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ill be posting Nov 2004 soon today INSHALLAH so i wish another person post the June 2005 so we can progress faster..8 days only left..thnx in advance
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As physics p1 MCQS YEARLY ONLY.
- Thread starter fatima 007
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I agree with you!
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Can you please help us with 2005 and 2006??
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
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.
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YOU ARE VERY SWEET. THANKS FOR HELPING ALL OF US. YOU WILL SUCCEED WITH OUR DUAS AND YOUR HARDWORK. THANKS ALOT.
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OK I will be posting June 2005 InshAllah today.
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thnx very much..i don' have any words to say except that we all succeded inshallah
- Messages
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- Reaction score
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thnk you very much for your cooperation
- Messages
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- Reaction score
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- 103
June 2005
=========
Q1. C.
Fact, basic stuff.
Q2. C
Rearrange the equation to make k the subject, so k = F/rv.
= (kgms^-1)/(m * ms^-1)
= kgm^-1s^-1
Q3.B
A reasonable estimate for an athelete running a 100m race is approximately 10 seconds.
K.E = 1/2 * 80 * 10^2
= 4000 J
Q4. C
Principle speed value = 16
Uncertainty = (0.1/40) + (0.05/2.50) * 16 = 0.36
We always round off the uncertainty value to one s.f., so it becomes 0.4
So the answer is 16 +/ 0.4
Q5. A
The length of the pulse is how long you see the change in the Y-axis. It's for 2cm, meaning 2 μs.
Q6. D
Easy stuff. To find acceleration you take the gradient of a velocity-time graph.
Q7. B
The acceleration during the motion of a falling ball will always be constant, i.e. 9.81 ms^-2. Since they told us to take upwards as positive and the gravitational force acts downwards, this is actually -9.81 ms^-2.
Q8. D
Distance is area of the graph. Class 4 stuff.
Q9. A
Acceleration doesn't act in the horizontal direction so it's value is therefore 0.
B is wrong because the object has velocity throughout the motion.
C is wrong because of B, the resultant velocity will be non-zero because of the horizontal velocity being non-zero.
D is just nonsense.
Q10. A
B, C and D are Newton's 2nd, 3rd and 1st law respectively.
Q11. A
Momentum is always conserved so we use that formula. Also when the objects stick on impact, the total mass will be the sum of the individual masses. Let 'm' be the mass of one of the objects.
60m + (40 * -m) = 2mx (where x is the speed of the masses after impact).
20m = 2mx
x = 10 because m cancel out.
Q12. C
Fact. D is wrong because it gravity is the point through which gravity APPEARS to act.
Q13. A
The forces are shown in this picture: http://www.xtremepapers.com/community/attachments/moe-png.12069/ (Thanks a ton to Unicorn for this).
(5*2) + (2*10) - (3*20) = 30 Nm anti-clockwise.
Q14. D
Resolve the horizontal 4N and vertical 3N component to get a 5N component parallel to the diagonal 4N component. Since the 5N force would be greater, the resultant force would be 1N in its direction.
Q15.B
K.E will be constant because the velocity and mass are constant (velocity beacause it says in the question).
P.E will start from a high value and decrease uniformly because the height is decreasing uniformly.
Q16. C
The gradient of an energy/time graph is power since P = E/t. So we are looking for the point where the gradient is the steepest. This is from 2s-3s, so the gradient there is (40-10)/1 = 30W.
Q17. B
P.E = mgh.
They have given the density and volume from which we can calculate the mass. g is 9.81 and h is 3.0m.
Q18. A
Fact.
Q19. C
Brownian motion, the molecules of liquid collide with the molecules of the pollen grains.
Q20. A
Let a regular extension be 'x'. In parallel, the extension is divided by the # of springs and the opposite for a 'series' extension.
Extension in X is e/2.
Extension in Y is e/2 + e/2 = e.
Extension in Z is e/2 + e = 1.5e.
The order is X -> Y -> Z.
Q21. D
You need to know these graphs. Brittle (glass) is just a straight steep line, rubber is like that of graph X (note that they don't obey Hooke's law) and Y is that of steel, a ductile material.
Q22. D
Let their Young Modulus be equal to 'E'. They have to have the same YM (same material), so..
E = FL/Ax (where x is extension).
F = EAx/L (E and x don't matter here because they're constant).
