As physics p1 MCQS YEARLY ONLY.

[bogus]

while applying formula for power do we consider internal resistance?????

 

[magnesium]

while applying formula for power do we consider internal resistance?????

if mentioned yes otherwise no

 

[bogus]

if mentioned yes otherwise no

but mcq 36 j 07 physics they mention it but dont use it :/ can u plz it to me magnesium ?

 

[magnesium]

while applying formula for power do we consider internal resistance?????

but mcq 36 j 07 physics they mention it but dont use it :/ can u plz it to me magnesium ?

u have to use internal resistance in calculating current in external circuit , using E = I(R+r) then use this current value and resistance of external circuit to get power ,P = V^2/R

 

[bogus]

u have to use internal resistance in calculating current in external circuit , using E = I(R+r) then use this current value and resistance of external circuit to get power ,P = V^2/R

why rnt we also using internal resistance while calculating power???

 

[h4rriet]

while applying formula for power do we consider internal resistance?????

The power of a battery is got by using P=IV only.

 

[bogus]

The power of a battery is got by using P=IV only.

Okay ! Got it ! Thanks a loooooot!!!

 

[Omar99]

Hi all, i've made this thread only for yearly papers starting from june 2002. We'll do 2-3 years each day. And anyone who wishes to help is most welcome. But please once again, this is only for YEARLY papers.

MOD EDIT

AsSalamoAlaikum Wr Wb!


Nov:2001 Answer Key

Here are few solved explanations for 9702 Physics Paper:1

June:2002

Nov:2002

June:2003

Nov:2003

June:2004

Nov:2004

June:2005

Nov:2005

June:2006

Nov:2006

June:2007

Nov:2007

June:2008

Nov:2008

June:2009

Nov:2011 # 1

Nov:2011 # 2

please help in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
question no. 6, 9,15,25,33

 

[Omar99]

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

can u please tell me how did u think that for question no.2 and can u please explain me Q 37 and Q 32. for varient 1

 

[Aishvarya.s]

hey can u plz solve mayjune 2010 variant 1

 

[kinxi winx]

are there any for the latest papers ?? like last 4 years

 

[abdullahaftab]

Woww What a thread this is. Can Someone plzz make a thread of yearly papers for Chemistry.... Just like this thread.... Or give me a link If its already somewhere

 

[bigboy]

A supermarket trolley, total mass 30 kg, is moving at 3.0 m s–1. A retarding force of 60 N is applied
to the trolley for 0.50 s in the opposite direction to the trolley’s initial velocity.
What is the trolley’s new velocity after the application of the force?

 

[A*(a*)]

The power of a battery is got by using P=IV only.

 

[A*(a*)]

The power of a battery is got by using P=IV only.

 

[h4rriet]

View attachment 28137

All you have to do is write down the balanced equations for the combustion of Mg, Al and S and then simplify them so that the mole ratio of Mg/Al/S to O2 is 1:x.

 

[h4rriet]

Ammonium compounds release ammonia when reacted with an alkali. Compound D is ammonium nitrate.

 

[A*(a*)]

All you have to do is write down the balanced equations for the combustion of Mg, Al and S and then simplify them so that the mole ratio of Mg/Al/S to O2 is 1:x.

Mg + 1/2 O2 --> MgO
Al + 3/4 O2 --> 1/2 Al2O3
S + O2 --> SO2
So moles are 0.5 , 0.75 and 1.
No graph shows them :/

 

[h4rriet]

Mg + 1/2 O2 --> MgO
Al + 3/4 O2 --> 1/2 Al2O3
S + O2 --> SO2
So moles are 0.5 , 0.75 and 1.
No graph shows them :/

0.5:1.5:1
1:3:2
Note: For Al, it's 3/2 O2.

 

[vishruthiit]

june 2006 question 34
guys plz tell me where to use P=VI or P=V²/R or P=I²/R

when resistance is given use the R formula or else use others...