solutions. Please try to make an effort to upload them when you have time. We really need them. Thank you in anticipation
take care

 

[leosco1995]

Yeah the detailed solutions. Please try to make an effort to upload them when you have time. We really need them. Thank you in anticipation
take care

I can't promise anything, but after I get all of my major things sorted out (SAT, university, etc.) like I said, I will try to upload solutions to those past papers. Hopefully someone else too will help as well.

 

[yumichikabyakuya renji]

November 2002

1) C;
micro: 10-⁶
nano: 10-⁹
pico: 10-¹²

2) C; uniform acc. means that the gradient for V-t graph wud be a diagonal straight line (velocity changes at a constt. rate).

3) C;
View attachment 12497

4) B; basic definition.

5) D; The values are way different from the actual that is 9.81 m/s² but are close to one another... are precise but not accurate.

6) C; systematic error causes deviation from the actual value, therefore the graph does not include X0. Random errors cause the result to be spreaded out, the graph comes wide. In C, the graph is narrow, and well away from actual.

7) B;
View attachment 12498
5 squares in the figure have in them 7 waves. One wave therefore occupies,
5 : 7
x : 1
x= 5/7 square.
One square represents 10ms, so Time period is 10 x 5/7 = 7.14 ms
Hence the frequency is 1/7.14 x 10-³ = 140Hz.

8) A; Horizontal velocity remains constant therefore the hor. component of acc. is zero.

9) D; s= ½ at2 + ut
where, ut =0.
Therefore a= 2h/time = 2h/(t22 - ti2)

10) D; In air resistance, the acceleration decreases from 9.81 to 0 ms-2. No other graph shows this.

11) A; for elastic collisions, e= -1
uA –uB = e( vA – vB)
uA - uB = -vA + vB

12) the masses are equal so suppose it to be *m*
Initial momentum = Final momentum
60m – 30m = 2mV
30m = 2mV
30m/2m = V
V= 15cm/s

13) A; equal and opposite forces form a couple.

14) C; upward force = tension in string = 20 x 9.81 = 196.2N
Sum of clockwise momentum = Sum of anti-clockwise momentum
Distance d x ( 50 x 9.81) = (100-40) x ( 196.2)
490.5d = 11772
D= 24 cm from the pivot.
This could either be at the mark of 16 cm or 44 cm.

15) A; equilibrium triangle.

16) C; P= Fv

17) D; constant speed down the hill therefore there is no change in kinetic energy.

18) D; F=W= mg
=1.3 x 109 x 9.81
=1.275 x 1010N
P = F x s /t
= (1.275 x 1010 x 2)/(60x60x24)
=295208
= 300kW

19) C; let m be the mass,
P.E. =mgh
=1962m.
60% of the energy left.
K.E = 60/100 of 1962m = 1177.2m
1177.2m = ½ m V2
V= 48.5m/s

20) C; Fact.

21) C; apply the formula p = ρgh
For the first liquid, pressure comes 3531.6
For the other liquid, it comes 7063.2 which is twice the previous one.

22) A; quite basic this is.

23) C; area under the graph within limits.

24) B; F= kx
F is constt., k is given, as stated, therefore if k is doubled x is halved.
2k would give x/2
Next, we know that energy = work done = ½ kx2
WP= ½ * 2k * (x/2)2
=kx2/4
WQ= ½ * k * x2
= kx2/2
So, WP= ½ * kx2/2 = ½ WQ

25) A; wavelengths are to be learned of the electromagnetic spectrum.

26)D; maximum displacement of a point is its amplitude. On x-axis is time, so the labeled part is T.

27) D; I1/I2 = (A1/A2)2
3/I = a2/4a2
3 x 4a2 = I x a2
12 = I

28) B; use the formula λ =ax/D so the wavelength is directly proportional to the fringe separation.

29) D; use the formula dsinθ = nλ
Where d is spacing of the lines on the grating, λ is 590 x 10-9m, θ = 43/2 o.