For P, F = A/l
For Q, F = 0.5A/2l
Ratio is 4:1.
Q23. A
Fact, all transverse waves travel at the same speed in a vaccuum.
Q24. B
You need to know a reasonable estimate of the wavelength of visible light, e.g. 500 nm.
# of wavelengths in ONE metre is 1/(500 nm) 2.0 * 10^6. This is in the order of 10^6, so B is right.
Q25. B
Use ratio of intensity and amplitude.
(I1/I2) = (a1/a2)^2
1/2 = (A/x)^2
x = √2A
Q26. B
Fact, sort of.
Q27. D
Distance between 2 maxima = 0.5λ.
So 1λ = 30 mm.
F = v/λ
F = 3.0 * 10^8 / 30 mm
F = 2.0 * 10 Hz.
Q28. B
Formula is x = λr/q
According to this equation, decreasing 'q' will increase 'x'. A has nothing to do with 'x'.
Q29. B
For 2nd order, d sin θ = 600 nm * 2, which is
d sin θ = 1200 nm
For 3rd order,
d sin θ = 3λ
Since d has to be the same and the angle is also the same, we can equate the 2 equations.
3λ = 1200 nm
λ = 400 nm.
Q30. D
E = V/d.
= 900 / (4 mm)
= 2.3 * 10^3 N/C
Q31. C
Fact.
Q32. C
The area is irrelevant to this question, because Q = I * t (there is nothing to do with area in this formula).
Q = 10 * 1 = 10 C
Since one electron has a charge of 1.6 * 10^-19 C, 10 C has 6.3 * 10^19 electrons.
Q33. D
Originally, R = ρL/A
Now, the length is doubled BUT the volume is the same. This means the area has to be halved. Mathematically proving this:
Volume is length * breadth * height, so:
2lbh = v
Since Area = lb
A = 2lb
lb = 0.5A
Anyway, new resistance will be 2ρL/0.5 = 4R.
Q34. D
A is wrong because Q is a thermistor/semi-conductor/etc.
B is wrong because the resistance decreases.
C is wrong because the resistances are the same at 1.9 (same V:I) ratio.
D is right because using I^2 * R proves this is correct.
Q35. B
Fact.
Q36. D
In parallel, voltage is the same so V2 = V3.
And terminal voltage V = V1 + V3
Rearranging this gives
V - V1 = V3
Q37. At X the voltmeter is connected directly so it gets the full 4V. At Y we use the potential divider formula to find the voltage:
V = 4 * (10/20) = 2V
B is the only graph that shows this correctly.
Q38. C
Easy stuff.
Q39. B
The range of α particles is approximately 0 - 5cm. In this question, they've given us values in mm, so we can say the range is 0 - 500 mm.
B is the safest maximum range.
Q40. C
The nucleon number decreases by (4+4+0) so it becomes 209.
The proton number decreases by (2+2-1) so it becomes 82.
=========
Q1. C.
Fact, basic stuff.
Q2. C
Rearrange the equation to make k the subject, so k = F/rv.
= (kgms^-1)/(m * ms^-1)
= kgm^-1s^-1
Q3.B
A reasonable estimate for an athelete running a 100m race is approximately 10 seconds.
K.E = 1/2 * 80 * 10^2
= 4000 J
Q4. C
Principle speed value = 16
Uncertainty = (0.1/40) + (0.05/2.50) * 16 = 0.36
We always round off the uncertainty value to one s.f., so it becomes 0.4
So the answer is 16 +/ 0.4
Q5. A
The length of the pulse is how long you see the change in the Y-axis. It's for 2cm, meaning 2 μs.
Q6. D
Easy stuff. To find acceleration you take the gradient of a velocity-time graph.
Q7. B
The acceleration during the motion of a falling ball will always be constant, i.e. 9.81 ms^-2. Since they told us to take upwards as positive and the gravitational force acts downwards, this is actually -9.81 ms^-2.
Q8. D
Distance is area of the graph. Class 4 stuff.
Q9. A
Acceleration doesn't act in the horizontal direction so it's value is therefore 0.
B is wrong because the object has velocity throughout the motion.
C is wrong because of B, the resultant velocity will be non-zero because of the horizontal velocity being non-zero.
D is just nonsense.
Q10. A
B, C and D are Newton's 2nd, 3rd and 1st law respectively.
