Physics: Post your doubts here!

[Zepudee]

Even I'm so confused! I believe the graph should have started from 20cm.

Yeah, i think so too!

 

[sagar65265]

please can someone explain the following questions as soon as possible. thnx in advance and please can u draw diagrams where necessary. thnk u very much.

1) 4b iv from papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w11_qp_23.pdf

2) 4b iii from papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w12_qp_21.pdf

3) 5b and 5c from papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w13_qp_22.pdf


4) 4b ii and 4b iii from papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w13_qp_23.pdf

9) 2b and 2 c from papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s12_qp_22.pdf

please help!!!

1) - 4 b (iv):

1. Applying Kirchoff's Law, we get the value of the current to be 0.24 Amperes. So, the potential difference across a resistor with resistance R and current I flowing
through it is given by

P.D. = IV = 0.24 * 5.5 = 1.32 Volts. (However, since this is a drop, the change in potential should be -1.32 Volts)

2. The terminal P.D. across any cell is the Potential Difference across it's terminals, and this includes not only the cells that convert chemical energy to electric
potential energy, but the internal resistance. So, the terminal P.D. across cell A is the algebraic sum of the potential changes that occur as you go from end to end.

In other words, you start from the right side, this side being defined as the negative terminal of the battery, and you move across the battery. There are two
potential changes here - the drop across the internal resistance, and the increase across the cells.

So the potential drop across the internal resistance of Battery A = 0.24 * 2.3 = 0.55 Volts. Since this is a decrease in potential, we write it as -0.55 V.
In addition, the potential rise across the cells of Battery A = 4.4 Volts since that is the emf of the collection of cells. Since this is a rise in potential, we write it as
+4.4 V.

Adding these up, we get 4.4 - 0.55 = 3.848 = 3.85 Volts.

3. For this, we return to Kirchoff's Law, and we input our values. The equation from my page is
-2.3I + 4.4 - 2.1 - 1.8I - 5.5I = 0

Alternatively, this can be written as (Potential change across Battery A) + (Potential Change across Battery B) + (Potential change across Resistor R) = 0

From what we have calculated so far, we can write (3.85 Volts) + (Potential Change across Battery B) + (1.32 Volts) = 0
However, the one issue with this equation is that since the current is going from the positive terminal of Battery B to the negative terminal of battery B, the potential difference given by this equation will be (Potential of negative terminal) - (Potential of positive terminal).
By definition, this is the negative of the terminal PD, so while the value we get is -2.53 Volts, the terminal PD itself is 2.53 Volts.

2) 4 b (iii)

From the earlier part of the question, you will have obtained the value of 7 * 10^5 slits per meter. It can be assumed that this is the same value to be used in the next part, so we continue as follows:

Part b(ii) is talking about light of wavelength 625 nm, passing through the diffraction grating with (7 * 10^5) slits per meter, that has a second order maxima at an angle of 61.0° to the straight-through direction.

Part b(iii) wants us to find the wavelength of light that passes through a diffraction grating with (7 * 10^5) slits per meter, that has a maxima of some other order at 61.0° to the straight-through direction. So, we replace n = 2 in the previous question with either 1, or 3, or 4, etc.

However, we see first that the wavelength of line concerned should be in the visible part of the spectrum (this part ranges from about 400 nm to 800 nm) , and in the equation

nλ = dsin(61.0)

the right side is constant; the number of slits per meter (and therefore the distance between slits) is constant in both equations, and the sine function is also constant, since the angle remains the same in both situations.

Therefore, the value of nλ from the previous question is the same as that used here. So, since n was 2 in b(ii) and λ was 625 nm, we can write

1250 nm = nλ(2)

Supposing we put n = 1, we get λ = 1250 nm. This is not is the range of visible light, so we have to discard this.
Supposing we put n = 3, we get λ = 417 nm. This is indeed in the visible spectrum.
Supposing we put n =4, we get λ = 312.5 nm, which is too low to be visible.

So, our answer is 417 nm, since this is the only possible wavelength for which light with form a maxima at 61.0 degrees to the straight-through direction in these conditions.

I'll try out the rest in some time, just let me know if you've understood these.

Hope this helped!

Good Luck for all your exams!

 

[boxfire1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_43.pdf

Can someone please explain to me Q8 part bi? Why did they take the no. Of particles of uranium? Why not barium?

Someone please?

 

[_Ahmad]

anyone here ?

 

[omaaaar]

anyone here ?

