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Hey guys...Can anybody explain me q10 of s12 paper 12?
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Physics: Post your doubts here!
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf
Q35 How is D is correct?
Q35 How is D is correct?
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The resistivity of both wires is equal to ρ, since they are both made of the same material and resistivity is only a property of a material, not a sample.
The formula relating resistance to the physical properties of the sample concerned is
Resistance = ρl/A
In this case, there are two wires, so let's focus on each at a time.
(First thing to note, though, that the area of a circular cross-section can be given in terms of the diameter as well; inserting d/2 for r in the formula πr² gives us πd²/4)
Let's take P. This has a length of 2l, and a diameter of d. Therefore, the cross sectional area A will be equal to πd²/4.
So we can write the resistance equation as
R(P) = ρ(2l)/(πd²/4) = 2ρl/(πd²/4) = (2ρl * 4)/(πd²) = 8ρl/πd²
This is the resistance of P. Let's do Q.
The length of Q is l, the diameter of Q is 3d. Therefore, the cross sectional area A of Q will be equal to π(3d)²/4 = π(9d²)/4 = 9πd²/4
Therefore, the resistance of Q is given by
R(Q) = ρ(l)/(9πd²/4) = 4ρl/9πd²
This is the resistance of Q. However, the question doesn't ask for resistance ratios. It asks for current ratios.
So, assuming that no connecting wires have any resistance, we can see that the potential difference across P is the same as the potential difference across Q (since both are connected in parallel, and the potential difference across parallel branches is the same). Therefore, using V = IR,
I(P) = V/R(P)
I(P) = V/(8ρl/πd²) =πd²V/ 8ρl
Where V is a variable for the potential difference.
In Q, the same applies:
I(Q) = V/R(Q) = V/(4ρl/9πd²) =9πd²V/4ρl
So I(P)/I(Q) = [πd²V/8ρl]/[9πd²V/4ρl] = (4ρl * πd²V)/(8ρl * 9πd²V) = 4/(8 * 9) = 1/18 = D.
Hope this helped!
Good Luck for all your exams!
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how?..plx explain..
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
questions 29,34,36,37,38,39 please!
questions 29,34,36,37,38,39 please!
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Since it is transverse wave, the speed same throughout and displacement varies.The option C about K.E is purely rubbish. While the option D is correct because it is a fact.
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I had solve this question, go few pages back or gimme paper link
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Actually the particle is changing its direction. Thus, from moving downwards to moving upwards, there is a change in direction. I am not sure about the speed here, but the acceleration would be high due to the change in direction, as acceleration has both magnitude and direction
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29)
c = f * λ
We know f = 1 / t
so c = 1 / t * λ
t = λ / c
now it took 3 wavefronts to reach XY to P
so t = 3λ / c
34)
36)
Its obviously D :
As Light intensity increases, its resistace decreases, so as R = V / I , R α V, so As R decreases, V would also decrease across LDR.
37)
current is passing through both lamps so they will glow dimly. dimly because the current will be less as now there are two compared to one light bulbs.
38)
What's hard in this ?
Find resistance in each case
3 would be highest with 0.7Ω
and 1 would be least with 0.4Ω
39)
Theory part maat pucho, chapter parho and then na aye to phucho
Last edited:
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9)
First find the time, using S = Ut + 0.5t^(2) ; S = 1.25 , U = zero, t = ?
That is 1.25= 0.5 x 9.81 x t^(2), t = 0.5 seconds.
Now for the Velocity, we know it equals Displacement / time taken ----> V = 10 / 0.5 = 20 m/s so in this type of question either of 2 quantities are given, we just need to find 3rd and input them all in second eqn to get the final result we are asked.
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V1= 5000/ (5000+5000) x 2 = 1V then V2= 3000/ (3000 + 2000) x 2 = 1.2 so V1 - V2 = -0.2 so C
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i'm sooo sooo sorry i pasted the wrong link x_x i it was supposed to be w12 not s12 ugh!29)
c = f * λ
We know f = 1 / t
so c = 1 / t * λ
t = λ / c
now it took 3 wavefronts to reach XY to P
so t = 3λ / c
34)
View attachment 44703
36)
Its obviously D :
As Light intensity increases, its resistace decreases, so as R = V / I , R α V, so As R decreases, V would also decrease across LDR.
37)
current is passing through both lamps so they will glow dimly. dimly because the current will be less as now there are two compared to one light bulbs.
38)
What's hard in this ?
Find resistance in each case
3 would be highest with 0.7Ω
and 1 would be least with 0.4Ω
39)
Theory part maat pucho, chapter parho and then na aye to phucho
For s12 my doubts are actually 7,10,17,28,30 ( and also 29,34,37 but you already answered them lol) soo sorry!