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As physics p1 MCQS YEARLY ONLY.

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ok :)
JAZAK ALLAH
Q21 so u find energy by using area under the traingle so as it is not a perfect triangle so make a straight line till P so force will become approximately 110 N so 1/2 x b x h=1/2 x 2x10^-3 x 110=0.11J
 
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ok i will do it and u will get ur answers today INSHALLAH
thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again
u mean all of them or just question 2
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again
Q3) since both waves have the same wavelenght frequency will b same so 1/50=0.02ms=20s
and for peak voltage count the boxes under the crest OR trough so it is 4V
 
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Finally...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
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again

Q12) u first find the tension at the top of the belt tension is a force so torque=one of the two forces x distance between then so force =3/100 x 10^-3=30N
total tension is 30 x 2 cuz u have to take one of the two forces to find torque but we need the total
then torque of P=30N(one of the two forces) x 150 x 10^-3=4.5
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again
Q13)ok over here u assume mass of ball = 2kg and initial speed =10m/s
now find E=1/2 x 2 x 10^2=100J
it is thrown at an angle of 45 degree so now u find the new velocity =sin 45 x 10 =7.07m/s
find new ke=1/2 x 2 x 7.07^2=50J
so 50/100=1/2 so ans=E/2
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again

q14)momentum b4 collision= momentum after collision
so b4 collision they were stationary so momentum =0
0= 2 x 2 + 1 x v
v=4m/s
so now v found the speed at wich the 1kg trolley moves
find KE OF BOTH 1/2 x 2 x2^2- (-1/2 x 2 x4^2)=12J
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again
Q21) u first find area i hope u noe dat=1.96 x 10^-7m^2
then YOUNG MODULUS=stress/strain
find strain from the given values u get 5.1 x 10^-4 this is equal to e/l of which we have to find the % so just multiply by 100 u get the answer
 
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thanks a lot but I am extremely srry to say that my paper was varient 2 and 1...I suddenly realised when I went to check my answers....the questions r same in varient 1 but their order was one number back..i.e if i had mentioned question 3 was my doubt, that was question number 2 for varient 2.....if u have time, can u xplain them to me??! i know its my mistake....srry again
q25)10^-3/300 x sin 90 = 450 x 10 ^-9 x n so n =7
so # of maxima =7 x 2 +1 central maxima =15
 
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