AS Physics P1 MCQs Preparation Thread.

[Nibz]

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

Qs 9) my answers were dis: 2mv and mv2. The answer is quite diff
Qs 13) answer is coming wrong. Help please

For Q13. Weight of the ruler acts at 50 cm mark (10 cm to the right of pivot)
Clockwise moment = ( 100 x 10 ) + (20 x 60) = 2200 Ncm
Anticlockwise = 50 y (where y is the unknown distance)
50 y = 2200
y = 44 cm
Answer C

 

[leosco1995]

about the 2nd mcq we are asked to tell velocity at the lowest point so why isnt it C. At C the wave should be moving downwards right? At D the wave wouldnt be moving down it wud be in the middle of its motion no? ((

Since it's a velocity/time graph, when the mass is at its lowest and highest point the velocity will have to be zero so it can't be C. This means that only B or D can be correct, but D is at the lowest point of its motion. I think you are confusing it with a displacement-time graph.

 

[leosco1995]

Read the question again. It says treat the region 'XY as a straight line'.

But his question (and mine too, sort of) is that why you can't take out the entire area by treating the whole thing as a triangle (beacause they said treat XY as a straight line, so wouldn't the whole line be straight?). Is my reasoning still wrong? Because I too want to know why you have to take out the areas separately.

 

[Nibz]

Cam anyone answere my 2 mcq questions?? I did not get them.....for the first one, dont we just take the area under the whole graph??(bigger triangle)

But his question (and mine too, sort of) is that why you can't take out the entire area by treating the whole thing as a triangle (beacause they said treat XY as a straight line, so wouldn't the whole line be straight?). Is my reasoning still wrong? Because I too want to know why you have to take out the areas separately.

No, because its gradient is less than that of the other line. So, when treating it as a straight line, it would be slightly bent, i.e more towards x-axis (smaller gradient).

 

[leosco1995]

No, because its gradient is less than that of the other line. So, when treating it as a straight line, it would be slightly bent, i.e more towards x-axis (smaller gradient).

Then I was right. I told him you can't because the gradient is different.
Thanks for the confirmation though.

 

[kinglynx]

The sides are always balanced! Look at any typical AS level nuclear equation and you will see that the proton number and nucleon number is always the same on both sides. Mass-energy is also conserved but you can't tell that from just looking at the diagram. But the neutron number isn't always conserved. For example, see this alpha decay equation:

The neutron number of Ra is 134 and 132 on Rn. They aren't the same!

but the number of neutrons are balanced on each side!!! and so is the nucleon number....

 

[raamish]

Since it's a velocity/time graph, when the mass is at its lowest and highest point the velocity will have to be zero so it can't be C. This means that only B or D can be correct, but D is at the lowest point of its motion. I think you are confusing it with a displacement-time graph.

oh ok so in parabolic motion the velocity will not be 0 at the lowest position right like it happened in a wave? i thought that i a wave the velocity is 0 only at the highest point and not the lowest point

 

[leosco1995]

oh ok so in parabolic motion the velocity will not be 0 at the lowest position right like it happened in a wave? i thought that i a wave the velocity is 0 only at the highest point and not the lowest point

It will be zero at the centre. And maximum at the highest/lowest point.

 

[leosco1995]

GUYS June 2005 has been posted in the other thread for anyone who still has doubts in that paper.

http://www.xtremepapers.com/community/posts/321571

 

[XPFMember]

AsSalamoAlaikum Wr Wb!

Here are few solved explanations for 9702 Physics Paper:1

June:2002

June:2004

June:2005

 

[fathimaash]

pls help with q 28

 

[fathimaash]

one more question. q 22.. pls ppl Allah will bless u

 

[sagar65265]

​

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_1.pdf Nov02 q7,10,18,24,32,36​

​

Okay, here goes:

Q7: The frequency is 1/time period, which is the duration for 1 cycle to complete. The diagram shows it as about 0.7 cm, so:
if 1 cm = 10 ms, 0.7 cm = 7 ms
Frequency should then be equal to 1/(7/ 1000) which should be around 140 Hz. Since it is only asking for an estimate, you could also average out the no. of wavelengths - about 8.5 - and find the frequency for that. The answer should be 140 Hertz.

Q10: Now, in a vacuum, there is NO air resistance, so the acceleration is constant since the force on the object (it's weight) is constant and there is no air resistance acting, so the gradient of the v/t graph shows constant acceleration. Good so far. On Earth, where air resistance does show, the maximum acceleration possible is equal to 9.81 ms^-2, since the maximum resultant force on the object is equal to that of gravity - so now you know that acceleration of a FREELY falling body on earth can NEVER be greater than 9.81 ms^-2 - in fact, on EARTH, vacuum or not there shouldn't be any possibility of the acceleration to be greater than 9.81, so B and D are ruled out. However, air resistance does catch up quite fast, so the acceleration should not stay at 9.81 for long - the only graph that works is "D".

