AS Physics P1 MCQs Preparation Thread.

[Dukeofwin]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
Someone please help me with qn no. 5.. an explanation will be appreciated.

To find maxima you use dxSin90= (no. of maxima) x (wavelength)
-To find 'd' we use 1/no of lines
-so 1/300= 3.3x10^-3
-Since 'd' is in mm change wavelength to mm as well which becomes 6.9x10^-4
- No put the values in eq: (3.3x10^-3) x (1) = (No. of maxima) x (6.9x 10^-4)
And you get 4.87, round it off to 5

Hope I helped!

 

[Obsidian Fl1ght]

Bcoz C is not the MOST suitable.
The lower k spring will be stretched quite a lot as compared to A.
In A, the load (Strain) on the lower k spring will be quiet reduced.
Make sense?

 

[Kyusam]

The uncertainty (accuracy) in this result is 3%
First calculate it: It's 9.8298
U must remmbr that uncertainties are ALWAYS written correct to 1 SF!
So, the above correct to 1 SF is:
10
Uncertainty will be 10 m/s
Also, remmbr that the calculated answer (speed 327.66 u have), is DEPENDENT on the uncertainty
I mean that the decimal places (or accuracy in this case) of the speed is determined by the uncertainty.
Uncertainty (accuracy - and let's refer them to be the same thing here for now), is 10.
So speed will be correct to 10m/s

ie 330m/s
That is all.
SO:
Find uncertainty first.
Make it correct to 1 sf
Write calculated value correct to the uncertainty

eg:
X = 34.5444 m
Uncertainty = 0.00498 m
Make uncertainty 1sf = 0.005m
Write X correct to the uncertainty ie 3dp
X = (34.544 +/- 0.005) m

Tadaa

Plzz can u explain Q 30 and 35 for me in may June 2012 variant 12

 

[stareaj]

Somebody please helppppppppp these are my problems don't have any link pls
Physic As oct 2010 / 12 mcq 22 and 34
May 2011/12 mcq 15,16,24
May 2011/ mcq 14,15
Oct2011/13 mcq 17,25
Oct 2011/12 mcq 26
May 2012/12 mcq 5

 

[stareaj]

Plzz can u explain Q 30 and 35 for me in may June 2012 variant 12

Mcq 35 is V equal to V1 plus V2 or V3
As in parallel so same thing soooo take V3 now V3 plus V1 gives total V

 

[Sajeeb Chowdhury]

This question

this is because it absorbs an orbital electron of the element..... so the electron number decreases...... 28-1=27 find the 1 with proton number 27 and mass number 59

 

[stareaj]

Bcoz C is not the MOST suitable.
The lower k spring will be stretched quite a lot as compared to A.
In A, the load (Strain) on the lower k spring will be quiet reduced.
Make sense?

Thanks

 

[Obsidian Fl1ght]

I'll ask again, anyone who can kindly summarize all they know abt stationary waves?
Including the waves formed in open-ended (one side % both sides) tubes?
It'll help me loads.

 

[Nidz-Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf can anyone help with this Q 30 .34,36, 37,39?

 

[Obsidian Fl1ght]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf can anyone help with this Q 30 .34,36, 37,39?

Apply d*sin thetha = n*lamida
d = 1/n where n is no. of line/m. Metre not millimetre.
Grating's got 300lines/mm. So this means it has 300*1000 lines/m
(Coz 1m = 1000mm)
d = 1/n ie 1/300,000
U wanna see all the maximas possible, so to get max. no. of maximas on one half of the screen, u put thetha = 90
lamida is given. Convert it into m from nm.

U get n = 4
This means 4 maximas on one half of the screen.
On whole (both halves) of screen, maximas are = 4*2 = 8

BUT, u also have to consider the central maxima.
So total maximas are 8+1 = 9

Central is in middle.
4 maximas above and 4 below it.

 

[Martee100]

Could someone help !!!

 

[salvatore]

I always get confused when solving vector addition/subtraction questions. Could anyone pleaseeee give me a good explanation on how to solve such type of questions?
Here are two examples:

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf
Qn no 3

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
Qn no 12

Waiting for your reply..

I'm reposting this.. anyone??