Q11. A
Momentum is always conserved so we use that formula. Also when the objects stick on impact, the total mass will be the sum of the individual masses. Let 'm' be the mass of one of the objects.
60m + (40 * -m) = 2mx (where x is the speed of the masses after impact).
20m = 2mx
x = 10 because m cancel out.
Q12. C
Fact. D is wrong because it gravity is the point through which gravity APPEARS to act.
Q13. A
The forces are shown in this picture: http://www.xtremepapers.com/community/attachments/moe-png.12069/ (Thanks a ton to Unicorn for this).
(5*2) + (2*10) - (3*20) = 30 Nm anti-clockwise.
Q14. D
Resolve the horizontal 4N and vertical 3N component to get a 5N component parallel to the diagonal 4N component. Since the 5N force would be greater, the resultant force would be 1N in its direction.
Q15.B
K.E will be constant because the velocity and mass are constant (velocity beacause it says in the question).
P.E will start from a high value and decrease uniformly because the height is decreasing uniformly.
Q16. C
The gradient of an energy/time graph is power since P = E/t. So we are looking for the point where the gradient is the steepest. This is from 2s-3s, so the gradient there is (40-10)/1 = 30W.
Q17. B
P.E = mgh.
They have given the density and volume from which we can calculate the mass. g is 9.81 and h is 3.0m.
Q18. A
Fact.
Q19. C
Brownian motion, the molecules of liquid collide with the molecules of the pollen grains.
Q20. A
Let a regular extension be 'x'. In parallel, the extension is divided by the # of springs and the opposite for a 'series' extension.
Extension in X is e/2.
Extension in Y is e/2 + e/2 = e.
Extension in Z is e/2 + e = 1.5e.
The order is X -> Y -> Z.
Q21. D
You need to know these graphs. Brittle (glass) is just a straight steep line, rubber is like that of graph X (note that they don't obey Hooke's law) and Y is that of steel, a ductile material.
Q22. D
Let their Young Modulus be equal to 'E'. They have to have the same YM (same material), so..
E = FL/Ax (where x is extension).
F = EAx/L (E and x don't matter here because they're constant).
For P, F = A/l
For Q, F = 0.5A/2l
Ratio is 4:1.
Q23. A
Fact, all transverse waves travel at the same speed in a vaccuum.
Q24. B
You need to know a reasonable estimate of the wavelength of visible light, e.g. 500 nm.
# of wavelengths in ONE metre is 1/(500 nm) 2.0 * 10^6. This is in the order of 10^6, so B is right.
Q25. B
Use ratio of intensity and amplitude.
(I1/I2) = (a1/a2)^2
1/2 = (A/x)^2
x = √2A
Q26. B
Fact, sort of.
Q27. D
Distance between 2 maxima = 0.5λ.
So 1λ = 30 mm.
F = v/λ
F = 3.0 * 10^8 / 30 mm
F = 2.0 * 10 Hz.
Q28. B
Formula is x = λr/q
According to this equation, decreasing 'q' will increase 'x'. A has nothing to do with 'x'.
Q29. B
For 2nd order, d sin θ = 600 nm * 2, which is
d sin θ = 1200 nm
For 3rd order,
d sin θ = 3λ
Since d has to be the same and the angle is also the same, we can equate the 2 equations.
3λ = 1200 nm
λ = 400 nm.
Q30. D
E = V/d.
= 900 / (4 mm)
= 2.3 * 10^3 N/C
Q31. C
Fact.
Q32. C
The area is irrelevant to this question, because Q = I * t (there is nothing to do with area in this formula).
Q = 10 * 1 = 10 C
Since one electron has a charge of 1.6 * 10^-19 C, 10 C has 6.3 * 10^19 electrons.
Q33. D
Originally, R = ρL/A
Now, the length is doubled BUT the volume is the same. This means the area has to be halved. Mathematically proving this:
Volume is length * breadth * height, so:
2lbh = v
Since Area = lb
A = 2lb
lb = 0.5A
Anyway, new resistance will be 2ρL/0.5 = 4R.
Q34. D
A is wrong because Q is a thermistor/semi-conductor/etc.
B is wrong because the resistance decreases.
C is wrong because the resistances are the same at 1.9 (same V:I) ratio.
D is right because using I^2 * R proves this is correct.
Q35. B
Fact.
Q36. D
In parallel, voltage is the same so V2 = V3.
