Physics: Post your doubts here!

[gary221]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

15) When the force = F, and distance = s, work done = Fs.
Since work done = K.E. gain → Fs = 8-4 = 4J
Fs = 4J

So, when force = 2F, and distance = 2s, work done = 2F * 2s →4Fs.
Since Fs = 4J, 4Fs = 4 * 4 = 16 J.
So gain in K.E. = 16 J.
And since the kinetic energy already equals 4 J....the new K.E. = 4 + 16 → 20 J
Ans = B

Hope u get it!
All credit to Nibz.

 

[gary221]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

16) As i already said, work done = amount of energy gained.
So, since the K.E. of the vehicle = 4.5 * 10^5 J
and we know that work done = Fs.
So, Fs = 4.5 * 10^5
s = (4.5 * 10^5)/6000
s = 75 m.

Ignore the mass, thts just to confuse u.
Hope u gt it!
All credit to Nibz.

 

[Manobilly]

M/j 2007 q13.

 

[gary221]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

26) As shown in the fig, at time = 18 s, phase diff btw the 2 waves = 180° or π.
1 wavelength = 360°
So, 1/8 of a wavelength = 360°/8 → 45°.

Phase diff = 180° at 18 s.
So, phase diff = 45° at x.
x = (18 * 45)/180
x = 4.5 s

Ans = B

Hope u get it!
All credit to Nibz.

 

[gary221]

M/j 2007 q13.

ppr link pls?

 

[Manobilly]

ppr link pls?

Can't upload the paper ,here is th question though.

 

[white rose]

26) As shown in the fig, at time = 18 s, phase diff btw the 2 waves = 180° or π.
1 wavelength = 360°
So, 1/8 of a wavelength = 360°/8 → 45°.

Phase diff = 180° at 18 s.
So, phase diff = 45° at x.
x = (18 * 45)/180
x = 4.5 s

Ans = B

Hope u get it!
All credit to Nibz.

THANNNKK UU verrry much....May Allah grant u success in ur life..

 

[gary221]

THANNNKK UU verrry much....May Allah grant u success in ur life..

Aameen..wa iyyaki.
np...

 

[gary221]

Can't upload the paper ,here is th question though.

http://www.s-cool.co.uk/a-level/phy...uilibrium/remember-it/s-cool-revision-summary
Torque = force acting perpendicular to an object * horizontal dist btw the 2 forces.
Now the force acting on the beam is at an angle of 60° → 30° from the normal.
Now we have to find the force acting on the beam NORMALLY.
remember → cosΘ = base/hypotenuse.
cos(30°) = normal force/8
force acting normally = cos 30 * 8 → 6.9 N

So, torque = force * dist btw the forces.
torque = 6.9 * 0.6 → 4.2 Nm
Ans = B

Hope u get it.
All credit to Nibz.

 

[white rose]

26) As shown in the fig, at time = 18 s, phase diff btw the 2 waves = 180° or π.
1 wavelength = 360°
So, 1/8 of a wavelength = 360°/8 → 45°.

Phase diff = 180° at 18 s.
So, phase diff = 45° at x.
x = (18 * 45)/180
x = 4.5 s

Ans = B

Hope u get it!
All credit to Nibz.

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf
can u also clear these ques. to me
ques. 8, 12, 24 & 30

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
can u also clear these ques. to me
ques. 8, 12, 24 & 30

8)okay, the equation we consider here is → s = 1/2 * g * t^2
When time = T, the dist = s.
However when the time = 0.5 T →time^2 = (0.5 T)^2 →0.25 T
So, the dist is proportional to the square of time.
Thus, when the time reduces to half its original value, dist reduces to 1/4th its original value.
So, the dist covered in time = 0.5T = 0.25L.
Ans = B

Hope u get it!
All credit to Nibz.

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
can u also clear these ques. to me
ques. 8, 12, 24 & 30

24)http://www.s-cool.co.uk/a-level/phy...finitions-of-stress-strain-and-youngs-modulus
We know that E is inversely proportional to the area, and directly proportional to the length.
Now, when we increase the length to 3L, the extension also increases 3 times → 3x.
But becoz we increase the diameter to 2D, we have increased the area by 4 times. A = π(d/2)^2 → A is now 4A.
And since extension is INVERSELY proportional to area, the extension now falls → x/4

So, the new wire's extension = 3x/4
Ans = B

Hope u gt it!
All credit to Nibz

 

[Manobilly]

Jza

http://www.s-cool.co.uk/a-level/phy...uilibrium/remember-it/s-cool-revision-summary
Torque = force acting perpendicular to an object * horizontal dist btw the 2 forces.
Now the force acting on the beam is at an angle of 60° → 30° from the normal.
Now we have to find the force acting on the beam NORMALLY.
remember → cosΘ = base/hypotenuse.
cos(30°) = normal force/8
force acting normally = cos 30 * 8 → 6.9 N

So, torque = force * dist btw the forces.
torque = 6.9 * 0.6 → 4.2 Nm
Ans = B

Hope u get it.
All credit to Nibz.

JazakAllah both of you

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
can u also clear these ques. to me
ques. 8, 12, 24 & 30

30)http://www.s-cool.co.uk/a-level/physics/diffraction/revise-it/diffraction-from-a-diffraction-grating
nλ = dsinΘ
Now, n = 1/(1 * 10^-6) → 1 * 10^6
Also, the angle of 70° is btw the 2 maxima so, the angle from the centre = 70/2 → 35°
Applying the formula,
λ = sin 35/(1 * 10^6)
λ = 574 nm

Ans = C
Hope u gt it!
All credit to Nibz.

 

[Manobilly]

M/j 2005 q29

 

[HongYue]

M/j 2005 q29

Since angle and diffraction grating is the same. Therefore, n(lamda) = n(lamda)
so 3(x) = 2(600nm)

 

[HongYue]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
Question 8, 9, 11, 12,
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Question 18, 24, 28

 

[HongYue]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
Question 5 , 8, 11, 12, 14, 15 ,20, 25
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf
Question 9

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdfQuestion 16, 28 29, 30,I still dontunderstand all the graph for terminal velocityExample S-T graph, V-t Graph and A-T graph with and without air resistance.Anybody mind explaning ?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
Question 4, 21, 26

 

[Tkp]

help me in oct nov 11
q-26
ans is b
Nibz

 

[Monojit Saha]

hi guys i have many problems to discuss i am writing all the problems below

firsly i have seen many mcqs about how many maximas minimas or bright fringes or observed in differaction grating and number of line per mm is D or N. i remeber one mcq in past paper plzz solve it oct/nov 2012 paper 12 que 30

second problem mcqs 13 from oct/2012
oct/nov 2011 paper variant 12 q26,31 why c is not write in q 31
oct/nov 2011 variant 11 q20 may june 2011 varient 11 q 27

Firstly, if there are 300 lines in 1mm then there would be 300 x 1000 lines in 1meter .. therefore D is 1/300000.
taking lamda as @..
n@= d sin&
put in the values of @ and d.
and take the value of sin@ as 1 because thats the value for the maximum angle
Find N.. it is somewhat 4.84
Thing is this was for only one side of the screen as angles are taken from the mid point in between the slits and thus the same number of maxima would be formed on the other side of the mid point between slits.. thus 4.76 x 2= 9.65 but as maximA OCCURS AT WHOLE NUMBER of N,, the maximum number of maxima is 9.