Physics: Post your doubts here!

[Thought blocker]

Paper 34 - I haven't practiced a single experiment. :'(
Help me someone :'(

 

[ShreeyaBeatz]

I'm still a little confused.

I made the language simple. Take a little bit more time.
step 1 and 2 you dont have to worry. It will already be in your previous answers. they will also make you find k1 and k2, in the previous question.

now just simply put (suppose you got 20%)
Solve k1+ (20%of k1) =
solve k1- (20% of k1)=

does your value of k2(already found in previous question) lie between this?
If yes, it satisfies.

Tell me where you're confused. Take time to read

 

[Asad Moosvi]

I made the language simple. Take a little bit more time.
step 1 and 2 you dont have to worry. It will already be in your previous answers. they will also make you find k1 and k2, in the previous question.

now just simply put (suppose you got 20%)
Solve k1+ (20%of k1) =
solve k1- (20% of k1)=

does your value of k2(already found in previous question) lie between this?
If yes, it satisfies.

Tell me where you're confused. Take time to read

Does it matter if k1 is the smaller one or the larger one?

 

[Asad Moosvi]

I made the language simple. Take a little bit more time.
step 1 and 2 you dont have to worry. It will already be in your previous answers. they will also make you find k1 and k2, in the previous question.

now just simply put (suppose you got 20%)
Solve k1+ (20%of k1) =
solve k1- (20% of k1)=

does your value of k2(already found in previous question) lie between this?
If yes, it satisfies.

Tell me where you're confused. Take time to read

Can I just calculate the percentage difference between the two k's and if it's greater than the percentage uncertainty calculated earlier then it's not supported? And if it's less than or equal to percentage uncertainty calculated earlier, it is supported. Is this essentially the same thing?

 

[Asad Moosvi]

Paper 34 - I haven't practiced a single experiment. :'(
Help me someone :'(

How do you choose a sensible scale and draw the line of best fit? This always takes me a while.
Also, could you look at my previous post and tell me whether what I've said is correct or not. I'd really appreciate it.

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_11.pdf

I need your help in few (or many XP) questions. They are Q5,9,11,14 and 30.

 

[NIM]

guyzz need help!!!
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf
18, 19, 37
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
2, 6, 12, 15, 36
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf
2, 9, 13, 21, 26, 31, 33

 

[NIM]

help with question 7,33,34 and 40 plz!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

i m nt so sure abut 7 and 34
q 33
Sp = (5/10) * 2= 1 V
for voltage at Q = (2/5) * 2 = 0.8
for V2 = 2-0.8 = 1.2
V1 - V2
1 - 1.2 = -0.20V

q 40
for finding the acceleration with lowest speed you have to divide no of protons with atomic mass
A. H ... 1/1 = 1
B. He ... 2/4 = 0.5
C. Li ... 3/7 = 0.43
D. Be ... 4/9 = 0.44
C has the lowest speed so C would be the answer.

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_11.pdf

I need your help in few (or many XP) questions. They are Q5,9,11,14 and 30.

 

[ZaqZainab]

guyzz need help!!!
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf
18, 19, 37
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
2, 6, 12, 15, 36
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
2, 9, 13, 21, 26, 31, 33

i will help you with w07 and w06 i havent done 2009 yet
w007
Q18
Surface pressure=100kPa
Pressure=dgh
surface height = 100000/(1030*9.81)=9.897 meters
now at 450kPa
450000/(1030*9.81)=44.535
so below surface=44.535-9.897=34.6

Q19 you need to know this learn this!

Q37 nuclear process like alpha or beta decay the proton number changes no option saying proton number proton is a nucleon

W06
Q2 Energry=mgh=kgm^2 s^-2
lets see what the first option says
Ft=ma*t=kgms^-1 nope not correct
Fvt=mavt=kg*ms^-2*ms^-2*s= kgm^2s^-2 yup correct thats it B is the answer

Q6 V = IR
R =V/I = 8/1 = 8 ohms
change in R/R = change in V/V + change in I/I
change in R = 8(.2 + .4) = 2
Therefore (8+- 2) ohms

Q12 momentum always conserved
mv before=mv after
1000*5=10*x
x=500

Q15 since both the forces are 10
cos(30)*20 gives the upward force=17.32
now resultant 17.32-10=7.32

Q36 Kirchhoff first law
sum of current entering a point=sum of current leaving
I2=20.2+10.6=30.8
I1+10.6=10.8

 

[Kamihus]

Paper 34 - I haven't practiced a single experiment. :'(
Help me someone :'(

Dw you can get many marks even without touching the apparatus. In first question ask the supervisor to set-up the apparatus if you don't have any idea (you'll get 2 marks deducted for it though). Change the values and you'll get the readings. Draw a graph and calculate gradient and draw line of best fit, these are in every practical. Read some marking schemes and examiner reports and you'll get some idea. For the second question you'll get marks for just finding the percentage uncertainties and writing correct significant figures. If you have no idea how to do it just write something closer to the value given like if its written take 50 cm write something close to it. Write subsequent values accordingly that look sensible and do the following calculations. You can write the uncertainties and the improvements anyway and just read some ms to know when to write what.

