AS Physics P1 MCQs Preparation Thread.

[Peter Check]

How do we do this? do we use V^2=U^2 + 2as ???

 

[Peter Check]

x = Lambda x D/a

x and a are inverse, so the answer is either A or C. C cannot be the answer as the fringe separation cannot become zero. So that leaves only A.

You mean A or B right??

 

[sagar65265]

anyone can help please!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
Q no.17
here is Marking Scheme
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_12.pdf

Any elastic material that has been stretched/deformed will want to pull the supports that are keeping it stretched in such a way as to bring it back to it's original length and/or shape. Usually, you can say that the material will pull the support/load to it's centre. Thus, there is a force on Y and Z towards the trolley, i.e. towards to centre of the rubber band within the region XY and XZ respectively. There is similarly a pull towards the right of 4 N on the trolley. Check the uploaded file.

There are thus 2 forces on the trolley from either side of the rubber band. These forces are 4 N each and have a component in the direction of P.
The angle between the horizontal and the tension forces is theta in the diagram. This angle has a hypotenuse of 50 and an opposite of 40. Thus,

Sin (Theta) = 4/5 = 0.8. So Theta = 53.13 degrees.

Therefore the force = 2 * 4 * cos (53.13) = 4.8

Good Luck for your exams!!

 

[raihan1904]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Q32 with explanation please

 

[raihan1904]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
Q34 with working plz!! tnx in adv,

 

[raamish]

in the formula P=Fv is to be used and the forces are friction and the force applied will we add these forces up in the formula or will be substract them and take out the resultant force?

 

[Nibz]

Somebody solve these doubts...?

View attachment 12864View attachment 12865View attachment 12866View attachment 12867View attachment 12868View attachment 12869View attachment 12870View attachment 12871View attachment 12872
View attachment 12874

Q17.
Upward is taken as positive, as mentioned in the question. Since acceleration always acts downwards, the graph should be in the negative region. A and B automatically eliminated.
Now the time for ball to reach the ceiling and back to the hand is same. So the maximum peak (i.e greatest magnitude) for acceleration will be in the middle. So D.

Q15.
It's moving with constant velocity. So kv = ma
a can be taken as g. Again, mentioned in the question.
So kv = ma
from here, v = ma/k

Put this in K.E = 1/2 mv^2
You will get your required answer.

Q27.

x = λ D /d
x and D are in direct proportionality.
D has been increased BY 2m, i.e twice the original length.
x increases FROM 1 to 3, i.e by a factor of 3.
Find the ratio. 2/3 = 0.6


Q25.
Don't get fooled by the column heading.
The fundamental frequency is the one mentioned in the question, i.e 75 Hz.
The first higher note in the column heading is actually the 2nd mode. Frequency of second mode = 3 x frequency of first mode
so 3 x 75 = 225 Hz
Frequency of third mode = 5 x frequency of first mode
5 x 75 = 375 Hz


Q34.

R = ρL/A

V = A x L
A = V/L

R now becomes ρ L^2 / V

Length of cube = V ^1/3
put this in above equation.

R = ρ ( V^1/3)^2 / V
= ρ / V^1/3

Q. last

Horizontal distance = speed x time = v t

Vertical:

s = ut + 1/2 at^2
s = 1/2 gt^2

Ratio: v t / (1/2 gt^2)
= 2v / gt

 

[Iadmireblue]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_13.pdf question 26

 

[kinglynx]

https://fbcdn-sphotos-a.akamaihd.ne...720/560661_10151015152721535_1393681482_n.jpg

https://fbcdn-sphotos-a.akamaihd.ne...720/545100_10151015152391535_1594903873_n.jpg

 

[N.M]

How do we do this? do we use V^2=U^2 + 2as ???View attachment 12904

whats the answer?
Is it C??

 

[N.M]

View attachment 12897

HOw to do this??

Is C the correct answer?

Q=It
the question wants the answer to be in s^-1
so for that 1/t=I/Q
so 4.8/1.6x10^-19
the answer becomes 3x10^19
the current flows from positive terminal of the battery to the negative terminal of the battery, but we also know that the eletrons flow in the opposite direction to the conventional current so electrons flow from Y to X
answer becomes C
i suppose C is the correct answer, isn't it?

 

[mominzahid]

Can someone please answer this problem i have my exam tomorrow please help guys

http://www.xtremepapers.com/community/threads/physics-as-estimation-questions.17728/

Seriously guys...im feeling that there is no serious consideration about starting with the revision from 2003, if there will be nothing then lets leave any person post his doubts and the others answer!!!

I'll be glad to, but the thing is i wont be online for too long each day, nor can i promise to be consistently online, but if iam online i will try my best to help!

7 - if you resolve the vector you will see that the horizontal component is ucosα while there are two vertical components...taking in mind that they are acting in the opposite directions. First is the result of resolving the vector on the y axis : usinα and the second is the acceleration downwards. But you cant subtract these so you will take the acceleration in terms of the velocity:
since (a = (v-u) / t) therefore, delta v = at replace the a with g becuz they have the same value therefore the vertical component downwards is gt.
so the vertical component is usinα - gt .....the answer is C

10 - i dunno if this is the right way but this is how i did it....since x = m and y = 3m. Take the force as 1 F if you divide the force according to their mass....so x gets 1/4 F and Y gets 3/4F hence the force exerted on y by x is 3/4 F. so D

18 - use V2 = U2 + 2as
for the first case:
v= 0, u= 10 and s=10
so putting it into the equation:
0 = 100 + 2(10)(a)
a= -5
now the second case:
v = 0, u=30 a=-5 (since the force is the same)
so 0 = 900 - 10s
s = 90 m
and the answer is D