30) C; V=IR

31) D; resistance of a filament does increase with a rise in temperature.

32) A; at +1.0V, current I = 50mA
V=IR
1/50x10-3 = 20 Ω
The graph at -1.0V is a straight line downwards so is infinite.

33) A; fact, charge entering a point must leave that point.

34) B; By Kirchhoff’s Law, I =I1+I2
Using V= IR, we get I =V/R
V/R = V1/R1 +V2/R2
The voltage across each is same, V =V1 =V2
Hence, 1/R = 1/R1 +1/R2

35) D; more resistance across variable resistor means more p.d. across it leaving decreased p.d. across XY. Then more portion of the wire would be used to maintain the galvanometer at zero deflection.


36) A;
View attachment 12495

37) B; field lines enter the negative point and originate from the positive.

38) A; isotopes have same number of protons, and different nucleon nmbr.

39) A; Only those that hit the nucleus directly bounce back or those that go quite near the nucleus are deviated at such large angles. As nucleus occupies a very small portion of an atom, this proportion of alpha- particles is really small. For more explanation, kindly take reference from your book.

40) C; it absorbed a neutron so nucleon nmbr increases by one. It emits two beta particles, so the proton nmbr increases in total by 2.

hey u explained Q36 really well

bt i'v a doubt for Q10, u said the acceleration becomes 0, but A also shows shows this, cud u clear my doubt plz??

i also didn't get Q13, the force F at the centre will have no turning effect, but the other one will, then they aren't in equlibrium or am i wrong but i know the other ones too aren't correct, so do we proceed by elimination?

And Q35 , cud u explain it with the equation, i mean, for e.g. when R increases, V will increase something like that,,,

thXx

 

[yumichikabyakuya renji]

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.

hey good one this really helped me correct all my mcqs n understand them well
but for Q15: answer is A
when height fr0m gr0und is zer0 (i.e object is on ground) P.E is also zer0
but when height increases, pe also increase
B is showing that at max. height. p.e is zero which is wrong.. :O

 

[yumichikabyakuya renji]

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.

really helppful advice as well thXx very very much

 

[bamteck]

really helppful advice as well thXx very very much

You help us now for P5

 

[yumichikabyakuya renji]

You help us now for P5

hmm, well i wud like to, bt am not that good in P5, still i cn try 2

 

[yumichikabyakuya renji]

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

hey thxx can u explain the last one? y do we need 2 divide proton no. by nucleon no.? is it a specific definition?

 

[Prawesh Dahal]

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

I did not get the 10th Question !
F=mv-mu/t right !
So mv = - p2
mu = p1
so mv-mu = -p2-p1 ???????? so how come p1-p2 ???

 

[Prawesh Dahal]

A

November 2007

1-C
Use base units and equate both sides.

2-A
Learn the approximations.

3-B
Resultant force is always is opposite direction of its two components in the triangle of forces.

4-D
Precision is the closeness of measured values.
Accuracy is the closeness of measured values to actual value.

5-C
P = fv
f = ma and v = s/t

P = add up all the % uncertainties.
%P = 0.1 + 1 + 1.5 + .5
= 3.1 %

6-A
Due to zero error, the initial value of angle of deflection will be greater than zero at zero current.

7-A
Basic concept that g gravity is the acceleration of free fall.

8-D
The gradient of velocity-time graph is acceleration.


9-B

Area of distance above x axis - area of distance below x axis.
(0.5 * 3 * 30) - (.5 * 2 * 20)
= 25 m

10-D

Driving force - frictional force = ma
12- Fre = 0.6(4)
12- 0.6(4) = Fre
9.6 N = Fre

11- B
View attachment 12245

12-A
Since the parachutist is falling, height is decreasing over time.
Gradient of distance-time graph is speed.

13-C
Use head to tail method.

