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Physics: Post your doubts here!

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

In question paper 21 can anyone explain Q)5) especially part b)?:)

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
 
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(ii)According to Kirchhoff's second law: E1 = І2R2/2 + І1R2/2 + І1R1+ І1r1
As I2 and I1 both pass half through the metal wire their p.d. will be divided by two.
(iii) E2 and I3 is in negative direction. p.d. across R2 will be half as the other half is not included in the loop. I1 is common across R1, r1 and R2.
E1 – E2 = –І3r2 + І1 (R1 + r1 + R2 / 2)
 
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Messages
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Reaction score
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(ii)According to Kirchhoff's second law: E1 = І2R2/2 + І1R2/2 + І1R1+ І1r1
As I2 and I1 both pass half through the metal wire their p.d. will be divided by two.
(iii) E2 and I3 is in negative direction. p.d. across R2 will be half as the other half is not included in the loop. I1 is common across R1, r1 and R2.
E1 – E2 = –І3r2 + І1 (R1 + r1 + R2 / 2)
I don't get why are we not including I2R2/2 in iii
 
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in this diagram the seperation of lines represent how weak the strength is. when the lines are almost parallel as well as the distance is same they are uniform. so by diagram they are uniform at centre of rod.
and decreasing means the lines are spreading aways(their distance increase).
 
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