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As physics p1 MCQS YEARLY ONLY.

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and inshallah guys ill be posting June 2009 later today..so we need any 1 to do Nov 2009..and if possible another help from anothre 1 to do june 2010...i congratulate leosco1995 and arlery for their help so if they could do these papers too ill be grateful
 
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Tkp

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this thread is superb.all my problems are clear now.thanks lesco,arierl and aelwy for ur co-operation
 
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Hi! I was just going through this when I stumbled upon this mistake. :p For 37, the correct answer is A. This is because the plate separation is increasing so the field strength would decrease. I'm sure it was just a slip. The rest is all correct, Masha'Allah. :)
sorry for disturbing, but can you tell me that when should we use a straight horizontal line in the graph of E against x or V? thanks in advance:D
 
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sorry for disturbing, but can you tell me that when should we use a straight horizontal line in the graph of E against x or V? thanks in advance:D
The electric field strength between two parallel plates is always uniform as long as plate separation is unchanged. So you get a straight horizontal line in the graph of E against x, where x is the distance from one plate to any point between the two plates.
E and x have an inverse relationship if x is the plate separation. If you increase the plate separation,x or d, E decreases since E = V/x or V/d.
 
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May/June 2008

1) C; kilo(k): 10^3
milli(m): 10^-3
Mega(M): 10^6
nano ( n):10^-9

2) D; Force = mass x acc. = kgms-²
work done = force x displacement = kgm²s-²
e.m.f. = work done/ charge = kgm²s-²/As

3) C; speed= distance/time = 100/10 = 10m/s
the approximate mass of an athlete can be estimated between 60 to 85 kg, if 80 kg is taken, by K.E.= 1/2 m V ² = 1/2 x 80 x 10 ² = 4000J

4) C; Here, R= V/I = 1/0.5 = 2 ohms.
ΔR/R = ΔV/V + ΔI/I
ΔR/2 = 0.05/1 + 0.o1/0.5
ΔR = ± 0.14
%age uncertainty = (0.14/2)x 100 = 7.0%

5) D; precise and not accurate result would be very different from 9.81 but within close range of one another.

6) C The case is similar to that when an object moves in a circular path. The speed remains same, but the direction keeps changing (direction at each instant is tangent to the circle). The acc. is because the velocity changes, by the changing direction.

7) B; Acc. of free fall on this planet would be, 9.81/6 = 1.635 m/s^2
W=mg
= 30 x 1.635
= 49.05N

8) C; At the top when it is released, it has a maximum acc. of 9.81 m/s^2. After that it keeps decreasing and finally when terminal velocity is reached, acc. becomes zero. The graph is often confused with that of D, but remember that shape is for a V-t graph.

9) D; The principle states that for a closed system (in which no external forces act) the momentum is always conserved.

10) D; Relative speed of approach = Relative speed of separation. Also, in an elastic collision, total kinetic energy is conserved.

11) A; F=ma
2000-R = 750 x 2
R = 500N

12) A;
Upthrust.gif

The drag is a resisting force in fluids ( like air resistance in air). This is similar to friction.
Upthrust is only because of the pressure difference on top and bottom of an object when in a fluid.
Weight is the greatest of all.

13) A; Forces in equilibrium form a head to tail vector triangle (clockwise or anti-clockwise).

14) A; moment = force x distance
clockwise moment = anticlockwise moment
W x a + F x h = W x 2a

15) D; fact. the concept of ρgh.

16) A; for potential energy, vertical height is always taken, so distance would be s.
The field is towards the right side, so left is positive and the right negative, which means that the potential energy will decrease. energy = workdone, therefore, w.d. = F x s

17) C; linear momentum is Always conserved. B and D rejected.
The collision here is inelastic, so K.E. is not conserved. However, the Total energy is still conserved because of the law of conservation of energy.

18) B; The speed is constant, therefore the Kinetic energy is constt, which would give a straight line graph.
The ball is falling in a fluid of constt density. The potential energy varies linearly with the height. Ball is falling, h is decreasing so P.E. decreasing at a constt rate.

19) C; efficiency = (*useful* output/ total input) x 100

20) B; same old repeated question. fact.
A ignored cuz container's pressure is nowhere discussed.
C rejected bcuz the collisions are elastic.
D put down cuz the weight is downwards, and the statement is just wrong.

21) A; In the derivation of this formula, these formulae/principles are used, P = F/A, F=W, W=mg, m= ρv and v=area x height. Out of these, only mass= density x volume is given.



