A Level Physics paper 1 doubts

[krtcobain82]

Guys I need answers + explanations for the following questions.

1) Oct/nov 07
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf

Question no  17, 25 and 40.

2) Oct/Nov 06
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w06_qp_1.pdf

Question no  10 and 27

3) May/june 08
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s08_qp_1.pdf

Question no   14, 26 and 37.

[Malak]

Sorry, i dont do CIE
But it might help if you post ur doubts in this thread (specifically for CIE physics)
https://studentforums.biz/sciences-149/all-cie-physics-doubts-here-!!!/720/

[EMO123]

Guys I need answers + explanations for the following questions.

1) Oct/nov 07
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf

Question no  17, 25 and 40.

in 17
as such density=mass/volume but here the story seems something different
i dint get that

[Ghost Of Highbury]

Guys I need answers + explanations for the following questions.

1) Oct/nov 07
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w07_qp_1.pdf

Question no  17, 25 and 40.

2) Oct/Nov 06
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w06_qp_1.pdf

Question no  10 and 27

3) May/june 08
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s08_qp_1.pdf

Question no   14, 26 and 37.

1.17. The density of the substance (x cm^3) = Density of the same substance (of unit volume)
       Keeping this in mind, density = mass/volume

        Mass of an atom = Mp, there Np atoms in all

        Thus, mass of Np atoms = Mp*Np

        Volume = unit volume
Density of P = (Mp*Np)/unit volume
Repeat the same with Q, density = (Mq*Nq)/unit volume
As density of P > density of Q
Mp*Np > Mq*Nq
(Unit volume gets cancelled)

25. dsin theta = n*lambda
      we know, d sin 45 = 3*lamda
     d/lambda = 3/sin 45 = 4.24264.....

Now, highest order diffraction, we take theta as 90, dsin 90/lamda = n = d/lamda * sin 90 = d/lambda = 4.24264...
Meaning, the 4th order diffraction is completely visible

40. Momentum = mass * velocity
     same p.d = same final velocity. Therefore, the one with the greatest mass has highest momentum = a-particle
--------
10. A body of mass m subject to a net force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma (newton's 2nd law)

27. (I'll use # for lambda)
      # = ax/D and d sin theta = n#
# is constant, thus, ax/D = dsin theta /n

x = dsin theta/n * D/a

a = 500d, D = 1m, theta = 30, n = 1

x = dsin 30/1 * 1/500d
  = 1/1000 = 1.0 * 10^-3

-----
14. Moments about Base. Forces creating torque at base are W (Mid-one) and F(wall).
equilibrium position, thus Fh - Wa = 0
Moments about point on the wall (Tip of ladder).
Fh+Wa - 2Wa = 0
Fh = Wa

Answer is A => Wa + Fh = 2Wa
Fh = Wa
26. I1 = Intensity at P, I2 = Intensity at Q
      I1 = k/r^2
      k = I1*r^2
     I2 = k/4r^2
     k = I2 * 4r^2
    I2 * 4r^2 = I1*r^2
I1/4 = I2

Also, I1 = kA^2
           = k*(8*10^-6)^2
Therefore I2 = k*(8*10^-6)^2 / 4 = 1.6*10^-11k
I2 = kA^2
A = sqrt(I2/k) = sqrt((1.6*10^-11k)/k) = 4*10^-6

37. I = V/R
    V across R2 = 50/150 * 6 = 2

    2/100kohms = 20*10^-6 A

[EMO123]

1.17. The density of the substance (x cm^3) = Density of the same substance (of unit volume)
       Keeping this in mind, density = mass/volume

        Mass of an atom = Mp, there Np atoms in all

        Thus, mass of Np atoms = Mp*Np

        Volume = unit volume
Density of P = (Mp*Np)/unit volume
Repeat the same with Q, density = (Mq*Nq)/unit volume
As density of P > density of Q
Mp*Np > Mq*Nq
(Unit volume gets cancelled)

25. dsin theta = n*lambda
      we know, d sin 45 = 3*lamda
     d/lambda = 3/sin 45 = 4.24264.....

Now, highest order diffraction, we take theta as 90, dsin 90/lamda = n = d/lamda * sin 90 = d/lambda = 4.24264...
Meaning, the 4th order diffraction is completely visible

40. Momentum = mass * velocity
     same p.d = same final velocity. Therefore, the one with the greatest mass has highest momentum = a-particle
--------
10. A body of mass m subject to a net force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma (newton's 2nd law)

27. (I'll use # for lambda)
      # = ax/D and d sin theta = n#
# is constant, thus, ax/D = dsin theta /n

x = dsin theta/n * D/a

a = 500d, D = 1m, theta = 30, n = 1

x = dsin 30/1 * 1/500d
  = 1/1000 = 1.0 * 10^-3

-----
14. Moments about Base. Forces creating torque at base are W (Mid-one) and F(wall).
equilibrium position, thus Fh - Wa = 0
Moments about point on the wall (Tip of ladder).
Fh+Wa - 2Wa = 0
Fh = Wa

Answer is A => Wa + Fh = 2Wa
Fh = Wa
26. I1 = Intensity at P, I2 = Intensity at Q
      I1 = k/r^2
      k = I1*r^2
     I2 = k/4r^2
     k = I2 * 4r^2
    I2 * 4r^2 = I1*r^2
I1/4 = I2

Also, I1 = kA^2
           = k*(8*10^-6)^2
Therefore I2 = k*(8*10^-6)^2 / 4 = 1.6*10^-11k
I2 = kA^2
A = sqrt(I2/k) = sqrt((1.6*10^-11k)/k) = 4*10^-6

37. I = V/R
    V across R2 = 50/150 * 6 = 2

    2/100kohms = 20*10^-6 A

thanx highburry