 

[Asad Moosvi]

Dw you can get many marks even without touching the apparatus. In first question ask the supervisor to set-up the apparatus if you don't have any idea (you'll get 2 marks deducted for it though). Change the values and you'll get the readings. Draw a graph and calculate gradient and draw line of best fit, these are in every practical. Read some marking schemes and examiner reports and you'll get some idea. For the second question you'll get marks for just finding the percentage uncertainties and writing correct significant figures. If you have no idea how to do it just write something closer to the value given like if its written take 50 cm write something close to it. Write subsequent values accordingly that look sensible and do the following calculations. You can write the uncertainties and the improvements anyway and just read some ms to know when to write what.

Hey I need help with a part from the second question. Where you have to show whether or not the relationship is supported by calculating two values of k. I've asked this before, but I'm still confused! Different people are telling me different things, this has just made it worse. You seem like you'd know this. Please help me!

Here, use this paper to explain. http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_34.pdf

 

[NIM]

i will help you with w07 and w06 i havent done 2009 yet
w007
Q18
Surface pressure=100kPa
Pressure=dgh
surface height = 100000/(1030*9.81)=9.897 meters
now at 450kPa
450000/(1030*9.81)=44.535
so below surface=44.535-9.897=34.6

Q19 you need to know this learn this!

Q37 nuclear process like alpha or beta decay the proton number changes no option saying proton number proton is a nucleon

W06
Q2 Energry=mgh=kgm^2 s^-2
lets see what the first option says
Ft=ma*t=kgms^-1 nope not correct
Fvt=mavt=kg*ms^-2*ms^-2*s= kgm^2s^-2 yup correct thats it B is the answer

Q6 V = IR
R =V/I = 8/1 = 8 ohms
change in R/R = change in V/V + change in I/I
change in R = 8(.2 + .4) = 2
Therefore (8+- 2) ohms

Q12 momentum always conserved
mv before=mv after
1000*5=10*x
x=500

Q15 since both the forces are 10
cos(30)*20 gives the upward force=17.32
now resultant 17.32-10=7.32

Q36 Kirchhoff first law
sum of current entering a point=sum of current leaving
I2=20.2+10.6=30.8
I1+10.6=10.8

Thnx a lot dude it did help me a lot

 

[Kamihus]

Hey I need help with a part from the second question. Where you have to show whether or not the relationship is supported by calculating two values of k. I've asked this before, but I'm still confused! Different people are telling me different things, this has just made it worse. You seem like you'd know this. Please help me!

Here, use this paper to explain. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_34.pdf

You subtract the smaller value of k from the bigger one, then divide them by the bigger one, although I'm not entirely sure which value to divide. If its less than the percentage uncertainty its supported if its not then simply write its not supported rather than it has an inverse relation.

 

[NIM]

need help
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
5, 12, 13, 16, 23, 24, 25, 28, 29

 

[Thought blocker]

Stationary waves summery :​

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

 

[Asad Moosvi]

You subtract the smaller value of k from the bigger one, then divide them by the bigger one, although I'm not entirely sure which value to divide. If its less than the percentage uncertainty its supported if its not then simply write its not supported rather than it has an inverse relation.

Which percentage uncertainty? Sometimes we calculate a percentage uncertainty earlier which isn't even part of the equation.

 

[Thought blocker]

i'm sooo sooo sorry i pasted the wrong link x_x i it was supposed to be w12 not s12 ugh!
For s12 my doubts are actually 7,10,17,28,30 ( and also 29,34,37 but you already answered them lol) soo sorry!

w12_12
29)
View attachment 44719
As you can see from the figure, there will be total difference b/w nodes = 3λ / 2 = L
Now we know the velocity, we can calculate λ so f = v / λ
λ = 0.2 m
f = 1650 Hz

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.

