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Umm alright I'll try my best to,
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As physics p1 MCQS YEARLY ONLY.
- Thread starter fatima 007
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thnx very much...between today and tommorwo ill be psting the 3 variants of the Nov 2011Umm alright I'll try my best to,
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
please explain Q9 ASAP
please explain Q9 ASAP
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For periodic motion, velocity at the highest/lowest points is always zero. Also, the lowest point would come after the end of a period with 'down' (negative) velocities. The lowest point is thus D.
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Some one plz post Nov 2005
JAZAK ALLAH KHAIR
JAZAK ALLAH KHAIR
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I'm in the middle of doing November 2005. Should be done within 30 minutes, InshaAllah.
Check the first page to see what papers are done.
Check the first page to see what papers are done.
heyyyy plz upload 2011 solved papers first it wl b so nice ov u
nd thnx 4 da rest of them God bless u...- Messages
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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.
=============
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 i think we dont need for variant 3. Because variant 1 and 3 have the same questions except that the order is different.
I can help too. InshaAllah
w 2010 if it aint booked by anybody
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http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf
qs 18, 19, 24 and 36 ???
qs 18, 19, 24 and 36 ???
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can someone please upload the examiner report for mj/01, on/11 and question paper for on/01
thanx soo much
thanx soo much
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Nov:2002, Nov: 2003
Can someone please upload the solved explanations for these urgently....thanx so much!
Can someone please upload the solved explanations for these urgently....thanx so much!
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I'd post Nov 2002, but that'd take about two hours.
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yes anybody can help here...and thnx, yes they are the same i didnt notice this!!Hey i think we dont need for variant 3. Because variant 1 and 3 have the same questions except that the order is different.
I can help too. InshaAllah
w 2010 if it aint booked by anybody
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ill post today inshallah the Nov 2011 P12....and then tommorw variant 3
- Messages
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owww please do itI'd post Nov 2002, but that'd take about two hours.
it would really benefit all of us soo much!
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NOV 2011 VARIANT 2
1: C....W=QV, so W which is = work done = energy..so C
2: D......basic stuff
3: D......again, basic stuff
4: D.....when we add or subtract numbers..we just add their uncertanities so 0.02 + 0.03 = + or - 0.05 so D
5: B....A is ofcourse wrong, C and D are systematic errors so B
6: C......area under the graph= 5 x 20 x 0.5 = 50
7: B..... terminal velocity, means no acceleration means no reusultant force so..acceleration is zero..then resultant force is zero..so the force of gravity or the weight must be = to the air resistance..and ofcourse weight is no zero, it is mg..so air resistance = weight = mg so no resultant force so B
8: D....C will not help to deduce anything from the graph..Ais wrong because acceleration is constant..and B is wrong because maybe when the d was zero..the time when it is zero,, they didnt start to count from 0 when it was stationary..maybe they started to count when the car started to move..D is correct because the line is not straight
9: D..... u can find the avg force..and the momentum..as the rate ofchange of momentum is Ft...and u can find the time of impact but u wont find the acceleration from the graph
10: B....using the conservation of momentum..when the sand is added the mass increases so the v decreases..then this v is constant even if the sand was removed again..so not because we remove the sand it will be faster..this is wrong..to make it faster again a forceneeds to be applied..so B
11: A....again, conservation of momentum...m1u1 + m2u2= (m1 + m2) v so (20 x 6) + (20 x -15) = 32v so v =1.9 ( NOTE: i used -15 because it was travelling in the opposite direction)
12: D...basic stuff
