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AS Physics P1 MCQs Preparation Thread.
- Thread starter mujtabashahnawaz
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I already posted them in the other thread..but looks like few ppl noticed them..i thought as this thread have more views and only 1 day before the exam..so ill post them here too..
NOV 2011 VARIANT 2
1: C....W=QV, so W which is = work done = energy..so C
2: D......basic stuff
3: D......again, basic stuff
4: D.....when we add or subtract numbers..we just add their uncertanities so 0.02 + 0.03 = + or - 0.05 so D
5: B....A is ofcourse wrong, C and D are systematic errors so B
6: C......area under the graph= 5 x 20 x 0.5 = 50
7: B..... terminal velocity, means no acceleration means no reusultant force so..acceleration is zero..then resultant force is zero..so the force of gravity or the weight must be = to the air resistance..and ofcourse weight is no zero, it is mg..so air resistance = weight = mg so no resultant force so B
8: D....C will not help to deduce anything from the graph..Ais wrong because acceleration is constant..and B is wrong because maybe when the d was zero..the time when it is zero,, they didnt start to count from 0 when it was stationary..maybe they started to count when the car started to move..D is correct because the line is not straight
9: D..... u can find the avg force..and the momentum..as the rate ofchange of momentum is Ft...and u can find the time of impact but u wont find the acceleration from the graph
10: B....using the conservation of momentum..when the sand is added the mass increases so the v decreases..then this v is constant even if the sand was removed again..so not because we remove the sand it will be faster..this is wrong..to make it faster again a forceneeds to be applied..so B
11: A....again, conservation of momentum...m1u1 + m2u2= (m1 + m2) v so (20 x 6) + (20 x -15) = 32v so v =1.9 ( NOTE: i used -15 because it was travelling in the opposite direction)
12: D...basic stuff
13: D..again basic stuff
14: B... F x d = F x d= ( 30KN x 10m) = F x 20m so F = 15Kn...i used F x 20 because the trailer causes a force at point X..but the opposite force exerted by the cab is on all the trailer..and the trailer is 20m long
15: sry..i got wrong
16: B...first we get v using K.E = 0.5 m v^2 so 4.5 x 10^5 = 0.5 x 1000 x v^2 so V = 30 then v^2 = u^2 + 2as and a = F/m so V^2/ 2 x ( F/m) = s then 30^2 / 2 x ( 6000 / 1000) = 75m
17: A...efficiency = ( useful output energy) / ( total input energy) so 8J is the useful output energy..and the total input energy is 100..because 92J is lost as heat so 92 + 8 = 100 so 8/100 = 0.08 x 100 = 8%
18: D...basic stuff
19: again basic stuff
20: A...Prssure = F/A so 290 / 0.036 =8056 PA..then P= pgh so h = 8056/ ( 930 x 9.81) = 0.88m
21: C...2 and 3 are definetley correct as the graph shows..but one is wrong..yes the extension of P is greater but not twice
22: A...for a force of 5N we had an extension of 3 cm...so for a load of 2N what will be the extension??? cross multiply and it will be (3cm x 2N) / 5N = 1.2 cm so the Total extension is 3 + 1.2 = 4.2cm
23: B...easy question, just compare the graphs
24: basic stuff
25: anotherbasic stuff THAT HAVE TO BE MEMORIZED
26: B...i made this question mentally so i dont know if my way is correct..but i got the correct answer
...at t = 18 the phase difference is 180 degrees....he wanted when is the phase difference will be 1/8 of the phase difference..so 180 x 1/8 = 22.5..so the phase difference is about 22.5..the graph is divided by an interval of starting from the left side where the amplitude of the wave is negative..i started to compare..at time 4s and 8s and 9s..the difference looked greater than 22.5 so i found 4.5 the closest one
27: B... simple question
28: A.....B is wrong....interference only occurs when there are two COHERENT SOURCES not uncoherent..C is wrong as light can not be polarised..and D is wrong as it have nothing to do with interference so A..as white light will produce 7 kinds of lights each with different frequency
29: D...E= v/d = 9/ 4 x 10^-3 so 2250 is about 2.3 x 10^3
30: A...D and c are wrong..electron is attracted towards the +ve plate which is upward..then the mass of the electron is 1/2000 of a proton..and an alpha particle will take the path B which is shown in the question..an electron is much lighter so it will be A as the force acting on it will be greater