And terminal voltage V = V1 + V3
Rearranging this gives
V - V1 = V3
Q37. At X the voltmeter is connected directly so it gets the full 4V. At Y we use the potential divider formula to find the voltage:
V = 4 * (10/20) = 2V
B is the only graph that shows this correctly.
Q38. C
Easy stuff.
Q39. B
The range of α particles is approximately 0 - 5cm. In this question, they've given us values in mm, so we can say the range is 0 - 500 mm.
B is the safest maximum range.
Q40. C
The nucleon number decreases by (4+4+0) so it becomes 209.
The proton number decreases by (2+2-1) so it becomes 82.
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This is Nov 2004
1: C...basic stuff
2: C...in A he said metre which is a unit, in D he said second which is also a unit..and in B the definition is wrong
3: D ... E/A x t= ( Kgm^2 s^-2 / m^2 x s ) so Kg s^-3 so D
4:A... the deflection was increasin constantly then starts to slow down till it becomes constant and in A the numbers start to get close to each other which will slow the deflection of the needle
5: B... T=1/F and one wavelength is 8 cm si 1/( 50 x 8) = 2.5 ms so B
6: D... A is wrong because it is accurate as it is close to 9.81..accuracy is how close my measurment to the true value..precision is how close my measured values to each other..in B they are not accurate or precise and also in C..in D it is accurate but precise so yes D
7: D...always the acceleration of free fall i the same in any vertical component
8:C..the V t graphshows constant velocity from o to t1..so A is wrong..then fromt1 to t2 a constant decrease in velocity..so D is wrong..C is corrct because the car had stopped in the V t graph..and the only distance time graph which shoes that the car stopped is C
9: D..use s=ut + 1/2 at^2 to find t...and v^2= u^2 + 2as to find V
10: B...always remember the example of a sky diver...resultant force is zero so no acceleration so terminal velocity...F is proportional to a so if there is acceleration so there is resultant force
11: B...elastic collision means that for example if 2 balls hit each other each one will go to the opposite direction they where going and each ball will take the velovity of the othre..ex if ball has has 30 ms and ball b has 50 after they collide..ball a will go to the other direction with 50 ms and ball b will go with 30 also in the opposite..so hear the ball went to the other dirction taking the momentum of the wall which is 2mv
12: C... a triangle is in equillibrium when all the sides doesnt meet each other
13: B..W is in the middle of the rod..so ( W x d=F x d) so W = (300 x 2)/1.25 = 480
14: B...electron is attracted toward the +ve side
15: A...use any imaginary number and try it...the increase from V^2 of the normal kinetic energy to the V^2 of the kinetic energy that is increased by 4 will be 2..ex 0.5 x 50 x 20^2 = 20000 and 0.5 x 50 x 40^2 = 40000
16: A... work done = gain in kinetic energy so = summaion of F x d = (90-50) x 6= 240
17: B...power = F x v and v = d/t so P=F x d/t so t = 180 x 4000 / 3000 = 240
18: A..( 9000 x 40) - ( 20000 x 12)= 120Kj
19: B..boiling have a fixed temperature while evaporation is a range
20: sry this was difficult
21: A...basic stuff
22: A...just by trial using the formula F/A
23: A...strain energy = 0.5 x F x e ...and if u draw a straight line from o to to the same point there will be a space between the line we draw and the graph found in the question, this space is loss of some energy so A
24: D...again basic stuff, sound waves are the only longitudnal waves
25: C...distance between 2 adjacent nodes or antinodes is half the wavelength
26: C...intensity is proportional to amplitude squared so 20^2 / 5^2= 16
27: C...also basic stuff
28: D ... the distance between the slits and the screen is indirectly proportional to the wavelength..so if the wavelength is decreased so the distance must increase and so D is the answer cause it is the only one greater than 1.00m
29: A...if the top plate is -ve so the down must be +ve and the field lines must go from +ve to -ve
30: D... A is wrong because the charge may change and so the force...B is ofcourse wrong..and C is wrong because we know it is 4m +ve to -ve not the opposite
31: C...W=QV...so V= W/Q...Q= It and then W/It..Power = W/t so P = P/I so C
32: B...by trial using the equation P= R I^2
33: C... R= pl/A = pl/(pie) (0.5d)^2..then p x 2l / (pie) x (2d/2)^2.....from 0.5d squared to 2d/2 squared is multiplication by 4 so R will be 0.5 then I=V/R so 2V/0.5R then 4I so V
34: C... I is proportional to V..and V increases as R increases..so A and B is wrong because it didnt start from the origin..C is the right answer as I will increase with V as the temperature is constant
35: D.... using kirchof's 1st law R=3 - 1= 2A..thn Q=It so 2 x 5 so 10 so D
36: B.... in the potential divider the minimum output voltage is 0 and the maximum must be less than the emf..so Voutput= (R1/R1 + R2) x 9 = 4.5 so B