13: D..again basic stuff
14: B... F x d = F x d= ( 30KN x 10m) = F x 20m so F = 15Kn...i used F x 20 because the trailer causes a force at point X..but the opposite force exerted by the cab is on all the trailer..and the trailer is 20m long
15: sry..i got wrong
16: B...first we get v using K.E = 0.5 m v^2 so 4.5 x 10^5 = 0.5 x 1000 x v^2 so V = 30 then v^2 = u^2 + 2as and a = F/m so V^2/ 2 x ( F/m) = s then 30^2 / 2 x ( 6000 / 1000) = 75m
17: A...efficiency = ( useful output energy) / ( total input energy) so 8J is the useful output energy..and the total input energy is 100..because 92J is lost as heat so 92 + 8 = 100 so 8/100 = 0.08 x 100 = 8%
18: D...basic stuff
19: again basic stuff
20: A...Prssure = F/A so 290 / 0.036 =8056 PA..then P= pgh so h = 8056/ ( 930 x 9.81) = 0.88m
21: C...2 and 3 are definetley correct as the graph shows..but one is wrong..yes the extension of P is greater but not twice
22: A...for a force of 5N we had an extension of 3 cm...so for a load of 2N what will be the extension??? cross multiply and it will be (3cm x 2N) / 5N = 1.2 cm so the Total extension is 3 + 1.2 = 4.2cm
23: B...easy question, just compare the graphs
24: basic stuff
25: anotherbasic stuff THAT HAVE TO BE MEMORIZED
26: B...i made this question mentally so i dont know if my way is correct..but i got the correct answer...at t = 18 the phase difference is 180 degrees....he wanted when is the phase difference will be 1/8 of the phase difference..so 180 x 1/8 = 22.5..so the phase difference is about 22.5..the graph is divided by an interval of starting from the left side where the amplitude of the wave is negative..i started to compare..at time 4s and 8s and 9s..the difference looked greater than 22.5 so i found 4.5 the closest one
27: B... simple question
28: A.....B is wrong....interference only occurs when there are two COHERENT SOURCES not uncoherent..C is wrong as light can not be polarised..and D is wrong as it have nothing to do with interference so A..as white light will produce 7 kinds of lights each with different frequency
29: D...E= v/d = 9/ 4 x 10^-3 so 2250 is about 2.3 x 10^3
30: A...D and c are wrong..electron is attracted towards the +ve plate which is upward..then the mass of the electron is 1/2000 of a proton..and an alpha particle will take the path B which is shown in the question..an electron is much lighter so it will be A as the force acting on it will be greater
31: A...B and D are wrong... C is wrong because in an electric field..an electron will not be attracted in a straight line..like the question 31...the force must be like a tangent to the field lines
32: C... I=Q/t so if V = IR then V = (Q/t) x R so ( 8 / 20) x 30 = 12
33: D...Basic stuff
34: A...R = x= pl/A = pl/ ( pie ) (d/2)^2 and Ry = p x 2l/(pie) (2d/2)^2..divide Rx by Ry and it will be 2/1
35: A....I= E/ ( R + r) and P=I^2 x R so P = [(E/ ( R+r)]^2 x R substitute the values given for the variable resistor..starting with 0.5 till 4 ex: for the variable resistor try 0.5 then 1 then 1.5 then 2 etc..ull find the power value increased then started to decrease similar to graph A
36: C...basic stuff..must be memorized
37: B..using the equation of a potential diviser..V output = [ R1/ ( R1 + R2)] x V = try when the variable resistor was o and also 4.5..as for minimum and maximum..ex 0/( 0 + 1) x 25 = 0 and 4/ ( 4 + 1) x 25 = 20 so B
38: A...the volt metre and the resistor are in parallel so their total resistance is 100 using the equation of the resistors in parralel..then we will substitute in the same equation i explained in the previous question..it will be (100 /( 400 + 100) x 60 = 12V
39: C..basic stuff..but note that B is wrong because he said helium atom...it must be helium nucleus
40: C..again simple question
1: C....W=QV, so W which is = work done = energy..so C
2: D......basic stuff
3: D......again, basic stuff
4: D.....when we add or subtract numbers..we just add their uncertanities so 0.02 + 0.03 = + or - 0.05 so D
5: B....A is ofcourse wrong, C and D are systematic errors so B
6: C......area under the graph= 5 x 20 x 0.5 = 50
7: B..... terminal velocity, means no acceleration means no reusultant force so..acceleration is zero..then resultant force is zero..so the force of gravity or the weight must be = to the air resistance..and ofcourse weight is no zero, it is mg..so air resistance = weight = mg so no resultant force so B
8: D....C will not help to deduce anything from the graph..Ais wrong because acceleration is constant..and B is wrong because maybe when the d was zero..the time when it is zero,, they didnt start to count from 0 when it was stationary..maybe they started to count when the car started to move..D is correct because the line is not straight
