# As physics p1 MCQS YEARLY ONLY.

#### SciGen

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
kindly explain me Q13 and Q19 ASAP

#### A.ELWY 7

bt v have both the values to 2 sig fig dn y nt the ans *confused*
sry i didnt well understand what u said..but all i can tell you is that he didnt ask for u to round to 2 significant figures..but to give an answer to the most appropriate significant figure...oohh now i understood what u mean ,,ok, see...the scale was given to 0.1 C so to 1 decimal place..so if ur answer was to 1 decimal place leave it like this..as we had hear..if it was more then round it to the nearest decimal place whish i helped

#### omg

sry i didnt well understand what u said..but all i can tell you is that he didnt ask for u to round to 2 significant figures..but to give an answer to the most appropriate significant figure...oohh now i understood what u mean ,,ok, see...the scale was given to 0.1 C so to 1 decimal place..so if ur answer was to 1 decimal place leave it like this..as we had hear..if it was more then round it to the nearest decimal place whish i helped
yes u did!
now im clear!
thanks

#### A.ELWY 7

kindly explain me Q13 and Q19 ASAP
Q13: moment is Fxd = fxd so F x 1.20 = 900 x 0.2 ( 0.20 is the radius given)...so F = 150..by the way i think it is Tkb..he just explained it to me..so all the credit goes for him Q 19: i explained it above but ill try again...here we all know that brittle materials, example glass..when heated they quiclk break...so B & C are wrong...still D and A..D is wrong because: when u have a 2 substances mixed to make 1 new single subtance..the new substance takes from the properties of the 2 substances mixed together..in this case steel take the properties of iron and carbon..( this is i think im correct as this is what remember from IG chemistry)..carbon is a non-metal so it have a high melting point...so steel ofcourse will have a higher melting point than aluminium because it have the , lets imagine it like this, the melting point of both iron and carbon combined..ofcourse much less than both combined but also higher than aluminum...Hope u understood and i helped

#### A.ELWY 7

yes u did!
now im clear!
thanks
welcome..i feel very happy when i help some 1

omg

#### SciGen

Nov 2008:
1: C...v=f x (lambda)..he said expressed as the number of the waves he meant the frequency..so speed of light is ( 3 x 10^8) so 3 x 10^8 / 600 x 10^-9 = 5 x 10^14
2: C...i dont know why they put a question of a specific heat capacity in AS level..but anyway u must know the equation of the specific heat capacity then derive it in the equation
3: A...use pythagoras theorm and try it for every choice..A will give the greater one
4: D..he said there is a Systematic uncertanity of 1% and the numbers show that there is another1% as it fluctuates between 1.98 and 2.02 so it is 2 +/- 0.02 so this is the second 1%...1% + 1% = 2% so answer is D
5: D...when the true current is 0.2 the false current > 0.2..when the true one is 0.4 the false one is >0.4...when the true current is 0.6 the false current is > 0.6..but at 0.8A ...they both meet at that point so D
6: D...just by using the area under the graph..there is area for triangle and area for rectangle
7: A...basic stuff
8: A ...also basic stuff..uniform increase in velocity means constant acceleration
9: B...the force of the ball is opposite and equal in magnitude to the ground force...not the weight
10: D..this is the usual equation used in a perfectley elastic collision
11: A... for the 2Kg box...because it is accelerated downward so the weight is greater than the tension, so W-T=ma...so ( 2 x 9.81) - T = 2a , then for the 8 Kg box...the tension is grater than the friction because the 2 kg box is pulling it downwards so, T-F=ma, so T - 6= 8a...add both equations..so [((2 x 9.81)-T= 2a) + (T - 6)= 8a] so T will cancel each other and it will be like this: (2 x 9.81) - 6= 10a so a = 1.4
12: D...friction is up the plane as it opposes the block motio..resultant force is zero because he said it is falling with uniform velocity so no acceleration so no resultant force
13: A..this is easy and doesn't need explanation
14: A...torque of couple is when 2 forces of equal magnitude but act on an object in the opposite directions of each other..so A
15: A...work is force in the direction of the motion x ditance...so F is the same direction as X so Fx...and PE is mgh so Wh
16: A...PE increases with h so B is wrong..D is wrong because the question said it is thrown upwards not downwards..C is wrong because he didnt say it fell again to the ground
17: C...mgh = 0.5 x m x v^2
18: D...by trial using the equation P=Fv
19: D...ductile because it can be drawn to a wire and plastic because it wont return to its original length and shape
20: C.... basic IG stuff
21: C...F is proportional to x...in A it is F/2 so x is x/2..in B it is F/4 so x/4 in D it is 4F/3 so 4x/3 means 1.33x but in C it is 3F/2 so 3x/2 means 1.5 so C is the answer
22: C...basic stuff
23: C...again basic stuff
24: D...wave y has half the amplitude but 3 times the frequency as the diagram shows so D
25: B.. sound can never be polarised but can have interference and reflection
26: A...I = a^2 so 3^2/1^2 so 9/1
27: C...v of light is 3 x 10^8 ....distance between two adjacent nodes or antinodes is lambda/2...so one lambda is 15mm x 2 then f= V/ lambda = ( 3 x 10^8) / (30 x 10^-3)= 1 x 10^10
28: D...A&B will increase the seperation and C will affect the intensity..but increasin the frequency decreases the wavelength so decreasing the seperation
29: B..obvious and mentally solved...
30: A...F=W so W=EQ and E = v/d so W = v/d x Q subtitute so the answer is A
31: D...I= V/R..and R = pl/a and A= pie r^2 so for P: I= V/ pl/(pie) x 1^2 and for Q: I=V/pl/(pie) x 0.5^2 then divide the equation of P by Q and it will be 4/1
32: B....the copper wires are in parrallel so total resistance in parrallel = 1/Rt = 1/R1 + 1/R2 etc... so 1/Rt= (1/10) x 6 = 3/5 so 1/3/5= 1.67 ...then also the steel core is parralell so 1/Rt= 1/1.67 + 1/100= 0.608...1/0.608 so 1.6 is the answer
33: B...basic stuff
34: D...i didnt understand it but what happened is that they divided 100KC/ (200 x 2) = 250
35: D...get the total resistance using R=V/I it will be 3.75 then use the equation of the resistors in parrallel..so 3.75= 1/10 + 1/10 + 1/x...so by trial, D is correct
36: B...7.5/15 = 0.50
37: B...we need the length and the EMF 1 to calculate the pd
38: D...basic stuff
39: A...same number of protons but different number of nutrons so different nucleon number so A
40: A...because as we all know alpha particle have proton number of 2 but beta particle have -1 and nucleon number of 0 while aplha have 4 so A
[;ease