31: A...B and D are wrong... C is wrong because in an electric field..an electron will not be attracted in a straight line..like the question 31...the force must be like a tangent to the field lines
32: C... I=Q/t so if V = IR then V = (Q/t) x R so ( 8 / 20) x 30 = 12
33: D...Basic stuff
34: A...R = x= pl/A = pl/ ( pie ) (d/2)^2 and Ry = p x 2l/(pie) (2d/2)^2..divide Rx by Ry and it will be 2/1
35: A....I= E/ ( R + r) and P=I^2 x R so P = [(E/ ( R+r)]^2 x R substitute the values given for the variable resistor..starting with 0.5 till 4 ex: for the variable resistor try 0.5 then 1 then 1.5 then 2 etc..ull find the power value increased then started to decrease similar to graph A
36: C...basic stuff..must be memorized
37: B..using the equation of a potential diviser..V output = [ R1/ ( R1 + R2)] x V = try when the variable resistor was o and also 4.5..as for minimum and maximum..ex 0/( 0 + 1) x 25 = 0 and 4/ ( 4 + 1) x 25 = 20 so B
38: A...the volt metre and the resistor are in parallel so their total resistance is 100 using the equation of the resistors in parralel..then we will substitute in the same equation i explained in the previous question..it will be (100 /( 400 + 100) x 60 = 12V
39: C..basic stuff..but note that B is wrong because he said helium atom...it must be helium nucleus
40: C..again simple question
1: C....W=QV, so W which is = work done = energy..so C
2: D......basic stuff
3: D......again, basic stuff
4: D.....when we add or subtract numbers..we just add their uncertanities so 0.02 + 0.03 = + or - 0.05 so D
5: B....A is ofcourse wrong, C and D are systematic errors so B
6: C......area under the graph= 5 x 20 x 0.5 = 50
7: B..... terminal velocity, means no acceleration means no reusultant force so..acceleration is zero..then resultant force is zero..so the force of gravity or the weight must be = to the air resistance..and ofcourse weight is no zero, it is mg..so air resistance = weight = mg so no resultant force so B
8: D....C will not help to deduce anything from the graph..Ais wrong because acceleration is constant..and B is wrong because maybe when the d was zero..the time when it is zero,, they didnt start to count from 0 when it was stationary..maybe they started to count when the car started to move..D is correct because the line is not straight
9: D..... u can find the avg force..and the momentum..as the rate ofchange of momentum is Ft...and u can find the time of impact but u wont find the acceleration from the graph
10: B....using the conservation of momentum..when the sand is added the mass increases so the v decreases..then this v is constant even if the sand was removed again..so not because we remove the sand it will be faster..this is wrong..to make it faster again a forceneeds to be applied..so B
11: A....again, conservation of momentum...m1u1 + m2u2= (m1 + m2) v so (20 x 6) + (20 x -15) = 32v so v =1.9 ( NOTE: i used -15 because it was travelling in the opposite direction)
12: D...basic stuff
13: D..again basic stuff
14: B... F x d = F x d= ( 30KN x 10m) = F x 20m so F = 15Kn...i used F x 20 because the trailer causes a force at point X..but the opposite force exerted by the cab is on all the trailer..and the trailer is 20m long
15: sry..i got wrong
16: B...first we get v using K.E = 0.5 m v^2 so 4.5 x 10^5 = 0.5 x 1000 x v^2 so V = 30 then v^2 = u^2 + 2as and a = F/m so V^2/ 2 x ( F/m) = s then 30^2 / 2 x ( 6000 / 1000) = 75m
17: A...efficiency = ( useful output energy) / ( total input energy) so 8J is the useful output energy..and the total input energy is 100..because 92J is lost as heat so 92 + 8 = 100 so 8/100 = 0.08 x 100 = 8%
18: D...basic stuff
19: again basic stuff
20: A...Prssure = F/A so 290 / 0.036 =8056 PA..then P= pgh so h = 8056/ ( 930 x 9.81) = 0.88m
21: C...2 and 3 are definetley correct as the graph shows..but one is wrong..yes the extension of P is greater but not twice
22: A...for a force of 5N we had an extension of 3 cm...so for a load of 2N what will be the extension??? cross multiply and it will be (3cm x 2N) / 5N = 1.2 cm so the Total extension is 3 + 1.2 = 4.2cm
23: B...easy question, just compare the graphs
24: basic stuff
25: anotherbasic stuff THAT HAVE TO BE MEMORIZED
26: B...i made this question mentally so i dont know if my way is correct..but i got the correct answer
...at t = 18 the phase difference is 180 degrees....he wanted when is the phase difference will be 1/8 of the phase difference..so 180 x 1/8 = 22.5..so the phase difference is about 22.5..the graph is divided by an interval of starting from the left side where the amplitude of the wave is negative..i started to compare..at time 4s and 8s and 9s..the difference looked greater than 22.5 so i found 4.5 the closest one 27: B... simple question