37: A...total resistance in 1st circuit is greater than the total resistance in the 2nd circuit so V1 >V2...and I = V/R so if V1 and Rt1 is greater so I1 must be also greater than I2
38: D...beta particle have o nucleon number and -1 proton number so D
39:basic stuff so C
40: D..also basic stuff...as explained in rutherford experiment so D
1: C...basic stuff
2: C...in A he said metre which is a unit, in D he said second which is also a unit..and in B the definition is wrong
3: D ... E/A x t= ( Kgm^2 s^-2 / m^2 x s ) so Kg s^-3 so D
4:A... the deflection was increasin constantly then starts to slow down till it becomes constant and in A the numbers start to get close to each other which will slow the deflection of the needle
5: B... T=1/F and one wavelength is 8 cm si 1/( 50 x 8) = 2.5 ms so B
6: D... A is wrong because it is accurate as it is close to 9.81..accuracy is how close my measurment to the true value..precision is how close my measured values to each other..in B they are not accurate or precise and also in C..in D it is accurate but precise so yes D
7: D...always the acceleration of free fall i the same in any vertical component
8:C..the V t graphshows constant velocity from o to t1..so A is wrong..then fromt1 to t2 a constant decrease in velocity..so D is wrong..C is corrct because the car had stopped in the V t graph..and the only distance time graph which shoes that the car stopped is C
9: D..use s=ut + 1/2 at^2 to find t...and v^2= u^2 + 2as to find V
10: B...always remember the example of a sky diver...resultant force is zero so no acceleration so terminal velocity...F is proportional to a so if there is acceleration so there is resultant force
11: B...elastic collision means that for example if 2 balls hit each other each one will go to the opposite direction they where going and each ball will take the velovity of the othre..ex if ball has has 30 ms and ball b has 50 after they collide..ball a will go to the other direction with 50 ms and ball b will go with 30 also in the opposite..so hear the ball went to the other dirction taking the momentum of the wall which is 2mv
12: C... a triangle is in equillibrium when all the sides doesnt meet each other
13: B..W is in the middle of the rod..so ( W x d=F x d) so W = (300 x 2)/1.25 = 480
14: B...electron is attracted toward the +ve side
15: A...use any imaginary number and try it...the increase from V^2 of the normal kinetic energy to the V^2 of the kinetic energy that is increased by 4 will be 2..ex 0.5 x 50 x 20^2 = 20000 and 0.5 x 50 x 40^2 = 40000
16: A... work done = gain in kinetic energy so = summaion of F x d = (90-50) x 6= 240
17: B...power = F x v and v = d/t so P=F x d/t so t = 180 x 4000 / 3000 = 240
18: A..( 9000 x 40) - ( 20000 x 12)= 120Kj
19: B..boiling have a fixed temperature while evaporation is a range
20: sry this was difficult
21: A...basic stuff
22: A...just by trial using the formula F/A
23: A...strain energy = 0.5 x F x e ...and if u draw a straight line from o to to the same point there will be a space between the line we draw and the graph found in the question, this space is loss of some energy so A
24: D...again basic stuff, sound waves are the only longitudnal waves
25: C...distance between 2 adjacent nodes or antinodes is half the wavelength
26: C...intensity is proportional to amplitude squared so 20^2 / 5^2= 16
27: C...also basic stuff
28: D ... the distance between the slits and the screen is indirectly proportional to the wavelength..so if the wavelength is decreased so the distance must increase and so D is the answer cause it is the only one greater than 1.00m
29: A...if the top plate is -ve so the down must be +ve and the field lines must go from +ve to -ve
30: D... A is wrong because the charge may change and so the force...B is ofcourse wrong..and C is wrong because we know it is 4m +ve to -ve not the opposite
31: C...W=QV...so V= W/Q...Q= It and then W/It..Power = W/t so P = P/I so C
32: B...by trial using the equation P= R I^2
33: C... R= pl/A = pl/(pie) (0.5d)^2..then p x 2l / (pie) x (2d/2)^2.....from 0.5d squared to 2d/2 squared is multiplication by 4 so R will be 0.5 then I=V/R so 2V/0.5R then 4I so V
34: C... I is proportional to V..and V increases as R increases..so A and B is wrong because it didnt start from the origin..C is the right answer as I will increase with V as the temperature is constant
35: D.... using kirchof's 1st law R=3 - 1= 2A..thn Q=It so 2 x 5 so 10 so D
36: B.... in the potential divider the minimum output voltage is 0 and the maximum must be less than the emf..so Voutput= (R1/R1 + R2) x 9 = 4.5 so B
37: A...total resistance in 1st circuit is greater than the total resistance in the 2nd circuit so V1 >V2...and I = V/R so if V1 and Rt1 is greater so I1 must be also greater than I2
38: D...beta particle have o nucleon number and -1 proton number so D
39:basic stuff so C
40: D..also basic stuff...as explained in rutherford experiment so D
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hey ppl can we discuss da M.J.01 papr.. da mark scheme isnt available..
