As physics p1 MCQS YEARLY ONLY.

[A.ELWY 7]

Hey I have a doubt in qs 8 paper 9702/11/O/N/11..
Pls help me out...
thx in advance
n gud luk to all...

ill be posting Nov 2011 P11 today inshallah

 

[Blur JJ]

Erm... when are u guys gonna post the papers for 2010?

 

[Sanis]

novemeber 2011 V2 ,, number 26 plz help

 

[applepie1996]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q21 answer is D

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_13.pdf
Q26 ans is D

culd sum 1 solv them plz

 

[leosco1995]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q21 answer is D

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_13.pdf
Q26 ans is D

culd sum 1 solv them plz

Q21)

Let the extension of a spring be 'e'.

In A) the extension is e/2 because there are 2 are in parallel.
In B) the extension is e/2 + e/2 = e
In C) the extension is e/2 + e = 1.5e
In D) the extension is e/3 + e = 1.33e

C should be the answer, not D. Check the MS again.

Q26)

d sin θ = nλ

d = 10^-3 / 500 = 2 * 10^-6

So maximum number of orders is,

2 * 10^-6 = n * 600 * 10^-9
n = 3.33
n = 3

Total number of orders = 3 + 3 + central max = 7

 

[applepie1996]

Q21)

Let the extension of a spring be 'e'.

In A) the extension is e/2 because there are 2 are in parallel.
In B) the extension is e/2 + e/2 = e
In C) the extension is e/2 + e = 1.5e
In D) the extension is e/3 + e = 1.33e

C should be the answer, not D. Check the MS again.

Q26)

d sin θ = nλ

d = 10^-3 / 500 = 2 * 10^-6

So maximum number of orders is,

2 * 10^-6 = n * 600 * 10^-9
n = 3.33
n = 3

Total number of orders = 3 + 3 + central max = 7

thnx genius

 

[A.ELWY 7]

novemeber 2011 V2 ,, number 26 plz help

it is already explained at the previous pages of this thread

 

[Rakayz]

ASSALAMU ÁLAIKUM
some1 plz post nov 2009. It was said that it will be posted but its not yet done I know u guys r doing a great work by solving other papers but it will be very grateful if some one posts all the papers from nov 2009 till nov 11

JAZAK ALLAH KHAIR

 

[Rakayz]

Guys didnt we agree that we must upload the explanation of Nov 2009 and june 2010..this is the second day and no one posted anything....i cant work alone i need help!! i know most or all of u are busy studying, but im busy too..and this is not the time to quit on the thread..ive posted Nov 2011 variant 2 and ill post variant 3 tommorow inshallah...so some backup pls.. and im thanking u mubaraka for posting Nov 2002 and covering what we missed earlier !!

ASSALAMU ÁLAIKUM
can u plz post nov 2009 if possible....it will be kind of u ??!
JAZAK ALLAH

 

[applepie1996]

ASSALAMU ÁLAIKUM
can u plz post nov 2009 if possible....it will be kind of u ??!
JAZAK ALLAH

culd u tell me specific question numbers and i'll try 2 solve them

 

[Rakayz]

culd u tell me specific question numbers and i'll try 2 solve them

hmmm...i had many doubts in this paper...is it possible for u to solve the whole paper and post it here like others? coz this will help others too

 

[Silent Hunter]

OK here's all of June 2002.

June 2002
==========




32. A

R = V²/R = 240²/100 = 576. However, this is when the filament is heated so it has increased by 16 times. The normal room temperature would therefore be 576/16 = 36.

Thanks alot for the effort one thing in this paper: in this specific question, isnt the 576 at room temperature and it gets increased from here ?

 

[donorsolutions]

November 2002

1) C;
micro: 10-⁶
nano: 10-⁹
pico: 10-¹²

2) C; uniform acc. means that the gradient for V-t graph wud be a diagonal straight line (velocity changes at a constt. rate).

3) C;
View attachment 12497

4) B; basic definition.

5) D; The values are way different from the actual that is 9.81 m/s² but are close to one another... are precise but not accurate.

6) C; systematic error causes deviation from the actual value, therefore the graph does not include X0. Random errors cause the result to be spreaded out, the graph comes wide. In C, the graph is narrow, and well away from actual.

7) B;
View attachment 12498
5 squares in the figure have in them 7 waves. One wave therefore occupies,
5 : 7
x : 1
x= 5/7 square.
One square represents 10ms, so Time period is 10 x 5/7 = 7.14 ms
Hence the frequency is 1/7.14 x 10-³ = 140Hz.

8) A; Horizontal velocity remains constant therefore the hor. component of acc. is zero.

9) D; s= ½ at2 + ut
where, ut =0.
Therefore a= 2h/time = 2h/(t22 - ti2)

10) D; In air resistance, the acceleration decreases from 9.81 to 0 ms-2. No other graph shows this.

11) A; for elastic collisions, e= -1
uA –uB = e( vA – vB)
uA - uB = -vA + vB

12) the masses are equal so suppose it to be *m*
Initial momentum = Final momentum
60m – 30m = 2mV
30m = 2mV
30m/2m = V
V= 15cm/s

13) A; equal and opposite forces form a couple.

