Physics: Post your doubts here!

[Dhaval Shah]

Oct/Nov 2004, Paper2, Question 1, Part B) (ii)

I got the percentage uncertainty of diameter of wire as (0.02 / 0.50) * 100 = 4 %

Now when I try to get the uncertainty of the area of cross-section of the wire, the answer doesn't come up as 8%.

This is how I do it:
Cross-sectional area = pi * (r^2)
Diameter = 0.50 mm / 2 = 0.25 mm
pi (0.25^2) = 0.196 ~= 0.2 mm^2
% uncertainty of cross-sectional area = (uncertainty of area / area)* 100
% uncertainty of cross-sectional area = (0.02 / 0.2) * 100 = 10 % ????

Do I have to separately calculate the uncertainty of cross-sectional area? If yes, how?

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_2.pdf

Thanks in advance.

All that working for the second part is completely unnecessary, its just for 1 mark! First of all you should know that the percentage uncertainty in r is the same as in the diameter (4%) as dividing or multiplying by a constant has no effect on uncertainty. Now Area = pie x (r^2), and pie is a constant so again that will have no effect on uncertainty. so we only need the r^2. Now when there are indices involved you have to multiply the uncertainty by the index (for example, uncertainty in r = 4%, so uncertainty in (r^2) is 4% x 2 = 8%, so if it was (r^3) like maybe volume or something, the uncertainty would be 4% x 3 = 12%) so in this case all we have to do is 4 x 2 which gives us 8%, that's it!

 

[mrnt3250]

May/June 2006 - QP2 - Q1) (c)(ii)

What is the procedures to solve this part of question?
I don't get it.

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_ms_2.pdf

Thanks in advance.

 

[queen of the legend]

I cannot understand these two question: 4b and 5 all from may/june 2009 first variant paper 2
some1 explain pls

 

[queen of the legend]

how do i do qs 2a(ii) The mark scheme is confusing me.
Link: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_2.pdf

since displacemnt is distance covered in particular direction, when a car travels suppose from point A to B and goes back to A its total distance is AB + BA, but the DISPLACEMET is zero from point A. i hope u understand now !

 

[queen of the legend]

May/June 2006 - QP2 - Q1) (c)(ii)

What is the procedures to solve this part of question?
I don't get it.

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_ms_2.pdf

Thanks in advance.

In part b air resistance is taken as negligible. since F=crv which is the resistive force calculated as 3.6x10^-5 N is very very small compared to wieght of the ball which is equal to 0.15N . this justifies neglectin air resistace in part b

 

[Dhaval Shah]

I cannot understand these two question: 4b and 5 all from may/june 2009 first variant paper 2
some1 explain pls

For 4b you should just think of the springs as resistors (like resistors in parallel and in series) the two springs in the first box of the question are in series and the two springs in the second box are in parallel. Now in series, you can say (just like how you find total resistance of 2 resistors in series) the extension is e + e = 2e (if you want to take it as resistors then you can say that resistance of each is e, so total resistance would be 2e). Its the same for all of them, just take them as resistors. Now to find k for each, you just use the equation F=kx, and use the respective extensions that you have calculated for each arrangement.

For number 5 it was a real challenge, especially part (b), but i think i got it (not sure if i am right though), i dont think i need to explain (a) right? Ok so part be, first calculate path difference, to do this first find the distance S2M (distance from S2 to M) using Pythagoras, then the path difference would be S2M - S1M = 128 - 100 = 28 (path difference is basically the difference that each wave has to travel, as the wave from S1 only travels 100 cm and wave from S2 travels 128 cm). Now for the intensity of sound to be 0 (a minima) the path difference should be (n +0.5)(lamda). So we have the speed (330) and 2 frequencies (1 kHz and 4kHz) so calculate 2 values of lamda using both these 2 frequencies, you will get 33 cm and 8.25 cm. Now after this part i am a bit less sure but i think our minima should be any point between 33 and 8.25. Remember where we said path difference and we said (n+0.5) x(lamda) right? Replace that n with any number starting from 1 alright, and equate that to your path difference (first it would be (1.5)(lamda) = 28) find lamda for each, now after that replace n with 2 then with 3 then with 4 and find lamda for each, now count how many of your lamdas that you found are between 33 and 8.25 (the two values we calculated before, you should get 2 values, that is the number of minima! But im not so sure if im right, but it seems like its right

Sorry for it being so long but that question was kinda tricky, hope you understand

 

[mrnt3250]

May/June 2007 - Question Paper 2 - Question 4 - Part (d)(i) and (ii)

I've tried till here, but I don't know how to get the angle ? Maybe my diagram is wrong?

(I know its not neat , it was, initially.. )

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_ms_2.pdf

Thanks in advance.

 

[queen of the legend]

question 7 b(iii) from may/june 2009 paper 2 first variant. i need explanation

 

[queen of the legend]

For 4b you should just think of the springs as resistors (like resistors in parallel and in series) the two springs in the first box of the question are in series and the two springs in the second box are in parallel. Now in series, you can say (just like how you find total resistance of 2 resistors in series) the extension is e + e = 2e (if you want to take it as resistors then you can say that resistance of each is e, so total resistance would be 2e). Its the same for all of them, just take them as resistors. Now to find k for each, you just use the equation F=kx, and use the respective extensions that you have calculated for each arrangement.

