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Physics: Post your doubts here!

Snow Angel

Snow Angel

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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!!!
 
B

Browny

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TheJDOG said:
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
Click to expand...
Thanks so much I posted this several times and I got the answer from only you.(y)
 
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sagar65265

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Snow Angel said:
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!!!
Click to expand...

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!
 
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sagar65265

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boxfire1995 said:
Someone please?
Click to expand...

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!
 
omaaaar

omaaaar

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

Q5 part (b),please!!!
Click to expand...
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
 
S

Snowysangel

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