Yeah me here

 

[_Ahmad]

Yeah me here

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Can you please help me with second varient Q7 b ii)

 

[sagar65265]

Someone please?

I'm not able to understand your question; hopefully this guess might help, if not i'm sorry, could you explain again?

Uranium-235, according to the given equation, can decay into Barium-141 and Krypton-92; each nucleus in the sample used will undergo this reaction, and since the equation is balanced for conservation of charge and mass, each nucleus of Uranium, when hit by a neutron, will produce 1 nucleus of Barium, another of Krypton and 3 neutrons, along with some energy.

So, when 1.2 grams of Uranium-235 decay, we need to find out how many nuclei of Barium-141 are produced. Since we have a one-one relation (i.e. one nucleus of
U-235 produces one nucleus of Ba-141) we can say that the number of Barium nuclei produced is equal to the number of Uranium nuclei that take part in the nuclear reaction. In this case, all of them react, so the number of atoms (and correspondingly, the number of nuclei) of U-235 is given by

Avogadro's Number * No. of moles

The number of moles of nuclei in the Uranium sample is given by

(Mass of Sample)/(Mass of 1 mole of element)

Substituting the values, we get

6.02 * 10^23 * 1.2/235 = 3.07 * 10^21 nuclei.

Hope this helped!

Good Luck for all your exams!

 

[immie.rose]

 

[immie.rose]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Q5 part (b),please!!!

 

[omaaaar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Q5 part (b),please!!!

Hey, for part b)

This is the key for this question: For minima to occur: Path difference= 0.5Lambda, 1.5 Lambda. 2.5 Lambda and so on...

If you look at the figure, you'll see that it's a right angled triangle, use paythagors theorem. You'll get S2M= 128 cm

Path diff= 128 - 100= 28 cm


Your range is from 1khz to 4 kHz

Use v= lambda x frequency
Speed = 330 m/s

Range of lambda= 8.25 cm to 33cm

Now, start trying, remember path diff= nlambda

So 28= 0.5 x Lambda, 56 not in allowed range
28= 1.5 x lambda, 18.6 allowed range
28= 2.5 x lambda , 11.2 allowed range

So 2 minima occur

 

[Snowysangel]

How was the exam guys? I think it was all right...not THAT easy as some people think. Some parts really required thinking

 

[omaaaar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Q5 part (b),please!!!

I just copy pasted it from one of the posts from last page

 

[omaaaar]

How was the exam guys? I think it was all right...not THAT easy as some people think. Some parts really required thinking

Have you done the exam

 

[omaaaar]

How was the exam guys? I think it was all right...not THAT easy as some people think. Some parts really required thinking

Can I inbox you to ask some information on what came and firstly which variant did you give

 

[immie.rose]

Hey, for part b)

This is the key for this question: For minima to occur: Path difference= 0.5Lambda, 1.5 Lambda. 2.5 Lambda and so on...

If you look at the figure, you'll see that it's a right angled triangle, use paythagors theorem. You'll get S2M= 128 cm

Path diff= 128 - 100= 28 cm


Your range is from 1khz to 4 kHz

Use v= lambda x frequency
Speed = 330 m/s

Range of lambda= 8.25 cm to 33cm

Now, start trying, remember path diff= nlambda

So 28= 0.5 x Lambda, 56 not in allowed range
28= 1.5 x lambda, 18.6 allowed range
28= 2.5 x lambda , 11.2 allowed range

So 2 minima occur

Thanks a bunchhhhhh!! JazakAllah kher.

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Can you please help me with second varient Q7 b ii)

You need to apply Kirchoff's Second Law for this question; if you put your finger at N, and move to M, then L, then K, and finally back to N, and sum up all the changes in potential energy of the current flowing through the wires then you should get zero (if you start at N, and return to N, then the potential has to be the same at the beginning and at the end, so the sum of changes has to be zero).

So, when you go from N to M, you can see that there is no resistance or impedance in the way (since the wires are assumed to have zero resistance). Therefore, no potential energy is lost or gained while going from N to M, so for this part we can write that the change in potential is = 0.

When you go then from M to L, the trick here is to notice that you are going from the positive terminal of the battery to the negative terminal of the battery. Since the positive terminal is considered to be at a higher potential than the negative terminal, you are going from a region of high potential to a region of low potential.
Therefore, you can write that the change in potential is negative.
But what is the value? It is simply the EMF of the battery. In other words, the potential decreases by the EMF of the battery E1, i.e. the change in potential is = -E1.

