As physics p1 MCQS YEARLY ONLY.

[XPFMember]

yes it ll be ok and thankyou so much for your help
God bless

ummm cuz there's nt much time now...m too sleepy...cud u plz assist me by telling the questions that are hard....around 15 questions i posted above...some are simple ones.....so can u plz tell which questions u need help with...

the thing is m too sleepy...i cant therefore gurantee to do the whole paper...but will atleast try to clear ur doubts in that paper before i go to sleep
inshaAllah

 

[yuuki]

ummm cuz there's nt much time now...m too sleepy...cud u plz assist me by telling the questions that are hard....around 15 questions i posted above...some are simple ones.....so can u plz tell which questions u need help with...

the thing is m too sleepy...i cant therefore gurantee to do the whole paper...but will atleast try to clear ur doubts in that paper before i go to sleep
inshaAllah

itz ok if u cant post them...the ones you have posted are i guess enough ,they were of great help,i ll somehow manage the rest really appreciate your help
thankyou so much for your effort
May God bless you

 

[XPFMember]

itz ok if u cant post them...the ones you have posted are i guess enough ,they were of great help,i ll somehow manage the rest really appreciate your help
thankyou so much for your effort
May God bless you

oh...sorry though....

 

[yuuki]

oh...sorry though....

no not a problem at all
you did an effort and i appreciate that

 

[Oliveme]

June 2005
=========

Q1. C.

Fact, basic stuff.

Q2. C

Rearrange the equation to make k the subject, so k = F/rv.

= (kgms^-1)/(m * ms^-1)
= kgm^-1s^-1

Q3.B

A reasonable estimate for an athelete running a 100m race is approximately 10 seconds.

K.E = 1/2 * 80 * 10^2
= 4000 J

Q4. C

Principle speed value = 16
Uncertainty = (0.1/40) + (0.05/2.50) * 16 = 0.36

We always round off the uncertainty value to one s.f., so it becomes 0.4

So the answer is 16 +/ 0.4

Q5. A

The length of the pulse is how long you see the change in the Y-axis. It's for 2cm, meaning 2 μs.

Q6. D

Easy stuff. To find acceleration you take the gradient of a velocity-time graph.

Q7. B

The acceleration during the motion of a falling ball will always be constant, i.e. 9.81 ms^-2. Since they told us to take upwards as positive and the gravitational force acts downwards, this is actually -9.81 ms^-2.

Q8. D

Distance is area of the graph. Class 4 stuff.

Q9. A

Acceleration doesn't act in the horizontal direction so it's value is therefore 0.

B is wrong because the object has velocity throughout the motion.
C is wrong because of B, the resultant velocity will be non-zero because of the horizontal velocity being non-zero.
D is just nonsense.

Q10. A

B, C and D are Newton's 2nd, 3rd and 1st law respectively.

Q11. A

Momentum is always conserved so we use that formula. Also when the objects stick on impact, the total mass will be the sum of the individual masses. Let 'm' be the mass of one of the objects.

60m + (40 * -m) = 2mx (where x is the speed of the masses after impact).
20m = 2mx
x = 10 because m cancel out.

Q12. C

Fact. D is wrong because it gravity is the point through which gravity APPEARS to act.

Q13. A

The forces are shown in this picture: http://www.xtremepapers.com/community/attachments/moe-png.12069/ (Thanks a ton to Unicorn for this).

(5*2) + (2*10) - (3*20) = 30 Nm anti-clockwise.

Q14. D

Resolve the horizontal 4N and vertical 3N component to get a 5N component parallel to the diagonal 4N component. Since the 5N force would be greater, the resultant force would be 1N in its direction.

Q15.B

K.E will be constant because the velocity and mass are constant (velocity beacause it says in the question).

P.E will start from a high value and decrease uniformly because the height is decreasing uniformly.

Q16. C

The gradient of an energy/time graph is power since P = E/t. So we are looking for the point where the gradient is the steepest. This is from 2s-3s, so the gradient there is (40-10)/1 = 30W.

Q17. B

P.E = mgh.

They have given the density and volume from which we can calculate the mass. g is 9.81 and h is 3.0m.

Q18. A

Fact.