14) C; upward force = tension in string = 20 x 9.81 = 196.2N
Sum of clockwise momentum = Sum of anti-clockwise momentum
Distance d x ( 50 x 9.81) = (100-40) x ( 196.2)
490.5d = 11772
D= 24 cm from the pivot.
This could either be at the mark of 16 cm or 44 cm.

15) A; equilibrium triangle.

16) C; P= Fv

17) D; constant speed down the hill therefore there is no change in kinetic energy.

18) D; F=W= mg
=1.3 x 109 x 9.81
=1.275 x 1010N
P = F x s /t
= (1.275 x 1010 x 2)/(60x60x24)
=295208
= 300kW

19) C; let m be the mass,
P.E. =mgh
=1962m.
60% of the energy left.
K.E = 60/100 of 1962m = 1177.2m
1177.2m = ½ m V2
V= 48.5m/s

20) C; Fact.

21) C; apply the formula p = ρgh
For the first liquid, pressure comes 3531.6
For the other liquid, it comes 7063.2 which is twice the previous one.

22) A; quite basic this is.

23) C; area under the graph within limits.

24) B; F= kx
F is constt., k is given, as stated, therefore if k is doubled x is halved.
2k would give x/2
Next, we know that energy = work done = ½ kx2
WP= ½ * 2k * (x/2)2
=kx2/4
WQ= ½ * k * x2
= kx2/2
So, WP= ½ * kx2/2 = ½ WQ

25) A; wavelengths are to be learned of the electromagnetic spectrum.

26)D; maximum displacement of a point is its amplitude. On x-axis is time, so the labeled part is T.

27) D; I1/I2 = (A1/A2)2
3/I = a2/4a2
3 x 4a2 = I x a2
12 = I

28) B; use the formula λ =ax/D so the wavelength is directly proportional to the fringe separation.

29) D; use the formula dsinθ = nλ
Where d is spacing of the lines on the grating, λ is 590 x 10-9m, θ = 43/2 o.

30) C; V=IR

31) D; resistance of a filament does increase with a rise in temperature.

32) A; at +1.0V, current I = 50mA
V=IR
1/50x10-3 = 20 Ω
The graph at -1.0V is a straight line downwards so is infinite.

33) A; fact, charge entering a point must leave that point.

34) B; By Kirchhoff’s Law, I =I1+I2
Using V= IR, we get I =V/R
V/R = V1/R1 +V2/R2
The voltage across each is same, V =V1 =V2
Hence, 1/R = 1/R1 +1/R2

35) D; more resistance across variable resistor means more p.d. across it leaving decreased p.d. across XY. Then more portion of the wire would be used to maintain the galvanometer at zero deflection.


36) A;
View attachment 12495

37) B; field lines enter the negative point and originate from the positive.

38) A; isotopes have same number of protons, and different nucleon nmbr.

39) A; Only those that hit the nucleus directly bounce back or those that go quite near the nucleus are deviated at such large angles. As nucleus occupies a very small portion of an atom, this proportion of alpha- particles is really small. For more explanation, kindly take reference from your book.

40) C; it absorbed a neutron so nucleon nmbr increases by one. It emits two beta particles, so the proton nmbr increases in total by 2.

Ok....Why cant the answer to question 10 be A ?..i mean the acceleration is shown constant ( g) at first and then decreases to zero?...
And the answer to question 9 is C..please correct it. We square the time taken to cover height h. Not the individual times
Like its 2h/(t2-t1)^2 !

 

[donorsolutions]

D

Assalamu Alaikum
i had a confusion
the answer that u have mentioned for Q10 is A but the marking scheme says its B...
it would be very kind of you f u clear out this confusion
JazakAllahu khair
may Allah reward you for this

Do you have the marking Scheme For Nov 2002 (Official One ?)..If yes please post it here i really want it badly.
I would be grateful if you could share it here.

 

[leosco1995]

Thanks alot for the effort one thing in this paper: in this specific question, isnt the 576 at room temperature and it gets increased from here ?

The values of the filament given are the operating values of the power and voltage when its being heated up. This value is 576 Ohm and room temperature is 16 times lesser than this value.

 

[Rakayz]

ASSALAMU ÁLAIKUM
is any one gonna do nov 2009?! i am waiting from yesterday
JAZAK ALLAH

 

[Gémeaux]

And the answer to question 9 is C..please correct it. We square the time taken to cover height h. Not the individual times
Like its 2h/(t2-t1)^2 !

Kindly make sure that when you criticize someone else, you're correct yourself. Here's a link to its examiner report
http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_w02_er.pdf
The answer for 9th *is* D.

 

[Gémeaux]

Ok....Why cant the answer to question 10 be A ?..i mean the acceleration is shown constant ( g) at first and then decreases to zero?...

In A, the velocity initially stays proportional to t for quite along time which is not the case when air resistance is considered.

 

[dayeed]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
Please someone help me out with question 26 and 36 ......

 

[Rakayz]

culd u tell me specific question numbers and i'll try 2 solve them

here are the questions: 3, 12, 13, 14, 20, 21, 25, 26, 27 and 35
can u pls explain these to me?!
JAZAK ALLAH