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june 2006 question 34 please
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agreedDamn!
Hats off to you guys for being so helpful, cooperative and nice
Keep up this good work.
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Can we have a checklist on the first post of all the papers solved?
I'll be posting Nov 2006 in a while.
I'll be posting Nov 2006 in a while.
XPFMember
XPRS Moderator
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AsSalamoAlaikum Wr Wb!
That's been done already [if that's what u're speaking about]
Thanks anywaysjazakAllahu khairen
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Wa alaikum assalam Wr Wb!AsSalamoAlaikum Wr Wb!
That's been done already [if that's what u're speaking about]
jazakAllahu khairen
Yeah I just noticed now.
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November 2006
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
1- D
because milli * tera = 10^-3 * 10^12 = 10^9 which has the greatest magnitude.
2- B
Base units of energy = kgm^2 s^-2
In case you can't remember that ^ you can always find it using the formula mgh = kg* ms^-2 * m = kg m^2 s^-2
Option A is Ft whose base units are kgms^-2 * s = kgms^-1 hence its eliminated
Option B is Fvt = kgms^-2 * ms^-1 * s = kgm^2 s^-2
Hence B is the correct answer.
3-A
Since Electric field strength = F/Q = kgms^-2/C = kg ms^-2 C^-1
4- D
F= 50 Hz
Since T = 1/F hence time period of one wave is 1/50 = 0.02 s
10*10^-3 s for one box on the x axis
therefore for 0.02 s = 2 boxes. Hence we know that 2 boxes will be required to complete one waveform.
Amplitude is 5 V and since the Y gain is set at 5 V/ division therefore, the wave's max height should be one box long.
So the answer is D.
5- A
Basic concept of random error; it can be eliminated or reduced by taking average of several readings.
6- D
V = IR
R =V/I = 8/1 = 8 ohms
[Note I've denoted '#' for delta]
#R/R = #V/V + #I/I
#R = R(#V/V + #I/I)
#R = 8(.2 + .4) = 2
Therefore (8+- 2) ohms
7-C
because the velocity time graph is always a straight line when acceleration is uniform/constant
8- D
Area of X = 1/2 *b*h
= .5*5*5 = 12.5 m
Area of Y = area of trapezium = .5(2+3) * 5
=12.5 m
9-C
for vertical component: v = u +at
v = usina - gt
and in horizontal component v = u cos a
10-B
basic concept that accelaration and force are always in the same direction.
11-D
In elastic collisions, speeds are interchanged.
12-B
0 = (1000)(5) - 10u
10 u = 5000
u = 500 m/s
13- B
basic definition of upthrust.
14- B
moment = F * 2R
= 2 Fr
15- C
Since both horizontal forces of 10 N cancel each other out, R ^2 = (10sin 60 + 10sin60 -10)^2
R = 7.32 ~ 7.3 N
16- D
P.E. = mgh
2.5*6*10*80 = 12000 J
17-A
Efficiency = 7/100 *100 = 7%
18 - C
At 100% power - voltage = 25000 V
So at 80% power- voltage = 20 000V
P = VI
4 * 10^6 = 20 000 * I
I = 4 000 000/20 000
I = 200 A
19- C
Basic definition of density is that it is the mass per unit volume of a substance
20-B
The concept of evaporation is that it:
* involves a change in state from liquid to vapour
* occurs at any temperature
* involves a reduction in the average kinetic energy of remaining atoms since most energetic ones leave the surface.
