AS Physics P1 MCQs Preparation Thread.

[umarashraf]

I'll send you the link soon! ^_^

oka mate...

 

[Silent Hunter]

You know that,

I α a²

And,

I α 1 / r²

Using this relationship, can you make a link between a and r (as they are related in this question)?

(r2 / r1)² = (a1 / a2)²

Then just plug in the waves to get the answer.

got it Alhamdulilah . ThankS Alot

 

[USMAN Sheikh]

Any one pls i didnot get the middle part :S quest 33 pls ?

 

[umarashraf]

Any one pls i didnot get the middle part :S quest 33 pls ?

V=IR.. in the middle part, for a small change in voltage, large current was produced... which outlines that resistance must have decreased... hope u got this....

 

[USMAN Sheikh]

thanks alot for my help can u do these too pls q 10 33 36 and 40 i toghest i have faced help needed mates waiting for ur replies

 

[Oliveme]

questions 10, 15, 23, 26, 37 and 38. thank you very much.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

please answer the questions above. second time I'm posting.

 

[Silent Hunter]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf

Question 2 please...

 

[leosco1995]

thanks alot for my help can u do these too pls q 10 33 36 and 40 i toghest i have faced help needed mates waiting for ur replies

10. Force = rate of change of momentum.

Change in momentum = p2 - p1 (please note that it's p2 and not -p2 as stated on the graph)
Multiply by -1, you get - p2 + p1, or p1 - p2
= p1 - p2 / (t2 - t1)

B

33. In parallel, voltage in each parallel combination is the same.

Current in 5k and 5k ohm block = 2/10000 = 20 mA.

Voltage across PS = V1 = 20 mA * 5000 = 1.0 V (alternatively you can use a potential divider to get this value).

Current in 2k and 3k ohm block = 2/5000 = 0.8 = 40 mA.

Voltage at Q = 40 mA * 2000 = 0.8 V

Voltage at S = 2.0 V

V1 = 1.0 V
V2 = 2.0 - 0.8 = 1.2 V

V1 - V2 = 1.0 - 1.2 = - 0.20 V

C

36.

I = 3/6 = 0.5

Vt = E - Ir
Vt = 3 - 0.5(2) = 2.0 V

Output power = power across resistor = I^2 * R = 0.5^2 * 4 = 1.0.

C

40. In an electric field,

F = ma
and F = EQ, so

EQ = ma
a = EQ/m

Electric field strength is a constant, so a = Q/m

Find the charge (proton #) : mass (nucleon #) ratio. It's the lowest in option C.

 

[leosco1995]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf

Question 2 please...

What I don't like about this question is that they assume you know the units of specific heat capacity (which isn't in the syllabus, to my knowledge). From Chemistry, I learnt that q = m * c * ∆T

c = q / m * ∆T

Units of c = J . kg^-1 . K^-1
Which can be further be made by changing J into kg . m^2 . s^-2,

kg . m^2 . s^-2 . kg^-1 . K^-1

kg and kg^-1 cancel out, giving:

m^2 . s^-2 . K^-1

And in the question, b = c / T^3
b = m^2 . s^-2 . K^-1 / K^3

b = m^2 . s^-2 . K^-4

I hope this is the right answer. If not, I must have made a careless mistake somewhere.

 

[xxfarhaxx]

sum hlp needed !!
Q- 2, 31, 34
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_s04_ms.pdf

Q- 13, 16, 27, 32
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_ms_1.pdf

Q- 3, 9 , 11, 15, 20, 28
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_ms_1.pdf

 

[kinglynx]

no one has solved these two yet, please can you help me with them!!!

anyone please!!!

 

[leosco1995]

anyone please!!!

Tilt the images next time.

For the 1st one, tension always acts on the outer edge of a bar and compression always acts on the inner edge. So there will be tension at X and Z and compression at Y. Another way of solving is imagining what will happen to the forces when the weight is pulled.

For the 2nd one, the strings are in simple harmonic motion. Think of a pendulum. The speed will be maximum at the centre (as in point Q) and zero furthest away from the centre (as in point S and P). And the acceleration will be maximum furthest away from the centre. So D is right.

Look up some notes on basic SHM.