#### SciGen

Q13: moment is Fxd = fxd so F x 1.20 = 900 x 0.2 ( 0.20 is the radius given)...so F = 150..by the way i think it is Tkb..he just explained it to me..so all the credit goes for him Q 19: i explained it above but ill try again...here we all know that brittle materials, example glass..when heated they quiclk break...so B & C are wrong...still D and A..D is wrong because: when u have a 2 substances mixed to make 1 new single subtance..the new substance takes from the properties of the 2 substances mixed together..in this case steel take the properties of iron and carbon..( this is i think im correct as this is what remember from IG chemistry)..carbon is a non-metal so it have a high melting point...so steel ofcourse will have a higher melting point than aluminium because it have the , lets imagine it like this, the melting point of both iron and carbon combined..ofcourse much less than both combined but also higher than aluminum...Hope u understood and i helped
bro still cant get that why arent we taking 0.4m as the distance in the formula of torque in Q13.....and why all this melting point stuff in Q19 wch asks abt plastic deformation

#### arlery

Alright ermm so I'll solve November 2009 V11, but I'll only be able to post it at mid day.
I've checked V12 and its pretty much the same, so if someone could post Nov 05 and Nov 02, we'll be done with all the papers from 02-09. Then it'll only be 10 & 11 left!

#### SciGen

but diameter is the distance between the two torque forces of weight....isnt it?

#### A.ELWY 7

bro still cant get that why arent we taking 0.4m as the distance in the formula of torque in Q13.....and why all this melting point stuff in Q19 wch asks abt plastic deformation
y r u complicating it on urself..he said whatis the minimum force APPLIED TO EACH END OF THE LEVELER..so we multiply by the ditance between the 2 force....and the 0.2 is the ditance from the weight to the pivot so... F x 1.2 = 900 x 0.2
wish u understood this time
Q19: the melting point suff is just imaginary to help u understand..he said at room temperature, which will sustain the largest platic deformation.. so a force must be applied to each one...and i think u dont or didnt take chemistry..u must know that steel is stronger than aluminum...so A....again, wish u understood this 2

#### A.ELWY 7

Alright ermm so I'll solve November 2009 V11, but I'll only be able to post it at mid day.
I've checked V12 and its pretty much the same, so if someone could post Nov 05 and Nov 02, we'll be done with all the papers from 02-09. Then it'll only be 10 & 11 left!
can i please post the Nov 2011 3 variants..i dont have time to resolve from 2010 tojune 11..if u and any 1 who can help u, just post till june 2011 and ill post the 3 variants of Nov 2011 in the next 2 days