Q18: Okay, so the power output is equal to the energy transferred per unit time, i.e. per second. The energy transferred to the water, in our case, is assumed to be purely potential - any speed values are not given, so the water only gains potential energy = mgh
m = 1.3 X 10^9 kg
g = 9.81 ms^-2
h = 2 m

Therefore the energy transferred is equal to 1.3 X 10^9 X 9.81 X 2 = 25506000000 Joules of energy transferred.
The time taken is transferred over 1 day = 60 seconds * 60 Minutes * 24 hours = 86400 seconds.
Thus the energy transferred to potential energy per second = 25506000000/86400 = 295208.333 Js^-1
So it should be 300 kW, since 1 kW is equal to 1000 Js^-1 = D

Q24: Okay, so the formula for work done on a spring is equal to the area under a Force/Extension graph = 0.5*F*x = Fx/2.
In this case the springs obey Hooke's Law, so we are all set to go here:

Since the maximum force for each spring is the same, the final extension of one spring with constant K is twice that of the spring with constant 2K. This is equal to x, let us say (extension of P = x/2, for Q = x)
Now, for Wp:

Wp = 0.5*F*(0.5*x) = 0.25*Fx

And for Wq:

Wq = 0.5*F*x = 0.5*Fx

Now, Wp/Wq = 0.5, so Wq = 2Wp := Wp = Wq/2

Simply put, the extension of P is half of that of Q so the work done is half, since no squaring is involved, etc.

Q32: Man, is this scale tough to read! Anyways, 1 small box on the y-axis is equal to 5 mA and 1 small box on the x-axis = 0.2 V. Note that it is 5 mA, not 5 A.
Now, since R = V/I, you want 1/Gradient of the graph to get the resistance, not the gradient!
At V = 1.0 V, the current is approximately 50/1000 amps = 0.05 amps, so the resistance = 1/0.05 = 20 ohms.
At V = -1.0 V, the current is exactly zero - no flow of current despite the application of a potential difference, i.e there is an infinite resistance - the diode is in reverse bias.

Q36: Now, let's see: the voltage in each branch is the same since the circuit is a parallel circuit and it is equal to the EMF since there is no internal resistance, okay? Now, the potential drop across 1 5 ohm resistor is:

5/15 * 2 = 2/3, since the entry voltage is 2 volts and the voltage is equally dropped across the resistors because they are all similar.
(You can take the two 5 ohm resistors after that to be one 10 ohm resistor)

Now in the other branch, the current has flown through two 5 ohm resistors that you can take as one 10 ohm resistor, so the potential drop across these 2 resistors is equal to:

10/15 * 2 = 4/3 = 20/15

Now the difference = 20/15 - 10/15 = 10/15 = 2/3 = A

Hope this helped!

 

[sagar65265]

one more question. q 22.. pls ppl Allah will bless u

Okay, so there are 3 springs, so the weight of the object is shared between the three; in other words, the extension force on each spring = W/3. This produces an extension of "X".
Now, when the weight of 2W newtons is exerted on 2 springs (the middle one is removed) so the weight is divided into W and W; i.e each spring carries a load of W newtons.
If W/3 newtons causes an extension of X, W/(W/3) should produce an extension of 3X; 3 times the force = 3 times the extension according to hooke's law provided the elastic limit has not been passed - which has not happened. Mathematically:

F = kx , but F = W/3
W/3x = k
Now, when 1 spring is removed and load = W, F = W, and "Y" is equal to the extension:
W/Y = k
Equating "k", since this remains constant,
Y = 3X ! = D

Hope this helped (PS this was copied from an earlier post).

 

[fathimaash]

Okay, so there are 3 springs, so the weight of the object is shared between the three; in other words, the extension force on each spring = W/3. This produces an extension of "X".
Now, when the weight of 2W newtons is exerted on 2 springs (the middle one is removed) so the weight is divided into W and W; i.e each spring carries a load of W newtons.
If W/3 newtons causes an extension of X, W/(W/3) should produce an extension of 3X; 3 times the force = 3 times the extension according to hooke's law provided the elastic limit has not been passed - which has not happened. Mathematically:

F = kx , but F = W/3
W/3x = k
Now, when 1 spring is removed and load = W, F = W, and "Y" is equal to the extension:
W/Y = k
Equating "k", since this remains constant,
Y = 3X ! = D

Hope this helped (PS this was copied from an earlier post).

thanks alot..

 

[kinglynx]

but the number of neutrons are balanced on each side!!! and so is the nucleon number....

 

[SalamH]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
I 'm askin for d sec tym pls explain

Im postin dis fr d third tym pls reply
12/o/n/09- Q5, Q16, Q25
01/M/J/03-Q6, Q10, Q20, Q23, Q40

 

[DARK DRAGON]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w04_qp_1.pdf
Q33 answer is C plz explain

 

[rabiafarooq101]

Can someone please help me with the questions 11 and 32 of May/June 2003 Physics P1?

 

[Nibz]

Im postin dis fr d third tym pls reply
12/o/n/09- Q5, Q16, Q25
01/M/J/03-Q6, Q10, Q20, Q23, Q40

M/J 2003

Q.6 True value = 895
Not accurate because the readings are not close to the true value ( readings not withing 1mm of the true value)
Precise because all the readings are withing 1mm of one another.

Q10: Total mass = m + 3m = 4m

F = m a
a = F / 4m

For Block Y
F = mass x acceleration (put acceleration = F/4m)
F = 3m x F/4m
= 3F/4

Q 20.
0.1 Po = ρ g h
0.1 Po / ρg = h -> this can also be written as 1Po/10ρg = h

Q23. All transverse waves can be polarised. Fact.

Q40. Since charge on proton is +ive e, the sum of both 'up quarks' and 'down quarks' should equal to 1.
2e/3 + (-e/3) = 1e
e = 3

Now -3/3 = -1 (1 is for quark. Negative is for down. So 1 down quark)
and 2 (3)/3 = +2 ( 2 is for quarks. +ive is for up)