21- D
17.5 *10^6 = 830*g*h + 1000*g*h
17.5 * 10^6 = 8142.3x + 9810(2000 - x)
17.5 *10^6 = 8142.3x + 19 620 000 - 9810x
-2 120 000= -1667.7x
1271.2 = x
~ 1270 m (3 s.f.)
22- B
The spring constant is always the gradient of a force-extension graph.
23-C
E = Fl/Ae
N* m /m^2*m
Nm^-2
24 -B
Basic definition: Frequency is the no. of complete oscillations in one second.
25- C
Since one wavelength ends at 2pii
therefore X is at 0.5pii
and Y is at 3.5pii
Hence phase difference is 3.5pii - .5 pii = 3 pii and n =3
26- D
The smaller the width of gap, the higher the diffraction.
27-C
n*lamda = dsin theta
1 lamda = d sin 30
1 lamda = .5 *d
slit spacing = 500 d
fringe spa5cing = lamda* D/a
x = .5 d * 1/ 500d
x = 1 *10^-3 m
28 - A
The easiest way to produce a stationary wave is to blow air over the top of an empty bottle.
29- B
The electric field pattern always moves away from the positive charge and towards the negative charge.
30 - B
The electric field strength between two parallel plates is always uniform.
31- A
Q = It
I = Q/t
For one complete rotation charge = 4Q
t = 1/f
hence I = 4Q/1/f
= 4Qf
32- D
Learn the shape of I-V graph of filament lamp.
33- A
So this is divided into 3 sections, the graph.
Notice that this is an I-V graph hence its gradient will be the reciprocal of resistance that is 1/R
so R = 1/gradient
therefore greater the gradient, lower the resistance and lower the gradient and higher the resistance.
For the first part, its a straight line hence resistance is constant.
For the second part, XY, the gradient seems to be increasing so resistance will be decreasing
For the last part, YZ, the gradient is decreasing so resistance will be increasing.
34- A
Min value of potentiometer = O ohms
So V = 0/40 * 12 = 0 V
Max value of poetentiometer = 12 ohms
So V = 12/40 * 12 = 6 V
Hence range = 0-6 V
35- D
Basic idea of a thermistor.
36 - B
I1 + 10.6 = 10.8
I1= 0.2 mA
I2 = 20.2 + 10.6
I2 = 30.8 mA
38- C
A is incorrect because it has the charge of +2e
B is incorrect because it has the nucleus of a helium atom.
C is the only correct property.
39- D
Basic concept of an atom is that the protons and neutrons are inside the nucleus while the electrons orbit the nucleus.
40-B
222- 214 = 8
Therefore 2 alpha particles
and 86- 4 = 82
so if one beta particle was emitted then the proton no will be 83.
Phew that was one long post! *Wipes sweat off forehead*
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
1- D
because milli * tera = 10^-3 * 10^12 = 10^9 which has the greatest magnitude.
2- B
Base units of energy = kgm^2 s^-2
In case you can't remember that ^ you can always find it using the formula mgh = kg* ms^-2 * m = kg m^2 s^-2
Option A is Ft whose base units are kgms^-2 * s = kgms^-1 hence its eliminated
Option B is Fvt = kgms^-2 * ms^-1 * s = kgm^2 s^-2
Hence B is the correct answer.
3-A
Since Electric field strength = F/Q = kgms^-2/C = kg ms^-2 C^-1
4- D
F= 50 Hz
Since T = 1/F hence time period of one wave is 1/50 = 0.02 s
10*10^-3 s for one box on the x axis
therefore for 0.02 s = 2 boxes. Hence we know that 2 boxes will be required to complete one waveform.
Amplitude is 5 V and since the Y gain is set at 5 V/ division therefore, the wave's max height should be one box long.
So the answer is D.
5- A
Basic concept of random error; it can be eliminated or reduced by taking average of several readings.