22) B; The metal stretches and then returns to its original shape. there is an extension, therefore, elastic behaviour. It follows the *same* curve when it contracts, therefore no plastic deformation.


23) A; The spring is being compressed, therefore there is a decrease in its length. The original length was 100mm. When the l is 70mm, the *change* in length = 30mm
so, the energy stored is area under the graph between 70 and 100 mm.
i.e. 1/2 x 30x10³ x 6 = 0.09J

24) C; the young modulus is same for a material.

25) B; p is not the height of the highest point on the wave so is not the amplitude. on x-axis is time, therefore q is time taken by one complete wave to pass through a point, time period.

26) D; I is inversely proportional to x² so doubling x means I would be divided by 4. I is directly proportional to A², so √(8²/4) = √16 = 4.0 μm
This cal also be simplified to this,
I1/ I2 = A1² / A2²
1/(1/4) = 8² /x²
4/8² = 1/x²
x²= 16
x= 4.0 μm

27) D; V= f λ where speed is c and wavelength is x+x+x+x =4x (as x is the distance between an antinode and a node, and one wave has three nodes and two antinodes).

28) B; d= 1/N
dsinθ = nλ
(1/N)sinθ = 3λ

29) A; λ = ax/D
re-arrange it to give, x = λD/a
this clearly shows that fringe spacing "x" is directly proportional to wavelength and inversely proportional to "a" (slit separation).
when λ is halved, x decreases, when a is doubled, x again is halved, so its like (3/2) /2 = 3/4 = 0.75 mm

30) A; the horizontal component of v is vcosθ
and u is in the same direction as vcosθ
re-arranging this gives u/cosθ = v

31) C;
E = V/d = F/Q
5000/0.8x10-² = E = 625000 N/C
now, Q=ne (where e is elementary charge)
625000 = F/5(e)
625000 x 5 x elementary charge = 5 x 10-¹³N
the drop has *gained* five electrons so is now negatively charged and will be attracted to the positive plate i.e. will experience an upward force.

32) B; P = I² x R
[(1/2)² x 2] / [1 x 1]
= 1/2

33) C; P= work done/ time
12 = w.d. / 50
w.d. = 600 J
v = work done/ charge
V= 600/100
V= 6.0V

34) C; R= ρL/A
which can also be written as R = 4ρL/πd²
the volume stays the same. L is increased four times, so the diameter is halved.
X has R1 = 4ρL/πd²
Y has R2 = 4ρ4L/π(d/2)² = 16ρL/(πd²/ 4)
ratio comes, (ignoring the constants) (16 x 4)/4 = 16.

35) C; e.m.f. = w.d./charge
w.d. = 12 x 4 = 48 J
P = w.d. /t
t = w.d. /P
t = 48/24 = 2 sec.

36) D;
basic concept of resistance of thermistor. More the resistance of a component, more will the voltage across it be. No other changes are made therefore those with LDRs rejected. when the temp. is reduced, resistance of thermistor increases and so does the p.d.

37) A; net resistance of R2 and voltmeter = 50kΩ
effecctive R of circuit = 150 kΩ
Total voltage across R2 and voltmeter is 1/3 rd of 6 V (same as ratio of resistances i.e. 50/150) = 2 V
V= IR
2= I x 100 x10³
I = 2 x 10-A

38) B; simple set up for a potentiometer.

39) C; estimations we're supposed to learn.

40) B; By beta decay, the proton nmbr increases by one and there is no change on the nucleon nmbr. After alpha emission, N decreases by 4 and Z decreases by 2.
On these graphs, x-axis is labelled Z and y-axis N.
can u please explain qs 34 a bit more???
 
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Here is the June 2009.. oh it was a tough one!!