26 - the time period (T) is the time taken for one wave, which is 2.5 x 10^-3 x 4 = 0.01
F = 1/T
= 1/0.01 = 100 Hz so the answer is B

gimme sometime till i figure out 5
this is what the examiner report says about 5:
A and B were common responses. Perhaps candidates did not see the significance of the fact that 3% of 330 is 10 but just tried to give rounding to 3 or 4 significant figures.

when blowing the air over the bottle it means the same as blowing into it because air disturbed at the top would cause wave formation in the bottle. (the concept of resonance) stationary waves are produced by interference of two waves exactly same but in opposite direction... (like at 180°)
making a loud sound near a mountain would cause reverberation to occur, or the waves would echo off into different directions, so there isn't a surety of stationary waves forming.
that's what i understood of the question hope it helps, otherwise there's always someone else with a better explanation

Was your acceleration -1.7?

 

[N.M]

As Salamu Alaikum
can somebody plz explain the calculations related to CRO n effects of changing Y-sensitivity n time-base settings?
plz
May Allah reward for this
JazakAllahu Khair

 

[A.ELWY 7]

As Salamu Alaikum
can somebody plz explain the calculations related to CRO n effects of changing Y-sensitivity n time-base settings?
plz
May Allah reward for this
JazakAllahu Khair

to find the peroid T...u count how many squares does 1 wavelength occupy . ex: 4..so 4 multiply by the time base setting..then to find the frequency it is 1/T...for the y sensitivity..u just multiplyit by the squares the amplitude occupy,,,or if he gave u the amplitude then the amplitude multiplies by the y sensitivity

 

[N.M]

Can someone please answer this problem i have my exam tomorrow please help guys

http://www.xtremepapers.com/community/threads/physics-as-estimation-questions.17728/

Estimations


audible frequency of sound: 20Hz - 20 KHz

speed of sound in air : 300-330 m/s

time taken for sound to travel 1km in air;
T=1000/300
t= 3.33 sec

volume of an adult's head:
4/3pie(r^3)
take r from 6-9 cm
so acceptable values 900-6000 cm^3

mass of an adult: 80kg

time taken for an athlete to cover 100m: 10 sec

K.E of that same athlete: 1/2 mv^2
v=dist/time
100/10
v=10
0.5 x 80 x 10^2
K.E=4000J

wavelength of red light: 650nm

wavelength of violet light: 450nm

frequency: v=f (lambda)
v= 3 x 10^8
so calculate it for any given wavelength of light

density of water; 1g/cm^3
1kg/m^3

density of air: 1.3kg/m^3

weight of 1000cm^3 of water:
vol=1000cm^3
d= 1 g/cm^3
M=density*volume
M= 1000 g
m=1kg
W=mg
W= 1 x 10
W= 10N

mass of an apple: 200g or 250g

mass of a protractor; 30g

mass of a 30cm ruler: 50g

pressure due 10m depth of water:
P=pgh
(1000)(10)(10)
P=10^5 Pa

P.S: these are just rough estimates examiner accept a range of values

 

[Anonymous']

Assalamualaikum.. can someone help me how to do question 34 june 2010 paper 12, and question 17 nov 2010 paper 13..

 

[mominzahid]

Estimations


audible frequency of sound: 20Hz - 20 KHz

speed of sound in air : 300-330 m/s

time taken for sound to travel 1km in air;
T=1000/300
t= 3.33 sec

volume of an adult's head:
4/3pie(r^3)
take r from 6-9 cm
so acceptable values 900-6000 cm^3

mass of an adult: 80kg

time taken for an athlete to cover 100m: 10 sec

K.E of that same athlete: 1/2 mv^2
v=dist/time
100/10
v=10
0.5 x 80 x 10^2
K.E=4000J

wavelength of red light: 650nm

wavelength of violet light: 450nm

frequency: v=f (lambda)
v= 3 x 10^8
so calculate it for any given wavelength of light

density of water; 1g/cm^3
1kg/m^3

density of air: 1.3kg/m^3

weight of 1000cm^3 of water:
vol=1000cm^3
d= 1 g/cm^3
M=density*volume
M= 1000 g
m=1kg
W=mg
W= 1 x 10
W= 10N

mass of an apple: 200g or 250g

mass of a protractor; 30g

mass of a 30cm ruler: 50g

pressure due 10m depth of water:
P=pgh
(1000)(10)(10)
P=10^5 Pa

P.S: these are just rough estimates examiner accept a range of values

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..

 

[Anonymous']

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf question 34

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_13.pdf question 17

 

[Wanzi21]

Is C the correct answer?

Q=It
the question wants the answer to be in s^-1
so for that 1/t=I/Q
so 4.8/1.6x10^-19
the answer becomes 3x10^19
the current flows from positive terminal of the battery to the negative terminal of the battery, but we also know that the eletrons flow in the opposite direction to the conventional current so electrons flow from Y to X
answer becomes C
i suppose C is the correct answer, isn't it?

yeap thats the correct answer and good explanation! ^^ thanks

 

[Nikesh]

How do we do this? do we use V^2=U^2 + 2as ???View attachment 12904

its C
--> for first case when the sphere is dropped
u^2 = 0 + 2as (let s be total height of the fall)
i.e. u = (2as)^1/2.........................(i)
-->2nd case for bounce
0 = v^2 - 2a(s/2) (since the height is half of total height this time)
or, as = v^2
i.e. v = (as)^1/2
now,
v\u = (as/ 2as)^ 1/2
= (1/2)^1/2