 

[Nidz-Ahmed]

Could someone help !!!

find the "2" of the upward vertical component to balance the downward component .
downward component is 4.2*9.81=41.20
since there are two upward component =41.20/2 this the upward component of one side
sin 25 * x =20.6
x= tension is equal too 20.6/sin 25 =48.7 ~ that 49

 

[Lyfroker]

Q#4, 5 & 15
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf

 

[Lyfroker]

Q#9(explain the k.e. part),
13(when we are taking the distance between pivot and the force in calculation, how can 44 be the mark on the rule n not the distance? :S)
15, 24, 27, 31 & 34.
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

 

[Obsidian Fl1ght]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf can anyone help with this Q 30 .34,36, 37,39?

34
Let's go choice by choice shall we...:
B: It's said both are made from same material. So resistivity is same.
Reject B

A:
R = (resistivity * L)/A
Both are same material so we ignore resistivity for now.
So we have: R = L/A
We have to compare their cross-section areas.
A = L/R

For X:
When R = 20, L = 0.6
A = .6/20 = 0.03A

For Y:
When R = 10,
L = 0.6,
A = 0.6/10 = 0.06A

Compare 'em.
2X = Y
Is this what dear (NOT!) A is suggesting?
No. It says the opposite in fact.
So A's a no-no.

C.
Take both lengths equal. I took them 0.6m
For X, R = 20
For Y, R = 10

P = VI
SInce they're in series, I will be same for both.

For X,
P = 20I
For Y,
P = 10I

Compare 'em away.
Power of Y *2 = Power of X
WHich is exactly what C says.
So C's our choice.

But to be clear, let
s have a look at D

In parallel, current in one of the two branches, in equal to ratio of R/total of OTHER branch.

For X, (R=20) and R = 10 for Y. Total = 30
For current in X, therefore, it's 10/30 ie 1/3 I

For Y,
20/30 I
or 2/3 I

This is clearly NOT what's in D.
So D goes down.

 

[Obsidian Fl1ght]

I'm reposting this.. anyone??

The best way (well, according to me) to do this, is make an equation with all the vectors on ONE SIDE.

First example:
q3
X - Y ... u have to find the resultant vector for this which is Z.
So Z = X - Y
Now, according to the "best way", X - Y = Z will be X - Y - Z = 0 !
My Sir taught me this.
Now for all those "triangles", just follow the above equation... with your finger!
Seriously.
First, consider + to be one direction. Eg arrow heads point clock-wise when the vector is POSITIVE.
So this means when vector is NEGATIVE (But this is for the equation above!), they'll face the opposite way.
The triangle in which X points one way (eg clock-wise), and Y and Z point against it, (eg anti-clockwise) is THE triangle.

And u know, following 'em arrows with your finger does help loads.

HelpfuL?

 

[Obsidian Fl1ght]

Stationary waves
Stationary waves
Someone end this agony and summarise everything abt these waves to me...

 

[salvatore]

The best way (well, according to me) to do this, is make an equation with all the vectors on ONE SIDE.

First example:
q3
X - Y ... u have to find the resultant vector for this which is Z.
So Z = X - Y
Now, according to the "best way", X - Y = Z will be X - Y - Z = 0 !
My Sir taught me this.
Now for all those "triangles", just follow the above equation... with your finger!
Seriously.
First, consider + to be one direction. Eg arrow heads point clock-wise when the vector is POSITIVE.
So this means when vector is NEGATIVE (But this is for the equation above!), they'll face the opposite way.
The triangle in which X points one way (eg clock-wise), and Y and Z point against it, (eg anti-clockwise) is THE triangle.

And u know, following 'em arrows with your finger does help loads.

HelpfuL?

Thanks a lot for the explanation. It is surely the best way.. lol

Helpful indeed!

 

[salvatore]

Stationary waves
Stationary waves
Someone end this agony and summarise everything abt these waves to me...

Main concepts:

• do not appear to propagate
• are produced by the interference of two waves traveling in opposite directions with the same frequency and amplitude
• Positions on a standing wave:
  • node: a point where the amplitude is zero or a minimum
    (always form at fixed ends)
  • antinode: a point where the amplitude is a maximum
    (always form at free ends)

Another important point to note:

The distance between a node and an antinode is always 1/4 of the wavelength.
The distance between 2 adjacent nodes/antinodes is 1/2 of the wavelength.

Not in detail, but these are the important points that you should be knowing for MCQs