For number 5 it was a real challenge, especially part (b), but i think i got it (not sure if i am right though), i dont think i need to explain (a) right? Ok so part be, first calculate path difference, to do this first find the distance S2M (distance from S2 to M) using Pythagoras, then the path difference would be S2M - S1M = 128 - 100 = 28 (path difference is basically the difference that each wave has to travel, as the wave from S1 only travels 100 cm and wave from S2 travels 128 cm). Now for the intensity of sound to be 0 (a minima) the path difference should be (n +0.5)(lamda). So we have the speed (330) and 2 frequencies (1 kHz and 4kHz) so calculate 2 values of lamda using both these 2 frequencies, you will get 33 cm and 8.25 cm. Now after this part i am a bit less sure but i think our minima should be any point between 33 and 8.25. Remember where we said path difference and we said (n+0.5) x(lamda) right? Replace that n with any number starting from 1 alright, and equate that to your path difference (first it would be (1.5)(lamda) = 28) find lamda for each, now after that replace n with 2 then with 3 then with 4 and find lamda for each, now count how many of your lamdas that you found are between 33 and 8.25 (the two values we calculated before, you should get 2 values, that is the number of minima! But im not so sure if im right, but it seems like its right

Sorry for it being so long but that question was kinda tricky, hope you understand

thanks a lot ! i appreciate it

 

[mrnt3250]

Another one..
Oct/Nov 2010 - QP 22 - Question 1 - part (b).

I have no idea about doing questions related to horizontal components, the ones with tan, etc.
Could someone give me a complete explanation of how to deal with such questions?

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_ms_22.pdf

Thanks.

 

[VelaneDeBeaute]

May/June 2007 - Question Paper 2 - Question 4 - Part (d)(i) and (ii)

I've tried till here, but I don't know how to get the angle ? Maybe my diagram is wrong?

(I know its not neat , it was, initially.. )

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_ms_2.pdf

Thanks in advance.

If you need to get the angle that you've question marked, simply subtract it from 90 degrees.

 

[bamteck]

For 4b you should just think of the springs as resistors (like resistors in parallel and in series) the two springs in the first box of the question are in series and the two springs in the second box are in parallel. Now in series, you can say (just like how you find total resistance of 2 resistors in series) the extension is e + e = 2e (if you want to take it as resistors then you can say that resistance of each is e, so total resistance would be 2e). Its the same for all of them, just take them as resistors. Now to find k for each, you just use the equation F=kx, and use the respective extensions that you have calculated for each arrangement.

For number 5 it was a real challenge, especially part (b), but i think i got it (not sure if i am right though), i dont think i need to explain (a) right? Ok so part be, first calculate path difference, to do this first find the distance S2M (distance from S2 to M) using Pythagoras, then the path difference would be S2M - S1M = 128 - 100 = 28 (path difference is basically the difference that each wave has to travel, as the wave from S1 only travels 100 cm and wave from S2 travels 128 cm). Now for the intensity of sound to be 0 (a minima) the path difference should be (n +0.5)(lamda). So we have the speed (330) and 2 frequencies (1 kHz and 4kHz) so calculate 2 values of lamda using both these 2 frequencies, you will get 33 cm and 8.25 cm. Now after this part i am a bit less sure but i think our minima should be any point between 33 and 8.25. Remember where we said path difference and we said (n+0.5) x(lamda) right? Replace that n with any number starting from 1 alright, and equate that to your path difference (first it would be (1.5)(lamda) = 28) find lamda for each, now after that replace n with 2 then with 3 then with 4 and find lamda for each, now count how many of your lamdas that you found are between 33 and 8.25 (the two values we calculated before, you should get 2 values, that is the number of minima! But im not so sure if im right, but it seems like its right

Sorry for it being so long but that question was kinda tricky, hope you understand

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf

please help me for no. 3(a)(ii) and b(i)

 

[bamteck]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf

Please someone help me for no. no. 3(a)(ii) and b(i)

 

[Dhaval Shah]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf

please help me for no. 3(a)(ii) and b(i)

For 3)a)ii)1. where the line stops and the curve starts, that's the point because Hooke's law is obeyed when extension is proportional to the load (force). and for 2. you know the equation relating F, k and x is F = kx, so to find k it would be the inverse of the gradient of the graph starting from 0 and only uptil the point that you marked in 1.

Now b(i)1., the extension of the wire will be the distance the pulley has rotated ok? so we have to calculate the arc made by the angle 6.5 degrees on the pulley. You will find that to be 0.17 cm. For 2. all you need to is calculate the strain caused by the extension of the wire by 0.17 cm, strain is extension/length, so it would be (0.17 cm)/(250 cm) = 0.00068, this value calculated is also the increase as the initial strain was 0 because there was 0 extension.

 

[bamteck]

For 3)a)ii)1. where the line stops and the curve starts, that's the point because Hooke's law is obeyed when extension is proportional to the load (force). and for 2. you know the equation relating F, k and x is F = kx, so to find k it would be the inverse of the gradient of the graph starting from 0 and only uptil the point that you marked in 1.

Now b(i)1., the extension of the wire will be the distance the pulley has rotated ok? so we have to calculate the arc made by the angle 6.5 degrees on the pulley. You will find that to be 0.17 cm. For 2. all you need to is calculate the strain caused by the extension of the wire by 0.17 cm, strain is extension/length, so it would be (0.17 cm)/(250 cm) = 0.00068, this value calculated is also the increase as the initial strain was 0 because there was 0 extension.

But how do i calculate the arc ?

 

[Dhaval Shah]

But how do i calculate the arc ?

it would be ((6.5)/(360) * 2(pi)(r), where r is the radius

 

[bamteck]

it would be ((6.5)/(360) * 2(pi)(r), where r is the radius

Thanks

 

[raamish]

hey guys need help in qs 2d) can some1 tell me the procedure to do this question. Link:
www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf

 

[Muhammad Talha]

Plz solve this ques: Q.1 part (b) of this ppr http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_22.pdf

No one is here to help me...?

 

[ZohaibAsad]

No one is here to help me...?

Use displacement/time at each of the times 20,40,60,80,100 keep in mind that after 50s velocity is negative. Plot values of velocity, draw your graph. Done