#### A.ELWY 7

[;ease
For Q 4...he said that there is a systematic uncertanity of 1%..but this question was tricky as there was another 1% uncertanity he didnt say..he wantedus to figure it out..which is when he gave us that the reading fluctuates between 1.98 and 2.02..if u think about it ull find that if it fluctuates between 1.98 and 2.02..so it is 2+0.2=2.02 or 2 - 0.2= 1.98...so this is another uncertanity which is 1%....so 1% + 1% = 2%..so 0.04 is the uncertanity

#### SciGen

June 2007

1: C...basic stuff
2: D..also basic stuff
3: im to0 sry i couldnt do this one and i would appreciate if some 1 can help me with it
4: C...one wavelength is 4 squares..so 4 x 2.5 = 10...amplitude is about 3.5 so 3.5 x 5= about 17
5: A...V=IR...1.2V / 0.48A= 2.5
6: D...again basic stuff
7: C...v is the vector sum of u and x and v=u +at so X is at so C
8: A...s= ut + 0.5at^2...get the time when the distance is 40m ..then when the distance is 30 m...then T1 - T2 = 0.38s
9: B..basic stuff
10: D...F=( mv - mu)/t ...so substituting into the equation it is ( p1 - p2)/ ( t2 - t1)
11: B... (20000 x 20) - (900 x 30) = 373000Ns
12: B...1st get the final V by: m1u1 + m2u2 =(m1 + m2)v so (2 x 4) + (4 x 1)= 6V..V = 2 then he said it stuck together so 0.5 x m x v^2 so 0.5 x 6 x 2^2= 12J
13: B...torque of a couple is defined as the 1 force x perpendicular distance between the 2 forces..so geth the perpendicular force by drawing a straight line perpendicular to the force..then use adj/hyp = cos( theta) so adj= 0.6 sin (30) so 0.52...0.52 x 8 = 4.2
14: C...basic stuff
15: A...P=pgh...so 100000/ (13.6 x 10^3) x 9.81 = h... so h = 0.75m
16: C... A is wrong because liquids doesnt have a very big seperation..B is wrong because it describes a gas..D is wrong because liquid atoms are not in fixed positions
17: C...copper is not brittle..and when drawn to a wire it will not return to its original shape again so plastic only
18: B... the whole area under the graph is the strain energy...and the unhighlited part which is between the 2 drawn lines is the heat enerfy..so the shaded are is B
19: A...1/2 F X so 0.5 x 2 x (0.90 - 0.50)= 0.40J
20: B...X is in tension as it is opposing the weight of the horizontal bar and the foce W...Y is aso in tension as it is opposing W..Z is in compression as it is between a force from X and a force of Y and W
21: C...A & B are wrong because no longitudnal waves are polarises..D is wrong because sound is longitudnal not transverse so C
22: A...I is proportional to a^2..try ti solve tis example by giving the intensity I of y a magnitude of 3600 and the intensity of X a magnitude greater than y by 10^12 then find the amplitude for both..it will be A the answer
23: D...im weak inthis kind of a question but i know that at Q it is not always zero at and R it is not entirley kinetic and at p the speed is not at maximum..so D
24: D..you must memorize this
25: C...basic stuff
26: D...A is wrong because decreasing the distance between the slits and the screen will decrease the fring seperation, B is wrong because this will also decrease the seperation..C is wrong because this will affect the intensity not the seperation..so increasing the frequency decreases the wavelength so seperation increase so D
27: A..stationary wave is when 2 owaves travelling opposite to each other ( so B and C is wrong) then he said 2 nodes and antinodes so a have 2 points shown so 2 nodes and anti nodes
28: D..basic stuff
29: A...again basic stuff to ne known..electron is attracted to +ve plate
30: B...E= v/d...so 4v / 2d so 2E
31: D..just memorize ohm's law..and becareful it is not A because he said they are equal not proportional
32: A...Q=It...(8 x 10^-3) x 0.020 = 0.16 mC
33: C....V1= 5000/ (5000+5000) x 2 = 1V then V2= 3000/ (3000 + 2000) x 2 = 1.2 so V1 - V2 = -0.2 so C
34: A...V was 4...then light increases so R decreases so V decreases and the only number less than 4 in the answers is 3V so A
35: C...W= QV so V = W/Q and I = Q/t
36: C... V= IR, i= v / r...3/ 2 + 4= 0.5..then pd= 0.5 x 4=2 then I= E/ (R+r) sp P = RI^2 so P= (E/R+r)^2 x R so (3/ ( 4+ 2)^2 x 4 = 1 so C
37: C...R = pl/A he siad they have same length..and they are of the same material so same resistivity..then he said of same volume and volume = A x h so same area so same Resistance so C
38: A...basic stuff
39: D..again B particle is 0 nucleon number and -1 is the proton number so D
40: C .... divide the proton number by the nucleon numbre for each element lithium will give the smaleest number so c

#### A.ELWY 7

am sry...but i think i wrote that im the one who needs help with it..so how can i explain it to u??