1: D...basic stuff
2: C...opp/adj = tan ( theta) so opp=adj tan(theta) = 20tan( 30) =11.5
3: B...the readings are -6.5 and 14...so 14- (-6.05) so 20.5
4: B...P=VI= 1.2 x 0.48
5: D..he siad terminal velocity means velocity is constant and so d must have a constant increase..all other 3 graphs have d starts to decrease so D is correct
6: C...obvious from the graph
7: A...basic stuff...
8: C...air resistance is neglected so constant horizontal velocity and constant vertical acceleration
9: B...rate of changeof momentum= mv-mu, so (0.1 x -30) - (0.1 x 20)= -5....we used - 30 because it was travelling in the opposite direction
10: C...the momentum of body m must equal the momentum of body 2m..this means that if body m has half the mass so it must have double the velocity so ratio of kinetic energy of m /2m = ( 0.5 m x 2v^2)/ (0.5 x 2m x v^2) so it will be 2/1
11: A..the ball will not fall diagonaly..so S must equal to Q...according to the question..he didnt say if the ball will flought orwill sink so i thought that the upthrust is the difference of the pressure betwwn the water and the weight of the object so R > than P
12: D...he said it is falling with uniform velocity so no acceleration so no resultant force..and he said it is still falling horizontally so no resultant torque..so the conditions for equillibrium are found
13: ooh ill blow from this one and i may need some 1 to explain it to me..thnx in advance
14: C...K= F/v^2 means F/v x v and F = P/v so K = P/v/ (v x v) so K=(P/v) x v^-2 and then K = P x v^-1 x v^-2 so P x v^-3 means P/v^
15: B...container X lost half of the water means it lost half the height and half THE MASS so m/2 g h/2 so 2 x 2=4 so ( mgh )/4
16 and 17 are basic and simple stuff
18: D...at first the pressure was pgh..then the height was raised by another h so pgh + h = pg2h
19: A...B and C are wrong because they are brittle so they will break..steel is a mixture of 2 substances so it have the melting point of 2 substances which are iron and carbon...so aluminium is a single metal susbtance and so it will melt first
20: D...young modulus is ( Fl)/ (Ae) he said they are the same material so same young modulus..and same extension..then tension is the F so F= A/l ( no need for the young modulus and the extensoin so ...( l/A) /( 2l /0.5A) continue and u will get 4 / 1
21: A...if u draw a straight line from the origin to the same point the drawn graph reaches and took 0.5 x 17 x 0.30 it will equal 2.7..and the graph shows that thre is much spce between it and the line we drew so it must be < 2.7..so C and are wrong..B is wrong because it is about or approximatley the same as 2.7 so it alo must be less than 2.6 so A
22 and 23 are also basic and simple stuff
24: B...the distance between 2 adjacent nodes orantinodes is half a wavelength..so the distance between a node and an anti node is the quarter
25: A..N lines per metre means 1/N...then substituting into the equation it will be (1/N) sin (theta)= n ( lambda)
26: B...as shown 1.5 lambda = 2.1m so one wavelength is 1.4..then v=f lambda = 80 x 1.4 = 112
27: A...both are attracted to the +ve plate so both are -ve
28: B...E is proportional to V so doubling V doubles E......E is inverseley proportional to d so halfing d doubles E so B
29: D...v^2 = 2as..a=F/m so v^2 = 2 x (F/m) s and F= EQ means Ee as in the question..so v^2/2 = (Ee/m) s continue and u will get D
30: A...I=Q/t..simple question
31: D...W=QV so Q= 7.2 x 10^4 / 12 = 6000
32: A...basic EMF definition
33: B....I = E1 - E2 / R so ( 3 - 1.2) / 9 = 0.2
34: D..simple question..total resistorsin series = R1+R2+R3..in parrallel...1/Rt= 1/R1 + 1/R2 + 1/R3
35: D...light intensity and temperature increases so resistance in the LDR and thermistor decrease so the other resistor resistance increases
36 and 37 are simple questoins
38: C...just subtract the proton number from the nucleon number of each
39: C...simple question
40: D...he said a neutron so proton number must be zero and nucleon number is one
 
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The electric field strength between two parallel plates is always uniform as long as plate separation is unchanged. So you get a straight horizontal line in the graph of E against x, where x is the distance from one plate to any point between the two plates.
E and x have an inverse relationship if x is the plate separation. If you increase the plate separation,x or d, E decreases since E = V/x or V/d.
thx :D
 

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AsSalamoAlaikum Wr Wb!
Here's a comment from the examiner:
Since there is considerable time pressure on candidates to complete 40 questions within the time limit of one hour, there is pressure to ignore the checking process. It is all the more important, therefore, to read the question carefully in the first place. In particular, when answering verbal questions, candidates should check all four options, rather than looking at just one or two of them. This is because one option may look correct, but a further option may trigger a different thought for the candidate; something that may not have been thought about when reading the first, apparently correct, option. Powers of t en are a source of many errors with numerical questions. These arise not just by using a calculator incorrectly, but also with the metric system prefixes.
Candidates must use space on the paper for working: on leaving the examination room their paper should be covered with writing.
Source: Nov:2008 Examiner Report
 

Tkp

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Here is the June 2009.. oh it was a tough one!!