Q19. C

Brownian motion, the molecules of liquid collide with the molecules of the pollen grains.

Q20. A

Let a regular extension be 'x'. In parallel, the extension is divided by the # of springs and the opposite for a 'series' extension.

Extension in X is e/2.
Extension in Y is e/2 + e/2 = e.
Extension in Z is e/2 + e = 1.5e.

The order is X -> Y -> Z.

Q21. D

You need to know these graphs. Brittle (glass) is just a straight steep line, rubber is like that of graph X (note that they don't obey Hooke's law) and Y is that of steel, a ductile material.

Q22. D

Let their Young Modulus be equal to 'E'. They have to have the same YM (same material), so..

E = FL/Ax (where x is extension).
F = EAx/L (E and x don't matter here because they're constant).

For P, F = A/l
For Q, F = 0.5A/2l

Ratio is 4:1.

Q23. A

Fact, all transverse waves travel at the same speed in a vaccuum.

Q24. B

You need to know a reasonable estimate of the wavelength of visible light, e.g. 500 nm.

# of wavelengths in ONE metre is 1/(500 nm) 2.0 * 10^6. This is in the order of 10^6, so B is right.

Q25. B

Use ratio of intensity and amplitude.

(I1/I2) = (a1/a2)^2

1/2 = (A/x)^2

x = √2A

Q26. B

Fact, sort of.

Q27. D

Distance between 2 maxima = 0.5λ.
So 1λ = 30 mm.

F = v/λ
F = 3.0 * 10^8 / 30 mm
F = 2.0 * 10 Hz.

Q28. B

Formula is x = λr/q

According to this equation, decreasing 'q' will increase 'x'. A has nothing to do with 'x'.

Q29. B

For 2nd order, d sin θ = 600 nm * 2, which is

d sin θ = 1200 nm

For 3rd order,

d sin θ = 3λ

Since d has to be the same and the angle is also the same, we can equate the 2 equations.

3λ = 1200 nm
λ = 400 nm.

Q30. D

E = V/d.
= 900 / (4 mm)
= 2.3 * 10^3 N/C

Q31. C

Fact.

Q32. C

The area is irrelevant to this question, because Q = I * t (there is nothing to do with area in this formula).

Q = 10 * 1 = 10 C

Since one electron has a charge of 1.6 * 10^-19 C, 10 C has 6.3 * 10^19 electrons.

Q33. D

Originally, R = ρL/A

Now, the length is doubled BUT the volume is the same. This means the area has to be halved. Mathematically proving this:

Volume is length * breadth * height, so:

2lbh = v

Since Area = lb
A = 2lb
lb = 0.5A

Anyway, new resistance will be 2ρL/0.5 = 4R.

Q34. D

A is wrong because Q is a thermistor/semi-conductor/etc.
B is wrong because the resistance decreases.
C is wrong because the resistances are the same at 1.9 (same V:I) ratio.
D is right because using I^2 * R proves this is correct.

Q35. B

Fact.

Q36. D

In parallel, voltage is the same so V2 = V3.

And terminal voltage V = V1 + V3

Rearranging this gives

V - V1 = V3

Q37. At X the voltmeter is connected directly so it gets the full 4V. At Y we use the potential divider formula to find the voltage:

V = 4 * (10/20) = 2V

B is the only graph that shows this correctly.

Q38. C

Easy stuff.

Q39. B

The range of α particles is approximately 0 - 5cm. In this question, they've given us values in mm, so we can say the range is 0 - 500 mm.

B is the safest maximum range.

Q40. C

The nucleon number decreases by (4+4+0) so it becomes 209.
The proton number decreases by (2+2-1) so it becomes 82.

Thank you so much. Jazakallah. You have no idea how many people must benefiting from these. May Allah bless you for this and help you achieve your goal, Inshallah.

 

[Firebender]

I know this is simple but can someone explain this to me?

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
Qs.10 and qs.15

...please?

 

[Sanis]

Can you tell mw, why you have to half the wavelength of the open ended tube compare with the first close ended tube?

that is the rule, try drawing the stationary wave inside the tube, the closed ended, will have a node at the closed end, and antinode at the open end... the distance between a node, and an antinode is 1/4 wavelength .... from an open ended tube, antinode is formed on both sides, so there must be a node in the middle, if u r goin 2 draw it, it look like twice the figure u drawn in the closed ended part, so twice the wavelength ...