6- D
V = IR
R =V/I = 8/1 = 8 ohms
[Note I've denoted '#' for delta
]#R/R = #V/V + #I/I
#R = R(#V/V + #I/I)
#R = 8(.2 + .4) = 2
Therefore (8+- 2) ohms
7-C
because the velocity time graph is always a straight line when acceleration is uniform/constant
8- D
Area of X = 1/2 *b*h
= .5*5*5 = 12.5 m
Area of Y = area of trapezium = .5(2+3) * 5
=12.5 m
9-C
for vertical component: v = u +at
v = usina - gt
and in horizontal component v = u cos a
10-B
basic concept that accelaration and force are always in the same direction.
11-D
In elastic collisions, speeds are interchanged.
12-B
0 = (1000)(5) - 10u
10 u = 5000
u = 500 m/s
13- B
basic definition of upthrust.
14- B
moment = F * 2R
= 2 Fr
15- C
Since both horizontal forces of 10 N cancel each other out, R ^2 = (10sin 60 + 10sin60 -10)^2
R = 7.32 ~ 7.3 N
16- D
P.E. = mgh
2.5*6*10*80 = 12000 J
17-A
Efficiency = 7/100 *100 = 7%
18 - C
At 100% power - voltage = 25000 V
So at 80% power- voltage = 20 000V
P = VI
4 * 10^6 = 20 000 * I
I = 4 000 000/20 000
I = 200 A
19- C
Basic definition of density is that it is the mass per unit volume of a substance
20-B
The concept of evaporation is that it:
* involves a change in state from liquid to vapour
* occurs at any temperature
* involves a reduction in the average kinetic energy of remaining atoms since most energetic ones leave the surface.
21- D
17.5 *10^6 = 830*g*h + 1000*g*h
17.5 * 10^6 = 8142.3x + 9810(2000 - x)
17.5 *10^6 = 8142.3x + 19 620 000 - 9810x
-2 120 000= -1667.7x
1271.2 = x
~ 1270 m (3 s.f.)
22- B
The spring constant is always the gradient of a force-extension graph.
23-C
E = Fl/Ae
N* m /m^2*m
Nm^-2
24 -B
Basic definition: Frequency is the no. of complete oscillations in one second.
25- C
Since one wavelength ends at 2pii
therefore X is at 0.5pii
and Y is at 3.5pii
Hence phase difference is 3.5pii - .5 pii = 3 pii and n =3
26- D
The smaller the width of gap, the higher the diffraction.
27-C
n*lamda = dsin theta
1 lamda = d sin 30
1 lamda = .5 *d
slit spacing = 500 d
fringe spa5cing = lamda* D/a
x = .5 d * 1/ 500d
x = 1 *10^-3 m
28 - A
The easiest way to produce a stationary wave is to blow air over the top of an empty bottle.
29- B
The electric field pattern always moves away from the positive charge and towards the negative charge.
30 - B
The electric field strength between two parallel plates is always uniform.
31- A
Q = It
I = Q/t
For one complete rotation charge = 4Q
t = 1/f
hence I = 4Q/1/f
= 4Qf
32- D
Learn the shape of I-V graph of filament lamp.
33- A
So this is divided into 3 sections, the graph.
Notice that this is an I-V graph hence its gradient will be the reciprocal of resistance that is 1/R
so R = 1/gradient
therefore greater the gradient, lower the resistance and lower the gradient and higher the resistance.
For the first part, its a straight line hence resistance is constant.
For the second part, XY, the gradient seems to be increasing so resistance will be decreasing
For the last part, YZ, the gradient is decreasing so resistance will be increasing.
34- A
Min value of potentiometer = O ohms
So V = 0/40 * 12 = 0 V
Max value of poetentiometer = 12 ohms
So V = 12/40 * 12 = 6 V
Hence range = 0-6 V
35- D
Basic idea of a thermistor.
36 - B
I1 + 10.6 = 10.8
I1= 0.2 mA
I2 = 20.2 + 10.6
I2 = 30.8 mA
38- C
A is incorrect because it has the charge of +2e
B is incorrect because it has the nucleus of a helium atom.
C is the only correct property.
39- D
Basic concept of an atom is that the protons and neutrons are inside the nucleus while the electrons orbit the nucleus.
40-B
222- 214 = 8
Therefore 2 alpha particles
and 86- 4 = 82
so if one beta particle was emitted then the proton no will be 83.
Phew that was one long post! *Wipes sweat off forehead*
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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.
33)
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
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.
33)
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