1: D...basic stuff
2: C...opp/adj = tan ( theta) so opp=adj tan(theta) = 20tan( 30) =11.5
3: B...the readings are -6.5 and 14...so 14- (-6.05) so 20.5
4: B...P=VI= 1.2 x 0.48
5: D..he siad terminal velocity means velocity is constant and so d must have a constant increase..all other 3 graphs have d starts to decrease so D is correct
6: C...obvious from the graph
7: A...basic stuff...
8: C...air resistance is neglected so constant horizontal velocity and constant vertical acceleration
9: B...rate of changeof momentum= mv-mu, so (0.1 x -30) - (0.1 x 20)= -5....we used - 30 because it was travelling in the opposite direction
10: C...the momentum of body m must equal the momentum of body 2m..this means that if body m has half the mass so it must have double the velocity so ratio of kinetic energy of m /2m = ( 0.5 m x 2v^2)/ (0.5 x 2m x v^2) so it will be 2/1
11: A..the ball will not fall diagonaly..so S must equal to Q...according to the question..he didnt say if the ball will flought orwill sink so i thought that the upthrust is the difference of the pressure betwwn the water and the weight of the object so R > than P
12: D...he said it is falling with uniform velocity so no acceleration so no resultant force..and he said it is still falling horizontally so no resultant torque..so the conditions for equillibrium are found
13: ooh ill blow from this one and i may need some 1 to explain it to me..thnx in advance
14: C...K= F/v^2 means F/v x v and F = P/v so K = P/v/ (v x v) so K=(P/v) x v^-2 and then K = P x v^-1 x v^-2 so P x v^-3 means P/v^
15: B...container X lost half of the water means it lost half the height and half THE MASS so m/2 g h/2 so 2 x 2=4 so ( mgh )/4
16 and 17 are basic and simple stuff
18: D...at first the pressure was pgh..then the height was raised by another h so pgh + h = pg2h
19: A...B and C are wrong because they are brittle so they will break..steel is a mixture of 2 substances so it have the melting point of 2 substances which are iron and carbon...so aluminium is a single metal susbtance and so it will melt first
20: D...young modulus is ( Fl)/ (Ae) he said they are the same material so same young modulus..and same extension..then tension is the F so F= A/l ( no need for the young modulus and the extensoin so ...( l/A) /( 2l /0.5A) continue and u will get 4 / 1
21: A...if u draw a straight line from the origin to the same point the drawn graph reaches and took 0.5 x 17 x 0.30 it will equal 2.7..and the graph shows that thre is much spce between it and the line we drew so it must be < 2.7..so C and are wrong..B is wrong because it is about or approximatley the same as 2.7 so it alo must be less than 2.6 so A
22 and 23 are also basic and simple stuff
24: B...the distance between 2 adjacent nodes orantinodes is half a wavelength..so the distance between a node and an anti node is the quarter
25: A..N lines per metre means 1/N...then substituting into the equation it will be (1/N) sin (theta)= n ( lambda)
26: B...as shown 1.5 lambda = 2.1m so one wavelength is 1.4..then v=f lambda = 80 x 1.4 = 112
27: A...both are attracted to the +ve plate so both are -ve
28: B...E is proportional to V so doubling V doubles E......E is inverseley proportional to d so halfing d doubles E so B
29: D...v^2 = 2as..a=F/m so v^2 = 2 x (F/m) s and F= EQ means Ee as in the question..so v^2/2 = (Ee/m) s continue and u will get D
30: A...I=Q/t..simple question
31: D...W=QV so Q= 7.2 x 10^4 / 12 = 6000
32: A...basic EMF definition
33: B....I = E1 - E2 / R so ( 3 - 1.2) / 9 = 0.2
34: D..simple question..total resistorsin series = R1+R2+R3..in parrallel...1/Rt= 1/R1 + 1/R2 + 1/R3
35: D...light intensity and temperature increases so resistance in the LDR and thermistor decrease so the other resistor resistance increases
36 and 37 are simple questoins
38: C...just subtract the proton number from the nucleon number of each
39: C...simple question
40: D...he said a neutron so proton number must be zero and nucleon number is one
fr the question no.13 u need to multiply the radius and the weightso it will be
900*0.2(radius of disc)=f*1.2so we will get 150N
 
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thnx very much....but why F x 1.2 it should 0.6 as the disc is in the middle ??
 