28: A.....B is wrong....interference only occurs when there are two COHERENT SOURCES not uncoherent..C is wrong as light can not be polarised..and D is wrong as it have nothing to do with interference so A..as white light will produce 7 kinds of lights each with different frequency
29: D...E= v/d = 9/ 4 x 10^-3 so 2250 is about 2.3 x 10^3
30: A...D and c are wrong..electron is attracted towards the +ve plate which is upward..then the mass of the electron is 1/2000 of a proton..and an alpha particle will take the path B which is shown in the question..an electron is much lighter so it will be A as the force acting on it will be greater
31: A...B and D are wrong... C is wrong because in an electric field..an electron will not be attracted in a straight line..like the question 31...the force must be like a tangent to the field lines
32: C... I=Q/t so if V = IR then V = (Q/t) x R so ( 8 / 20) x 30 = 12
33: D...Basic stuff
34: A...R = x= pl/A = pl/ ( pie ) (d/2)^2 and Ry = p x 2l/(pie) (2d/2)^2..divide Rx by Ry and it will be 2/1
35: A....I= E/ ( R + r) and P=I^2 x R so P = [(E/ ( R+r)]^2 x R substitute the values given for the variable resistor..starting with 0.5 till 4 ex: for the variable resistor try 0.5 then 1 then 1.5 then 2 etc..ull find the power value increased then started to decrease similar to graph A
36: C...basic stuff..must be memorized
37: B..using the equation of a potential diviser..V output = [ R1/ ( R1 + R2)] x V = try when the variable resistor was o and also 4.5..as for minimum and maximum..ex 0/( 0 + 1) x 25 = 0 and 4/ ( 4 + 1) x 25 = 20 so B
38: A...the volt metre and the resistor are in parallel so their total resistance is 100 using the equation of the resistors in parralel..then we will substitute in the same equation i explained in the previous question..it will be (100 /( 400 + 100) x 60 = 12V
39: C..basic stuff..but note that B is wrong because he said helium atom...it must be helium nucleus
40: C..again simple question
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And this is...NOV 2011 variant 1
1: D...basic stuff
2: B.. ke is 0.5 m v^2...100m race can be run with a speed of about 10ms-1 so 0.5 x 80 x 10^2 = 4000..or any number which is in the thousands
3: C..basic stuff
4: B...lets try having P to be m^2 substitute in the equation..it will be m ( 3+ ( m^2/m^2) = Q x s^2 so Q = ms-2
5: B...simple questoin..just substitute the values in the equation..and try with the uncertanities..ex..when u put F..put it 19.63 then 19.61..ull find that d causes the largest change
6: A...it was having an increase in velocity..then terminal velocity..then after it collided with the ball INELASTICALLY, it moved in the opposite direction so V must be -ve and constant with a value less than the one before collision..then as it goes up the inclined plane..v starts to decrease uniformly till it reaches zero..so A..because as the ball lost KE after the collision, v decreased so more time is needed to reach the plane
7: D..with air resistance..resultant force becomes zero means that a will start 2 decrease it reaches zero so D
8: B...the vertical component of a is always 9.81..he said in the quest that up is +ve..and we know that a will be +ve down so it will be -ve when going up
9: B..mass hape and density of the air affects the velocity so B
10: C...must be memorized
11: A..basic stuff
12: A..he said inelastic sollision..the momentum is constant until the collisoin occurs, then it decreases to the negative value as the hockey goes to the opposite direction
13: A...basic equillibrium triangle rule, B and C aewrong because Q and p will meet and D is wrong all te forces are made the opposite way
14: C..torque is F x perpendicular distance between the 2 forces..using the equation opp/hyp = sin theta.......so opp = 0.3sin(50) = 0.23 so 2 x 0.23 = 0.46
15: C...one ring of mass m is 7 scales away from the pivot...and 2 rings of mass 2m is 5 scales away from the pivot..so this means that 2 x 5 is 10...and the other ring is 7 scals away so 10 - 7 = 3 ..so the new ring must be 3 scales away from the pivot so 5
16: C.....m1u1 + m2u2 = ( m1 + m2) v so Y is satationary so u = o so , mu/2m = v so v is 1/2 so the kinetic energy is also halfed so C
17: A...how work is done against gravity when ur falling???