36)
Here first we'll find the power at two points of current provided, and as its not a fixed current we would take the mean of power we get.
So it follows as :¬
Power at 2 ampere = 4 * 100 = 400W
Power at -1 ampere = -1 * 100 = -100W
Now as I said take mean of these powers : 500 / 2 = 250W

37)
Its not D as no current will pass if both switch are open. Then its not even A as both are closed so current will be I. In B S2 is open and S1 is closed so current just flows in series so current is still I. In C current flows through parallel circuit as S1 is open and S2 closed and current is not equal in parallel so it will not be I

38)
remember when two resistor are in series, more voltage drop is at resistor with more value, so their a flashing light then R = 5 mega ohm when dark, then 5M is very large than 1 K of other resistor, then almost all voltage drop at 5M, hope you get this, it is a very simple concept.

39)
Whats the big deal here ?
You know that nucleon number in helium is 4 (helium = alpha particle)
Then from original substance 100 alpha get removed every second, so in alpha, 4 * 100 = 400 nucleon each second.

s12_12
7)
Its A, because according to velocity time graph, we can see that its negative after half the time. That means that guy has changed his direction. this describes The scenario that a man run stratis gradually increasing hi speed and then after certain reducing speed and an he changes is direction(moves backward) with same speed pattern.
Now considering this troy with displacement, that guy reaches the farthest point and then comes back i.e max displacement after halftime.

10)
Keep this thing in mind when you are asked for projectile thingy :

• horizontal component of velocity = constant
• vertical component of acceleration constant = constant
• At top most, vertical velocity is zero

Now look it practically,
you throw a ball, when it reaches max height (top) it stops for a while
KE cannot be zero
KE = 0.5mv^2
v = root(square(vx) + square(vy))
vy = 0
but vx = constant
so v is nonzeror and hence KE canot be zero
similarly momentum canot be zero
P=mv

17)
Base has h = 0, i.e is block 1 = zero * mgh
Now we have other 3 blocks.
block 2 = at height h so m*g*h = mgh
block 3 = at height 2h = 2mgh
block 4 = height 3h = 3mgh
Sum all this P.E = 6mgh

28)
View attachment 44722

30)
Given
λ/2=33cm
so λ is 66cm or 0.66m
f=330/0.66=500
T=1/500=2ms
so it means wave should be completed in 4 blockss as the time base is 0.50ms/cm
the only one is B

 

[Kamihus]

Which percentage uncertainty? Sometimes we calculate a percentage uncertainty earlier which isn't even part of the equation.

The one you calculate in the earlier part like the one in w in the paper you gave above. That is used to calculate A which is in the equation. Alternatively you can just say its within a sensible limit like 10% so its supported.

 

[Thought blocker]

need help
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
5, 12, 13, 16, 23, 24, 25, 28, 29

5)
Vo = (5*2*1) = 10 (principle volume value)
Vu = (0.01/5) + (0.01/2) + (0.01/1) * 10 = 0.17

So volume = 10 +/- 0.17
And mass = 25 +/- 0.1

Uncertainty in density = (0.17/10) + (0.1/25) * 2.5 = 0.05

12)
Clockwise = 20 * 0.4 = 8 Nm
Anti-clockwise = 10 * 0.6 + 100 * 0.1 = 16 Nm (don't forget the weight of the beam!)

Therefore we need a clockwise moment of 8 more Nm

(20 * x) = 8, x = 0.4m from the pivot, so D

13)
Resultant torque = 45 N and resultant force = 60 N to the right

16)
Tension = mg sin θ = 10^3 sin 30 = 500 N (note: weight was already given, so need to multiply by 9.81)

Work = force * distance moved in direction of force = 500 * 5 = 2500 J

23)
In A and C, the amplitude is marked incorrectly. In D, λ is actually the time period.

24)
I α a^2 and I α f^2.

Rather than doing all the math to do this, compare the amplitude and frequency of the waves and use the formula to figure out this stuff:

If Q's amplitude is twice as much, the intensity will be four times as much.
If Q's frequency is half that of P, the intensity will be one-fourth.

Net change = 0, so the intensity remains the same.

25)
λ in water = 1500/150 = 10m and λ in air = 300/150 = 2m


28)
x=lambda*D/a
(Lambda/2*D)/2a
x=1/4*Lambda*D/a
so answer should be.. 0.75. A

29)
Since the adjacent 1st orders are 60 apart, that means one 1st order from the undeviated beam is 30 apart.

1.15 * 10^-6 sin 30 = 1 * λ
λ = 575 nm