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Here is the June 2009.. oh it was a tough one!!

1: D...basic stuff
2: C...opp/adj = tan ( theta) so opp=adj tan(theta) = 20tan( 30) =11.5
3: B...the readings are -6.5 and 14...so 14- (-6.05) so 20.5
4: B...P=VI= 1.2 x 0.48
5: D..he siad terminal velocity means velocity is constant and so d must have a constant increase..all other 3 graphs have d starts to decrease so D is correct
6: C...obvious from the graph
7: A...basic stuff...
8: C...air resistance is neglected so constant horizontal velocity and constant vertical acceleration
9: B...rate of changeof momentum= mv-mu, so (0.1 x -30) - (0.1 x 20)= -5....we used - 30 because it was travelling in the opposite direction
10: C...the momentum of body m must equal the momentum of body 2m..this means that if body m has half the mass so it must have double the velocity so ratio of kinetic energy of m /2m = ( 0.5 m x 2v^2)/ (0.5 x 2m x v^2) so it will be 2/1
11: A..the ball will not fall diagonaly..so S must equal to Q...according to the question..he didnt say if the ball will flought orwill sink so i thought that the upthrust is the difference of the pressure betwwn the water and the weight of the object so R > than P
12: D...he said it is falling with uniform velocity so no acceleration so no resultant force..and he said it is still falling horizontally so no resultant torque..so the conditions for equillibrium are found
13: ooh ill blow from this one and i may need some 1 to explain it to me..thnx in advance
14: C...K= F/v^2 means F/v x v and F = P/v so K = P/v/ (v x v) so K=(P/v) x v^-2 and then K = P x v^-1 x v^-2 so P x v^-3 means P/v^
15: B...container X lost half of the water means it lost half the height and half THE MASS so m/2 g h/2 so 2 x 2=4 so ( mgh )/4
16 and 17 are basic and simple stuff
18: D...at first the pressure was pgh..then the height was raised by another h so pgh + h = pg2h
19: A...B and C are wrong because they are brittle so they will break..steel is a mixture of 2 substances so it have the melting point of 2 substances which are iron and carbon...so aluminium is a single metal susbtance and so it will melt first
20: D...young modulus is ( Fl)/ (Ae) he said they are the same material so same young modulus..and same extension..then tension is the F so F= A/l ( no need for the young modulus and the extensoin so ...( l/A) /( 2l /0.5A) continue and u will get 4 / 1
21: A...if u draw a straight line from the origin to the same point the drawn graph reaches and took 0.5 x 17 x 0.30 it will equal 2.7..and the graph shows that thre is much spce between it and the line we drew so it must be < 2.7..so C and are wrong..B is wrong because it is about or approximatley the same as 2.7 so it alo must be less than 2.6 so A
22 and 23 are also basic and simple stuff
24: B...the distance between 2 adjacent nodes orantinodes is half a wavelength..so the distance between a node and an anti node is the quarter
25: A..N lines per metre means 1/N...then substituting into the equation it will be (1/N) sin (theta)= n ( lambda)
26: B...as shown 1.5 lambda = 2.1m so one wavelength is 1.4..then v=f lambda = 80 x 1.4 = 112
27: A...both are attracted to the +ve plate so both are -ve
28: B...E is proportional to V so doubling V doubles E......E is inverseley proportional to d so halfing d doubles E so B
29: D...v^2 = 2as..a=F/m so v^2 = 2 x (F/m) s and F= EQ means Ee as in the question..so v^2/2 = (Ee/m) s continue and u will get D
30: A...I=Q/t..simple question
31: D...W=QV so Q= 7.2 x 10^4 / 12 = 6000
32: A...basic EMF definition
33: B....I = E1 - E2 / R so ( 3 - 1.2) / 9 = 0.2
34: D..simple question..total resistorsin series = R1+R2+R3..in parrallel...1/Rt= 1/R1 + 1/R2 + 1/R3
35: D...light intensity and temperature increases so resistance in the LDR and thermistor decrease so the other resistor resistance increases
36 and 37 are simple questoins
38: C...just subtract the proton number from the nucleon number of each
39: C...simple question
40: D...he said a neutron so proton number must be zero and nucleon number is one
for q.3, why is it nt 21?? (2 sig fig)
 

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because the answer was exactly 20.5 and he didnt say round to 3 sf...he meant the most appropriate which is 20.5
bt v have both the values to 2 sig fig dn y nt the ans *confused*
 
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