14-D
At midpoint, d = 0.8/2 = 0.4m
Moment = F*d
12 = F * 0.4
30 N = F

15-B
Initial K.E. - Final K.E.
(0.5* 1000 *25^2) - (0.5 * 1000 * 5^2)
300 000 J
300 kJ

16 - C
At max d, K.E. = minimum and elastic potential energy = maximum.

17-C
Density = mass/volume
Hence higher density means a higher mass, hence more no. of atoms.
So MpNp > MqNq

18-A
Height at surface =?
Pressure = density * g* h
100 000= 1030 * 9.81 * h
9.896 m = h1

Height at 450 kPa =?
450 000 = 1030 * 9.81 * h
44.535 = h2

Therefore height below surface = 44.535 - 9.896
=34.639 ~ 34.6m

19-B
Basic definition.

20- A
Strain energy = 1/2 * F * extension
= 0.5 * 25 * (0.4 - 0.2)
= 2.5 J

21- B
View attachment 12247

22- B
Learn the wavelengths.

23- B
Max speed = 2pii *a *f
f = ?
s = f* lamda
8 = f (50)
0.16 Hz = f

Max speed = 2pii * 2 * 0.16
=2 m/s
Max K.E. = .5 m* v^2 = 0.5 * 2 * 10^-3 * 2^2
= 4 mJ

24-D
x = lamda * D/ a

Hence increasing the lamda, increases x, that is the fringe separation.

25-B

d sin theta = n* lamda
At n = 3, theta = 45 degrees
d sin45 = 3*lamdaa
0.7 d = 3 lamda

and max angle = 90 , so n =?
d sin90 = n*lamdaa
d = n* lamda

Using the ratio method:
If 0.7d = 3 lamda
1 d =?
cross multiply and you get n as 4.25 hence rounding it off to 4th order.

26- C
Since electric field direction is from higher p.d./charge to lower p.d./charge the electron will be attracted towards the more positively charged plate (or direction) hence to the left.

27-D
Direction of electric field is away from positive charge.

28-B
E = V*Q
E/Q = V

29-D

P = V^2/R
P = 12^2/Rx and P = 6^2/Ry
Rx = 144/P and Ry = 36/P

Rx/Ry = 144/P divided by 36/P
Rx/Ry = 144/36 = 4

30-D

V = IR
6 = I (10 +10)
0.3 A = I

Q = It
0.3 * 60 = 18 C

31-A

View attachment 12248

32-A

Basic concept of L.D.R.; when light intensity increases, resistance decreases hence voltage decreases too.

33-D
V in = R1/T.R. * V out
4.8 = 10/25 * V out
4.8* 25/10 = V out
12 = V out

34-D
Using the ratio method:
If 1.1 V - 0.7 m
? - 0.9 m
x = 0.9* 1.1/ .7
x = 1.4 V

35-A
The readings won't change because the set up is the same.

36- C
Charge is same since proton no. is same, but mass differs since no. of neutrons differs.

37- C
Basic concept.

38-C
The alpha scattering experiment proves the small size of a gold nucleus.

39-B
In a beta emission, the nucleon no. remains same but the proton no. increases by 1
For 2 beta emissions, proton no will increase by 2 hence 40 +2 = 42

40-A
Momentum = mv, greater m = greater momentum.
hence the particle with the greatest mass from all the choices is A, an alpha particle.

Actually, there is a mistake in Question 7 ! g is not gravity ! Its given g is acceleration of free fall ! so W=mg and hence g is ratio of weight by mass ! ...Its C in the mark scheme too !

 

[AArmas]

I know this is meany to be multi choice but does any1 know if Q5 in nov 2002 long answer paper is actually wrong in the mark scheme?

 

[Abdul Moiz]

thanks

 

[syed1995]

How about completing with the rest of the papers til November 2012 ... ?

 

[magnesium]

help required in p1 2012 nov component 11 !!!!!!!PLZ

 

[syed1995]

help required in p1 2012 nov component 11 !!!!!!!PLZ

which question?