http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
please anyone explain Q25 ASAP
please anyone explain Q25 ASAP
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As physics p1 MCQS YEARLY ONLY.
- Thread starter fatima 007
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Hello people, i need to confirm something, for oct/nov 2002 the examiner report says for mcq no 13 that the answer is A (which does not make sense) however i have seen other people quoting the answer as C and others as D......so could i please have the link of a correct and verified marking scheme for that exam session....thanx soo much!
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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;
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;
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;
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.
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;
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;
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 4
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;
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.
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The answer *is* A. D is totally wrong because they are in a same direction. Confirm it with your clock... these are both going clockwise.
C is wrong because couple is of forces that are opposite And *Equal*. Though the forces here are opposite, one is twice the other.
B is obviously wrong.
A, for this one, it isn't necessary for the forces to be on the circumference of the circle. It does fulfill the requirement of being a couple of forces.. they're equal, they're opposite.
Hope it helps
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Mubarka thanx soo much for uploading the solved paper for on/02....really helps
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thanxThe answer *is* A. D is totally wrong because they are in a same direction. Confirm it with your clock... these are both going clockwise.
C is wrong because couple is of forces that are opposite And *Equal*. Though the forces here are opposite, one is twice the other.
B is obviously wrong.
A, for this one, it isn't necessary for the forces to be on the circumference of the circle. It does fulfill the requirement of being a couple of forces.. they're equal, they're opposite.
Hope it helps
i understood now
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Mubarka
you have done an amazing job MashaAllah! MAy God bless you
just a few doubts if you could clarify
qn 11....what does the 'e' you have mentioned in the eqn stand for?
qn 32.....at -1V the way i saw it the line was horizontal...it became vertical at -1.5V
qn 35....how does more portion of the wire maintain the galvanometer at zero deflection?
thanx soo much! God bless
you have done an amazing job MashaAllah! MAy God bless you
just a few doubts if you could clarify
qn 11....what does the 'e' you have mentioned in the eqn stand for?
qn 32.....at -1V the way i saw it the line was horizontal...it became vertical at -1.5V
qn 35....how does more portion of the wire maintain the galvanometer at zero deflection?
thanx soo much! God bless
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http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Q.14, 15, 22, 26, 28, pretty please?
Q.14, 15, 22, 26, 28, pretty please?
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Guys didnt we agree that we must upload the explanation of Nov 2009 and june 2010..this is the second day and no one posted anything....i cant work alone i need help!! i know most or all of u are busy studying, but im busy too..and this is not the time to quit on the thread..ive posted Nov 2011 variant 2 and ill post variant 3 tommorow inshallah...so some backup pls.. and im thanking u mubaraka for posting Nov 2002 and covering what we missed earlier !!
and im thanking u mubaraka for posting Nov 2002 and covering what we missed earlier !!http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
Please explain Q20 ....ASAP
Please explain Q20 ....ASAP
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okay the thing is in this question its quite obvious that since P takes double the height, so it would have half the density. Q has double the density of P. so if P is on 1, Q is 2.
- hope that helped
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http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Q.14, 15, 22, 26, 28, pretty please?
so ummm
Q 14. since it is at its highest point the vertical velocity would be zero and since only horizontal velocity remains the KE will be half that of its initial magnitude.
Q 15. tricky question - here the kinetic energy would equal the energy in the spring, the momentum of both trolleys would be equal so u have both the velocities and the masses.
trolley A - 2 kg n 2 m/s (given) (momentum is 4)
trolley B - 1 kg n 4 m/s (so momentum is also 4)
apply K.E = 1/2 mv^2 (on both)
so we have k.e of A = 4 and B = 8, add that its 12, so D
Q26.
here you do n(lambda)=dsin(theeta),
u put theeta as 90, and u get 8
Q28. well thing is if the drop has to float, the strength of the charge is higher than the gravity, hence the g/E.
and its negative bcuz if it was positive the charge would be repelled by the upper positive plate and not float (or whatever the word is)
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Hi peoplee, finally! Very happy with the progress!! Soo, i was thinking if anyone has all the graphs that we need to know compiled on one or two papers so we all can download and memorize them. Does anyone of you have it?????
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has the solved explanation for Nov 2003 been uploaded?? If not, can someeoone pleeeaasee upload it...i need it real bad...thanx so much! May God bless you!
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pleeeasseee help
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for polymeric materials ( ex rubber)....after the elastic limit and before the breaking point, a large increase in the Force cause a small increase in the extension...so A is the graph wich supports this idea..and besides u must know what is the shape of a force-extension graph for each type of material ( brittle, ductile, and polymeric)...HOPE I HELPED
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Soldier313 sorry forgot to mention. E in q.11 is the co-efficient of restitution. It is -1 for elastic collisions.
Q 32: yes u're right, im probably confused :S but this i know that semi-conductor diode lets passage of current in only one direction. For the other direction (negative in this case) the resistance is incredibly high abt infinite.
Q 32: yes u're right, im probably confused :S but this i know that semi-conductor diode lets passage of current in only one direction. For the other direction (negative in this case) the resistance is incredibly high abt infinite.
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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.
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.
Assalamu Alaikum
i had a confusion
the answer that u have mentioned for Q10 is A but the marking scheme says its B...
it would be very kind of you f u clear out this confusion
JazakAllahu khair
may Allah reward you for this