AS Physics P1 MCQs Preparation Thread.

[ibadsiddiqi]

^no prob!

 

[fatima 007]

Hello, people!! This is the 3rd time im posting this. can anyone pleaseeeeeeee help? ​

Help needed :
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_1.pdf J03 q17,22,28,30
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf Nov02 q7,10,18,24,32,36
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf Nov03 q4,5,7,22,28,31 ​

 

[Nibz]

AsSalamoAlaikum Wr Wb

need help with these questions, plz. Answers have been marked.

JazakAllahu khairen

View attachment 12090View attachment 12091View attachment 12092View attachment 12093

W.S.

21. I get ''4" as my answer. No idea if I'm wrong or the ms.

Young Modulus 'E' = F/A . L/e

So E for wire Q => E = F.L/A.e
Ee = FL/A

E for wire P => E = F.2L/0.5A
Ee = 4FL/A

Solve these simultaneously since Young Modulus for both is the same. (wires are of same material)

22. Since the thinner spring has a 'low k', there is a possibility that it would be affected by small external forces (wind etc). Therefore it should be kept in rigid box to avoid an error in calculating the mass. The thicker spring, having a larger k, would need a much greater force to be extended. So A is the answer.

 

[Nirma]

In the Young's double slit expt, wat would happen to
1- fringe separation
2- position of fringe pattern

if one of the slits is covered with glass slide.Why?

Plz do answer. I'd be grateful.

 

[leosco1995]

Thats not true, in Beta decay, often one side will have one more proton than the other due to the nature of the decay (a proton and electron together) so charge/proton cant be, thus if the proton number isnt conserved neither can be the nucleon number. Also neutron number then isnt conserved either, it makes no sense!!!

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!

 

[freakybandi]

View attachment 12065

as you can see there are 7 periods in 5cm which is 50ms
so 1 period = 50ms/7 =1/140 and this is T

T= 1/f so f= 140

ohh i get it..thnkx!!

 

[raamish]

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

 

[Peter Check]

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), and for the second one, why is it not B??

 

[Peter Check]

Moment about Q = force * p.d. from pivot
3.0 = f * 50 mm
f = 60 N

(Q just provides an upward force, so there is no couple and hence we don't use the diameter. Don't be confused about a torque always involving a couple, torque just means turning effect and it could apply to a regular moment also.

You already have the answer with this, but anyway for P:

moment = force * distance from pivot
= 60 * 0.75 mm (force was calculated from Q)
= 4.5 Nm


How does it not make sense? Things which are always conserved in a nuclear reaction are mass-energy, charge/proton # and nucleon number. The number of neutrons aren't always the same.

thank u!

 

[ughani99]

23 The Young modulus of steel is determined using a length of steel wire and is found to have the
value E.
Another experiment is carried out using a wire of the same steel, but of half the length and half
the diameter.
What value is obtained for the Young modulus in the second experiment?
A
2
1
E B E C 2E D 4E

 

[umarashraf]

23 The Young modulus of steel is determined using a length of steel wire and is found to have the
value E.
Another experiment is carried out using a wire of the same steel, but of half the length and half
the diameter.
What value is obtained for the Young modulus in the second experiment?
A
2
1
E B E C 2E D 4E

E... young modulus is a ratio and is constant for a material...

 

[leosco1995]

23 The Young modulus of steel is determined using a length of steel wire and is found to have the
value E.
Another experiment is carried out using a wire of the same steel, but of half the length and half
the diameter.
What value is obtained for the Young modulus in the second experiment?
A
2
1
E B E C 2E D 4E

E because Young Modulus is the same for the same material, regardless of its dimensions.

 

[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), and for the second one, why is it not B??View attachment 12154View attachment 12155

In 1st question, area under the graph gives the work done.
There is one triangle and one trapezium ( the extension is in 'mm' so make sure you convert it to 'm').
Area of triangle = 1/2 (10 x 10^-3) x 500 = 2.5 J
Area of trapezium = 1/2 ( 500 + 550 ) x (2 x 10^-3) = 1.05 J

Add both the areas = 2.5 + 1.05 = 3.55J


Q2. Velocity is lowest on the x-axis (i.e v= 0) and when it is on the 'verge' of moving up. So that's point D.

 

[raamish]

In 1st question, area under the graph gives the work done.
There is one triangle and one trapezium ( the extension is in 'mm' so make sure you convert it to 'm').
Area of triangle = 1/2 (10 x 10^-3) x 500 = 2.5 J
Area of trapezium = 1/2 ( 500 + 550 ) x (2 x 10^-3) = 1.05 J

Add both the areas = 2.5 + 1.05 = 3.55J


Q2. Velocity is lowest on the x-axis (i.e v= 0) and when it is on the 'verge' of moving up. So that's point D.