18: B...Work done = ( Final mechanicl energy - initial mechanical energy) so W.D = ( Final K.E + Final PE) - ( initil KE + Initial PE) so wrok is done against friction so -10KJ and Q is 50KJ less than P so -10 = ( KE + [P - 50]) - ( 5 KJ + P) continue and ull get KE as 45KJ
19: C...efficiency = useful output power / TOTAL input power so 80% = 120 / X so X is 120 / 80% = 150 then the heat loss is 150 - 120 = 30
20: E= F x d = ma x d so a = E/m x d and the dirction is to the right as it loses PE and so gains Pe and so increasing the velocity so A
21: basic stuff
22: again basic stuff
23: B..he said same type of material so same young modulus
24: B..area under Force - extension graph is the energy stored in the material
25: C...F = kx
26: using the young modulus equation, Fl/Ae..F is inverselyproportional to l so B
27: basic stuff
28: D..covering one of the slits decreases the amplitude by half..amplitude^2 = Intensity so I decreases by 4
29: A..stationary wave can be formed in a closed or an open pipe
30: C...constant phase difference in the line Rs so no interference
31: A.....E=v/d = 50 / 5 x 10^-3 so 10000 and from +ve to -ve is from up to down
32: C....the resultant force is obviously zero..the resultant torque is anticlockwise because the -ve point charge will be attracted to the +ve one from right to left
33: B....R = pl/A so p = RA / l..in A C & D he didnt say the three quantities needed to get the reistivity..while in B..he did mention the Area Length and the Resistance
34: D..for A: P = I^2 R so ( Q/t)^2 x r not only Q to be squared so its wrong B: E/ t = Power and P = V^2 / R not V^2 / R^2 so it wrong and C: P= VI not Pt= VI..for D = Q = It so E I / P as E / P = t so It so D is correct
35: A...basic definitions..current is the RATE OF FLOW OF CHARGE
36: B..again basic stuff
37: D..justmentally think about it
38: B...I is inversely proportional to R ..so as R is doubled I is halfed etc...
39: B...V output= ( 5000/ 5000 + 5000) x 9 = 4.5..this is the maximum.. Vout and ofcourse0 is the minimum
40: D..beta is 0 nucleon number and - 1 proton number so when it is emmitted the proton number is increased by 1
1: D...basic stuff
2: B.. ke is 0.5 m v^2...100m race can be run with a speed of about 10ms-1 so 0.5 x 80 x 10^2 = 4000..or any number which is in the thousands
3: C..basic stuff
4: B...lets try having P to be m^2 substitute in the equation..it will be m ( 3+ ( m^2/m^2) = Q x s^2 so Q = ms-2
5: B...simple questoin..just substitute the values in the equation..and try with the uncertanities..ex..when u put F..put it 19.63 then 19.61..ull find that d causes the largest change
6: A...it was having an increase in velocity..then terminal velocity..then after it collided with the ball INELASTICALLY, it moved in the opposite direction so V must be -ve and constant with a value less than the one before collision..then as it goes up the inclined plane..v starts to decrease uniformly till it reaches zero..so A..because as the ball lost KE after the collision, v decreased so more time is needed to reach the plane
7: D..with air resistance..resultant force becomes zero means that a will start 2 decrease it reaches zero so D
8: B...the vertical component of a is always 9.81..he said in the quest that up is +ve..and we know that a will be +ve down so it will be -ve when going up
9: B..mass hape and density of the air affects the velocity so B
10: C...must be memorized
11: A..basic stuff
12: A..he said inelastic sollision..the momentum is constant until the collisoin occurs, then it decreases to the negative value as the hockey goes to the opposite direction
13: A...basic equillibrium triangle rule, B and C aewrong because Q and p will meet and D is wrong all te forces are made the opposite way
14: C..torque is F x perpendicular distance between the 2 forces..using the equation opp/hyp = sin theta.......so opp = 0.3sin(50) = 0.23 so 2 x 0.23 = 0.46
15: C...one ring of mass m is 7 scales away from the pivot...and 2 rings of mass 2m is 5 scales away from the pivot..so this means that 2 x 5 is 10...and the other ring is 7 scals away so 10 - 7 = 3 ..so the new ring must be 3 scales away from the pivot so 5
16: C.....m1u1 + m2u2 = ( m1 + m2) v so Y is satationary so u = o so , mu/2m = v so v is 1/2 so the kinetic energy is also halfed so C
17: A...how work is done against gravity when ur falling???