 

[Soldier313]

Can someone pleeeaaaaseee help with these?



^ why is ans A and not D?



^why is ans D and not C?



^how is ans c?



^how do you this????:S ans is C?

 

[XPFMember]

AsSalamoAlaikum Wr Wb!

Soldier313

^ why is ans A and not D?

cuz couple is two equal and opposite forces....that's only the case in A. In D, they are equal, but NOT opposite.

^why is ans D and not C?

If the resistance of variable resistor is increased, the pd across XY decreases than before [when it was showing zero deflection], so we need to increase the value of pd, and that can be done when we move the contact towards Y, that way it's resistance will increase [cuz it'll be longer], and therefore pd can be increased to the same value....=> when it shows zero deflection...

^how is ans c?

I also need to know this...

^how do you this????:S ans is C?

I, too, need help here... anyone?

 

[Oliveme]

Can someone pleeeaaaaseee help with these?

View attachment 12999

^ why is ans A and not D?

View attachment 13000

^why is ans D and not C?

View attachment 13001

^how is ans c?

View attachment 13002

^how do you this????:S ans is C?

can you please look at this post. they have all the answers and explanations here, if you're struggling. http://www.xtremepapers.com/community/threads/as-physics-p1-mcqs-yearly-only.17330/ hope this helps you.

 

[Tkp]

xpf fr the 1st 1 couple is two equal foces in opp direction so ans is a.

fr the 3rd one absolute uncertainity would always be in 2s.f.so ans is c.

and 4 th 1 is dne above

 

[Soldier313]

AsSalamoAlaikum Wr Wb!

Soldier313

^ why is ans A and not D?

cuz couple is two equal and opposite forces....that's only the case in A. In D, they are equal, but NOT opposite.


but how can a force which is not at the circumference form a couple?

^why is ans D and not C?

If the resistance of variable resistor is increased, the pd across XY decreases than before [when it was showing zero deflection], so we need to increase the value of pd, and that can be done when we move the contact towards Y, that way it's resistance will increase [cuz it'll be longer], and therefore pd can be increased to the same value....=> when it shows zero deflection...

so that means when the pd increases, in order to get zero deflection we have to increase it again?

 

[Soldier313]

can you please look at this post. they have all the answers and explanations here, if you're struggling. http://www.xtremepapers.com/community/threads/as-physics-p1-mcqs-yearly-only.17330/ hope this helps you.

thanx for this

 

[Soldier313]

I don't understand this path difference issue.....we were never exactly taught this at school
Tkp Oliveme XPFMember
sorry for botherin:/

 

[XPFMember]

Soldier313 that's not necessary..that circumference thing...i mean the condition is JUST 2 forces n opposite direction..

and for the emf....note that the pd across that XN...when adjusted should equal that of the cell E2

 

[XPFMember]

I don't understand this path difference issue.....we were never exactly taught this at school
Tkp Oliveme XPFMember
sorry for botherin:/

i need help in that tooo...

 

[Soldier313]

XPFMember thanx a lot! that really helped

 

[1357913579]

AsSalamoAlaikum Wr Wb!

Soldier313

^ why is ans A and not D?

cuz couple is two equal and opposite forces....that's only the case in A. In D, they are equal, but NOT opposite.


but how can a force which is not at the circumference form a couple?

^why is ans D and not C?

If the resistance of variable resistor is increased, the pd across XY decreases than before [when it was showing zero deflection], so we need to increase the value of pd, and that can be done when we move the contact towards Y, that way it's resistance will increase [cuz it'll be longer], and therefore pd can be increased to the same value....=> when it shows zero deflection...

so that means when the pd increases, in order to get zero deflection we have to increase it again?

for first one forces are qual and are in oppsoite direction i n A becuase there should be a common prependicular distance to then whihc is in A and also oppsote and equal

 

[Tkp]

which year

I don't understand this path difference issue.....we were never exactly taught this at school
Tkp Oliveme XPFMember
sorry for botherin:/

 

[Soldier313]

which year

mj/02 qn 28