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

 

[XPFMember]

W.S.

21. I get ''4" as my answer. No idea if I'm wrong or the ms.

Young Modulus 'E' = F/A . L/e

So E for wire Q => E = F.L/A.e
Ee = FL/A

E for wire P => E = F.2L/0.5A
Ee = 4FL/A

Solve these simultaneously since Young Modulus for both is the same. (wires are of same material)

u made a slight mistake there...
i get the answer now..jazakAllah
E for wire P => E = F. 2L/4A.e
Ee = F.L/2A.e

therefore A is correct..

 

[leosco1995]

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), and for the second one, why is it not B??View attachment 12154View attachment 12155

I am not sure about the 1st one, because I also got the answer wrong the first time (I did the same thing as you). I guess it is because the wire is undergoing elastic deformation from point X and Y so it doesn't have the same gradient as from 0 to X. Someone correct me if I'm wrong, though.

The 2nd one is just simple harmonic/wave motion. At point B, the speed is 0, but it is at the highest point of its motion, not lowest. Let me know if you are confused about this stuff.

 

[sagar65265]

Hello, people!! This is the 3rd time im posting this. can anyone pleaseeeeeeee help? ​

Help needed :​

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf J03 q17,22,28,30​

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​

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf Nov03 q4,5,7,22,28,31 ​

Okay, so as for June 2003:

Q17: The work done on the object of weight W is equal to the total energy transferred to it; you do work on an object if it gains energy from and vice versa, I guess; when the object is lifted, at the end of the lifting, the speed is zero and hence the K.E. is also zero. Note that the initial K.E. is also zero, so the change in K.E. is zero. The change in P.E. = mgh, but W = mg, so Change in P.E. =Wh, which is in this case "q" thus it is "Wq" = C

Q22: Now, 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

Q28: When the diffraction grating is used, the formula to use is:

dsin(theta) = n(lambda)

Note that theta = the angle between the central maxima and the maxima in question (between zeroth and maxima in question)
Now, if between the 1st maxima the angle is 70, the angle between the zeroth and the first order = 70/2 = 35 (since the pattern is symmetrical; i'm not sure but it makes sense - if i'm wrong please corrext me)

So, the space between each slit = 1X10^-6, so Lambda = (1X10^-6 X sin(35))/1 = 5.74 X 10^-7, but moving the dp to the right twice, = 574 X 10^-9 = 574 nm = C

Q30: I don't really get this question fully, but assuming uniform decrease of current, we can expect that a current of 90 mA will flow for 1 second, 80 mA will flow for 1 second, etc, in which case the total using Q = It gives us 90+80+70+60+50+40+30+20 = 440 which is close to C, but i'm pretty sure this method ain't right - i'm sorry!

I'll check out the other questions ASAP and keep you posted. Good Luck, Everybody!

Update: After I discussed with my dad, he told me that it is best to use the average current - we can do that since the current decreases UNIFORMLY - so 100 + 20 = 120, and 120/2 = 60 and 60 * 8 seconds = 480 mC!!

 

[Oliveme]

10) I didn't get that either. tell me too when you understand this one... n also 26th and 35th
15)The initial K.E is 4J in each case. for the force F, the change in K.E is 8-4= 4J
also, K.E = work done
so, 4J = (F x s)
Then, for the next part, with 2F and 2s,
the w.d, i.e the K.E = 2F x 2s = 4Fs which means four times (F x s)
i.e, 4 x 4 J = 16 J
now this is the *change* in K.E., therefore the new K.E would be 4 + 16 = 20J

Thank you very much.

 

[Nibz]

I am not sure about the 1st one, because I also got the answer wrong the first time (I did the same thing as you). I guess it is because the wire is undergoing elastic deformation from point X and Y so it doesn't have the same gradient as from 0 to X. Someone correct me if I'm wrong, though.

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

u made a slight mistake there...
i get the answer now..jazakAllah
E for wire P => E = F. 2L/4A.e
Ee = F.L/2A.e

therefore A is correct..

Oh right. Glad it helped

 

[raamish]

I am not sure about the 1st one, because I also got the answer wrong the first time (I did the same thing as you). I guess it is because the wire is undergoing elastic deformation from point X and Y so it doesn't have the same gradient as from 0 to X. Someone correct me if I'm wrong, though.

The 2nd one is just simple harmonic/wave motion. At point B, the speed is 0, but it is at the highest point of its motion, not lowest. Let me know if you are confused about this stuff.

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? ((