When you go from L to K, the same argument as N to M is made; the change in potential is = 0.

While going from L to N back again, you go from the negative terminal to the positive terminal, i.e. you go from a region of low potential to a region of higher potential. Therefore, the potential change is positive, and the magnitude is equal to the EMF of battery E2. In other words, the change in potential while crossing the battery E2 = +E2 Volts.

The last part of the journey from L to N involves going past the resistor; since you are going with the current, you are automatically going from a region of high potential to a region of lower potential, i.e. the potential difference decreases across the resistor (since current goes from a region of high potential to a region of low potential). The potential difference across the resistor, therefore, will be negative. The magnitude of this difference will be equal to I3 * r = RI3. Therefore, the change in potential is equal to -RI3.

Adding these up, we get -E1 + E2 - I3R = 0
Rearranging, we get E2 - E1 = I3R

Which is the answer.

Hope this helped!

Good Luck for all your exams!

 

[_Ahmad]

You need to apply Kirchoff's Second Law for this question; if you put your finger at N, and move to M, then L, then K, and finally back to N, and sum up all the changes in potential energy of the current flowing through the wires then you should get zero (if you start at N, and return to N, then the potential has to be the same at the beginning and at the end, so the sum of changes has to be zero).

So, when you go from N to M, you can see that there is no resistance or impedance in the way (since the wires are assumed to have zero resistance). Therefore, no potential energy is lost or gained while going from N to M, so for this part we can write that the change in potential is = 0.

When you go then from M to L, the trick here is to notice that you are going from the positive terminal of the battery to the negative terminal of the battery. Since the positive terminal is considered to be at a higher potential than the negative terminal, you are going from a region of high potential to a region of low potential.
Therefore, you can write that the change in potential is negative.
But what is the value? It is simply the EMF of the battery. In other words, the potential decreases by the EMF of the battery E1, i.e. the change in potential is = -E1.

When you go from L to K, the same argument as N to M is made; the change in potential is = 0.

While going from L to N back again, you go from the negative terminal to the positive terminal, i.e. you go from a region of low potential to a region of higher potential. Therefore, the potential change is positive, and the magnitude is equal to the EMF of battery E2. In other words, the change in potential while crossing the battery E2 = +E2 Volts.

The last part of the journey from L to N involves going past the resistor; since you are going with the current, you are automatically going from a region of high potential to a region of lower potential, i.e. the potential difference decreases across the resistor (since current goes from a region of high potential to a region of low potential). The potential difference across the resistor, therefore, will be negative. The magnitude of this difference will be equal to I3 * r = RI3. Therefore, the change in potential is equal to -RI3.

Adding these up, we get -E1 + E2 - I3R = 0
Rearranging, we get E2 - E1 = I3R

Which is the answer.

Hope this helped!

Good Luck for all your exams!

What if we go from N to K to L to M and then N

PLEASE HELP!!!!

 

[MiniSacBall]

But isn't the resistance of C decreasing with voltage? Why would the PD through it be greater than in R? Also, isn't the current through R constant? Why would there be a 'curve' for R?

Dude just jumping onto parts of questions will confuse you do the whole question,
Well first do b(i)
On Fig. 6.1, draw a line to show the variation with potential difference V of the
current I in resistor R.

Once you have drawn this curve, you will find that, for example take the I as 2 A. now take the P.D (Volts) of C and R from the graph! Example: At 2A, P.D for C is 28 while for R is 20.
So now recall the formula of Power = IV, Currently, we are only looking at I and V we don't need resistance, if you want to count resistance then count for each resistor its resistance at specific current and volts.
As P = IV ,
For C => 2 x 28 = 56
For R => 2 x 20 = 40
As we know Power is the rate of energy lost (dissipated)
Thus, C has higher rate of Energy dissipation, than R.

If you want to take resistance into account then, take a specific current, take readings of volts for both resistors, then calculate their resistance from V= IR.
Then, calculate the power using any power formula, there you go. You will find same results. Instead of guessing answers on the internet, just print and do it.
Hope that solves your problem!

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_23.pdf

Q1 a ii)
How to do it i mean apart from adding 0.02, why do we add 0.01, is it because of the +-1 digit.

Can Any one help me in this one.

And,
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_23.pdf
Q3 a)
How to calculate the area, by counting the squares right, i have got like 3600+ and the answer is only 39m
The heck!!

 

[princeali97]

HOW WAS THE EXAM?!