18: B...Work done = ( Final mechanicl energy - initial mechanical energy) so W.D = ( Final K.E + Final PE) - ( initil KE + Initial PE) so wrok is done against friction so -10KJ and Q is 50KJ less than P so -10 = ( KE + [P - 50]) - ( 5 KJ + P) continue and ull get KE as 45KJ
19: C...efficiency = useful output power / TOTAL input power so 80% = 120 / X so X is 120 / 80% = 150 then the heat loss is 150 - 120 = 30
20: E= F x d = ma x d so a = E/m x d and the dirction is to the right as it loses PE and so gains Pe and so increasing the velocity so A
21: basic stuff
22: again basic stuff
23: B..he said same type of material so same young modulus
24: B..area under Force - extension graph is the energy stored in the material
25: C...F = kx
26: using the young modulus equation, Fl/Ae..F is inverselyproportional to l so B
27: basic stuff
28: D..covering one of the slits decreases the amplitude by half..amplitude^2 = Intensity so I decreases by 4
29: A..stationary wave can be formed in a closed or an open pipe
30: C...constant phase difference in the line Rs so no interference
31: A.....E=v/d = 50 / 5 x 10^-3 so 10000 and from +ve to -ve is from up to down
32: C....the resultant force is obviously zero..the resultant torque is anticlockwise because the -ve point charge will be attracted to the +ve one from right to left
33: B....R = pl/A so p = RA / l..in A C & D he didnt say the three quantities needed to get the reistivity..while in B..he did mention the Area Length and the Resistance
34: D..for A: P = I^2 R so ( Q/t)^2 x r not only Q to be squared so its wrong B: E/ t = Power and P = V^2 / R not V^2 / R^2 so it wrong and C: P= VI not Pt= VI..for D = Q = It so E I / P as E / P = t so It so D is correct
35: A...basic definitions..current is the RATE OF FLOW OF CHARGE
36: B..again basic stuff
37: D..justmentally think about it
38: B...I is inversely proportional to R ..so as R is doubled I is halfed etc...
39: B...V output= ( 5000/ 5000 + 5000) x 9 = 4.5..this is the maximum.. Vout and ofcourse0 is the minimum
40: D..beta is 0 nucleon number and - 1 proton number so when it is emmitted the proton number is increased by 1
Thanx a bunch gal
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ok as it involves hieght we r gona take kinetic energy and potential energy too..
1/2 mu^2 = mgh
1/2 mv^2= (mgh /2) (cz its saya half hieght)
now divide them in da same form as given above...from left syd 1/2 cancels, m cancels out..from ryt syd m and g cancels out leaving us with u^2/ v^2 = 2/1
da question asks da ratio of v by u so invert da equation which gives v^2 / u^2 = 1/2 hence 2 remove da square root v/u=1/sq root of 2 which is C
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wow you really listed most of the estimates.. thanks dude u rock..
can u tell me some more? like the approximate diameter of a uranium atom or young modulus of a metal or i think its called the range of alpha particles.... and any other like these that come across ur mind.. thanks bud..
youngs modulus of materials are like to the power 10 or 11 (eg: 10^11 or 10^10)
you can find out the diameter of uranium atom by the formula of volume
4/3 pie (r^3)
mass of uranium is 238u
where u=1.67x10^-27
so the mass is around 10^-25
the range of alpha particles is like a few cms like 5-6cm in air n can be stopped by a piece of paper/cardboard
it travels with 10% speed of light, has greatest ionization effect
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Q 27 its D... it can either be A or D...it cant be A cz lines cant be parallel there
for 14 use Esin45..u dont have E so just use sin 45... and i dont exactly know hw 2 explain dis so sory
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
no 25...why is it A?
no 25...why is it A?
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yeah, how do we do this??
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is the ans D for 27?
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Q no 27 from 1 anti node to node will be lamda/4=x (x is da distance)
so lamda is 4x..then use da formula f=c/lamda hence u get c/4x which is D
Q no 34... w8 4 smtym
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Why are we assuming that the initial and final velocity are 0?? It dusn't say it is dropped from rest?
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Lambda=4x, as distance btwn 1 node and antinode =lambda/4
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thankyou so much that certainly cleared it up..
one more doubt.. i absolutely no idea of how a potentiometer works. what is its purpose and how does it works.. also the effects of making the changes like increasing/decreasing the length, increasing decreasing the voltage of the either batteries.. please explain all this to me and in as simple language as possible cuz in my pacific physics textbook i couldnt understand properly the concept of potentiometer.. help would be greatly appreciated thank you in advance.
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and is the ans D for q,34???