 

[Silent Hunter]

What I don't like about this question is that they assume you know the units of specific heat capacity (which isn't in the syllabus, to my knowledge). From Chemistry, I learnt that q = m * c * ∆T

c = q / m * ∆T

Units of c = J . kg^-1 . K^-1
Which can be further be made by changing J into kg . m^2 . s^-2,

kg . m^2 . s^-2 . kg^-1 . K^-1

kg and kg^-1 cancel out, giving:

m^2 . s^-2 . K^-1

And in the question, b = c / T^3
b = m^2 . s^-2 . K^-1 / K^3

b = m^2 . s^-2 . K^-4

I hope this is the right answer. If not, I must have made a careless mistake somewhere.

yes.... me too thought abut the syllabus thingy.... i mean we studied it in Olevels... and i was not able to recall that exactly... so thought that there might be another way of it according to the given syllabus

and thanks again

 

[USMAN Sheikh]

10. Force = rate of change of momentum.

Change in momentum = p2 - p1 (please note that it's p2 and not -p2 as stated on the graph)
Multiply by -1, you get - p2 + p1, or p1 - p2
= p1 - p2 / (t2 - t1)

B

33. In parallel, voltage in each parallel combination is the same.

Current in 5k and 5k ohm block = 2/10000 = 20 mA.

Voltage across PS = V1 = 20 mA * 5000 = 1.0 V (alternatively you can use a potential divider to get this value).

Current in 2k and 3k ohm block = 2/5000 = 0.8 = 40 mA.

Voltage at Q = 40 mA * 2000 = 0.8 V

Voltage at S = 2.0 V

V1 = 1.0 V
V2 = 2.0 - 0.8 = 1.2 V

V1 - V2 = 1.0 - 1.2 = - 0.20 V

C

36.

I = 3/6 = 0.5

Vt = E - Ir
Vt = 3 - 0.5(2) = 2.0 V

Output power = power across resistor = I^2 * R = 0.5^2 * 4 = 1.0.

C

40. In an electric field,

F = ma
and F = EQ, so

EQ = ma
a = EQ/m

Electric field strength is a constant, so a = Q/m

Find the charge (proton #) : mass (nucleon #) ratio. It's the lowest in option C.

thanks alot mate u r superb MASHALLAH i have one doubt that if examiner in 40th ques says fastest speed then answer will be A ??? :S

 

[Hassi123]

10... p2 is negative.. so overall change in momentum is -p2-p1.. which gives same numerical value as p2+p1... the answer is B

but the answer is p1-p2

 

[Hassi123]

10. Force = rate of change of momentum.

Change in momentum = p2 - p1 (please note that it's p2 and not -p2 as stated on the graph)
Multiply by -1, you get - p2 + p1, or p1 - p2
= p1 - p2 / (t2 - t1)

B

33. In parallel, voltage in each parallel combination is the same.

Current in 5k and 5k ohm block = 2/10000 = 20 mA.

Voltage across PS = V1 = 20 mA * 5000 = 1.0 V (alternatively you can use a potential divider to get this value).

Current in 2k and 3k ohm block = 2/5000 = 0.8 = 40 mA.

Voltage at Q = 40 mA * 2000 = 0.8 V

Voltage at S = 2.0 V

V1 = 1.0 V
V2 = 2.0 - 0.8 = 1.2 V

V1 - V2 = 1.0 - 1.2 = - 0.20 V

C

36.

I = 3/6 = 0.5

Vt = E - Ir
Vt = 3 - 0.5(2) = 2.0 V

Output power = power across resistor = I^2 * R = 0.5^2 * 4 = 1.0.

C

40. In an electric field,

F = ma
and F = EQ, so

EQ = ma
a = EQ/m

Electric field strength is a constant, so a = Q/m

Find the charge (proton #) : mass (nucleon #) ratio. It's the lowest in option C.

why do you multiply the change in momentum by -1?

 

[kinglynx]

Tilt the images next time.

For the 1st one, tension always acts on the outer edge of a bar and compression always acts on the inner edge. So there will be tension at X and Z and compression at Y. Another way of solving is imagining what will happen to the forces when the weight is pulled.

For the 2nd one, the strings are in simple harmonic motion. Think of a pendulum. The speed will be maximum at the centre (as in point Q) and zero furthest away from the centre (as in point S and P). And the acceleration will be maximum furthest away from the centre. So D is right.

Look up some notes on basic SHM.

I will tilt them next time thanks, Also thanks for the answer to the graph question that really helped me!!! However, the first question about the compression and tension, your answer is wrong, the answer is actually Tension at X AND Y, and compression at Z ... which doesnt follow what you said....

 

[Hassi123]

I will tilt them next time thanks, Also thanks for the answer to the graph question that really helped me!!! However, the first question about the compression and tension, your answer is wrong, the answer is actually Tension at X AND Y, and compression at Z ... which doesnt follow what you said....

i think he meant X and Y, it was a typo

 

[leosco1995]

thanks alot mate u r superb MASHALLAH i have one doubt that if examiner in 40th ques says fastest speed then answer will be A ??? :S

Thanks and you're welcome. And yes, A would be right if we had to find the acceleration with the fastest speed.

 

[AppleGreen]

MJ 2007 ques 23, help plz