9: D..... u can find the avg force..and the momentum..as the rate ofchange of momentum is Ft...and u can find the time of impact but u wont find the acceleration from the graph
10: B....using the conservation of momentum..when the sand is added the mass increases so the v decreases..then this v is constant even if the sand was removed again..so not because we remove the sand it will be faster..this is wrong..to make it faster again a forceneeds to be applied..so B
11: A....again, conservation of momentum...m1u1 + m2u2= (m1 + m2) v so (20 x 6) + (20 x -15) = 32v so v =1.9 ( NOTE: i used -15 because it was travelling in the opposite direction)
12: D...basic stuff
13: D..again basic stuff
14: B... F x d = F x d= ( 30KN x 10m) = F x 20m so F = 15Kn...i used F x 20 because the trailer causes a force at point X..but the opposite force exerted by the cab is on all the trailer..and the trailer is 20m long
15: sry..i got wrong
16: B...first we get v using K.E = 0.5 m v^2 so 4.5 x 10^5 = 0.5 x 1000 x v^2 so V = 30 then v^2 = u^2 + 2as and a = F/m so V^2/ 2 x ( F/m) = s then 30^2 / 2 x ( 6000 / 1000) = 75m
17: A...efficiency = ( useful output energy) / ( total input energy) so 8J is the useful output energy..and the total input energy is 100..because 92J is lost as heat so 92 + 8 = 100 so 8/100 = 0.08 x 100 = 8%
18: D...basic stuff
19: again basic stuff
20: A...Prssure = F/A so 290 / 0.036 =8056 PA..then P= pgh so h = 8056/ ( 930 x 9.81) = 0.88m
21: C...2 and 3 are definetley correct as the graph shows..but one is wrong..yes the extension of P is greater but not twice
22: A...for a force of 5N we had an extension of 3 cm...so for a load of 2N what will be the extension??? cross multiply and it will be (3cm x 2N) / 5N = 1.2 cm so the Total extension is 3 + 1.2 = 4.2cm
23: B...easy question, just compare the graphs
24: basic stuff
25: anotherbasic stuff THAT HAVE TO BE MEMORIZED
26: B...i made this question mentally so i dont know if my way is correct..but i got the correct answer
...at t = 18 the phase difference is 180 degrees....he wanted when is the phase difference will be 1/8 of the phase difference..so 180 x 1/8 = 22.5..so the phase difference is about 22.5..the graph is divided by an interval of starting from the left side where the amplitude of the wave is negative..i started to compare..at time 4s and 8s and 9s..the difference looked greater than 22.5 so i found 4.5 the closest one 27: B... simple question
28: A.....B is wrong....interference only occurs when there are two COHERENT SOURCES not uncoherent..C is wrong as light can not be polarised..and D is wrong as it have nothing to do with interference so A..as white light will produce 7 kinds of lights each with different frequency
29: D...E= v/d = 9/ 4 x 10^-3 so 2250 is about 2.3 x 10^3
30: A...D and c are wrong..electron is attracted towards the +ve plate which is upward..then the mass of the electron is 1/2000 of a proton..and an alpha particle will take the path B which is shown in the question..an electron is much lighter so it will be A as the force acting on it will be greater
31: A...B and D are wrong... C is wrong because in an electric field..an electron will not be attracted in a straight line..like the question 31...the force must be like a tangent to the field lines
32: C... I=Q/t so if V = IR then V = (Q/t) x R so ( 8 / 20) x 30 = 12
33: D...Basic stuff
34: A...R = x= pl/A = pl/ ( pie ) (d/2)^2 and Ry = p x 2l/(pie) (2d/2)^2..divide Rx by Ry and it will be 2/1
35: A....I= E/ ( R + r) and P=I^2 x R so P = [(E/ ( R+r)]^2 x R substitute the values given for the variable resistor..starting with 0.5 till 4 ex: for the variable resistor try 0.5 then 1 then 1.5 then 2 etc..ull find the power value increased then started to decrease similar to graph A
36: C...basic stuff..must be memorized
37: B..using the equation of a potential diviser..V output = [ R1/ ( R1 + R2)] x V = try when the variable resistor was o and also 4.5..as for minimum and maximum..ex 0/( 0 + 1) x 25 = 0 and 4/ ( 4 + 1) x 25 = 20 so B
38: A...the volt metre and the resistor are in parallel so their total resistance is 100 using the equation of the resistors in parralel..then we will substitute in the same equation i explained in the previous question..it will be (100 /( 400 + 100) x 60 = 12V
39: C..basic stuff..but note that B is wrong because he said helium atom...it must be helium nucleus
40: C..again simple question