Thursday, June 19, 2014

9702 November 2013 Paper 41 42 Worked Solutions | A-Level Physics

  • 9702 November 2013 Paper 41 & 42 Worked Solutions | A-Level Physics


Paper 41 and Paper 42 are similar


SECTION A

Question 1
(a)
Gravitational potential at a point is the work done in moving unit mass from infinity to that point.

(b)
(i)
gravitational potential energy = GMm /x
Energy = (6.67x10-11 x 7.35x1022 x 4.5) / (1.74x106) = 1.27 x 107J

(ii)
Change in gravitational potential energy = change in kinetic energy
½ x 4.5 x v2 = 1.27x107
v = 2.4 x 103ms-1

(c)
As the potential at Earth’s surface is not zero (less than zero), the Earth would attract the rock. Thus, the escape speed would be lower.



Question 2
(a)
(i)
N is the total number of molecule

(ii)
<c2> is the mean square speed/velocity

(b)
pV = 1/3 Nm<c2> = NkT
<c2> = 3kT / m
Mean kinetic energy = ½ m<c2> = (3/2)kT

(c)
(i)
Either
1 mol contains NA molecules = 6.02x1023 molecules
Energy required for 1 mol = (3/2)NAkT = (3/2) x 1.38x10-23 x 1.0 x 6.02x1023
                                          =  12.5J
Or
R = molar gas constant = kNA = 8.31 JK-1mol-1
Energy = pV = (3/2)RT = (3/2) x 8.31 x 1.0 = 12.5J

(ii)
The energy required to raise the temperature of the gas by 1K is now greater because energy is needed to push back (do work against) the atmosphere.



Question 3
(a)
(i)
Choose any 2 from: 0.3s, 0.9s, 1.5s

(ii)
Either
From graph,
Maximum displacement, x = 1.5x10-2m
Period, T = 1.2s
v = ωx and      ω = 2π/T
v = (2π/1.2) x 1.5x10-2 = 0.079ms-1

Or
Draw gradient clearly at either t = 0.3s or t = 1.5s.
Show the working clearly to give (0.08 ± 0.01)ms-1

(b)
(i)
The curve should be from (±1.5, 0) passing through (0, 25). The shape should be similar to the inverted curve for Ep.
Intersections should occur on both sides between y =12.0 to 13.0mJ.

(ii)
At maximum amplitude, potential energy is the total energy.
From graph, total energy = 4.0mJ.


Question 4
{Detailed explanations for this question is available as Solution 574 at Physics 9702 Doubts | Help Page 113 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-113.html}




Question 5
(a)
(i)
Field: right to left

(ii)
Lines are more spaced out at ends

(b)
The Hall voltage depends on angle between the field and the plane of the probe. (It is maximum when the field is normal to the plane or the probe and zero when the field is parallel to the plane of the probe.)

(c)
(i)
Faraday’s law of electromagnetic induction states the induced (electromotive force) e.m.f is proportional to the rate of change of (magnetic) flux (linkage)

(ii)
Choose any 3:
Move the coil towards/away from solenoid (not considered 2 different ways)
Rotate the coil
Vary current in solenoid
Insert an iron core into the solenoid



Question 6
{Detailed explanations for this question is available as Solution 713 at Physics 9702 Doubts | Help Page 144 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html}






Question 7
{Detailed explanations for this question is available as Solution 578 at Physics 9702 Doubts | Help Page 114 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-114.html}



Question 8
(a)
Nuclear binding energy is the energy needed to separate nucleons (in a nucleus) to infinity.

(b)
(i)
Type of nuclear reaction: Fission

(ii)


The peak corresponds to atomic mass of Iron which is about 56. You should refer to the nucleon number for positions.
U : near right-hand end of line
Mo : to right of peak, less than 1/3 distance to U
La : 0.4-0.6 of distance from peak to U


(iii)
1. 

Left-hand side, mass = 236.132u
Right-hand side, mass = 235.92185u (sum of all on right hand side)
Right-hand side, mass = 235.922u
Mass change = 0.210u


2.
Energy = mc2 = 0.210 x 1.66x10-27 x (3.0x108)2 = 3.1374x10-11J = 196MeV
 




SECTION B
Question 9
{Detailed explanations for this question is available as Solution 674 at Physics 9702 Doubts | Help Page 135 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-135.html}




Question 10
(a)
Main principles behind the use of ultrasound to obtain diagnostic information about the internal body structures:
A quartz / piezo-electric crystal produces a pulse of ultrasound that is reflected from the boundaries between the media present. The reflected pulse is detected by the ultrasound transmitter and the signal is processed and displayed. The intensity of the reflected pulse gives information about the boundary and the time delay gives information about the depth.

(b)
Advantage of use of high frequency ultrasound as compared with low frequency ultrasound for medical diagnosis:
High frequency ultrasound has a shorter wavelength than low frequency ultrasound, so smaller structures are resolved/detected.

(c)
(i)
I = I0exp(-μx)
Ratio = I/I0 = exp (- 23 x 6.4x10-2) = 0.23

(ii)
An ultrasound transmitter emits a pulse. When the signal from the pulse is processed, any signal received later at the detector is usually amplified more than received at an earlier time because:
The later signal has passed through a greater thickness of medium and so, has greater attenuation / greater absorption / smaller intensity




Question 11
(a)
Most significant bit of sample: underline left-hand bit

(b)
1010, 1110, 1111, 1010, 1001

(c)
There are significant changes in detail of changes in V between the times at which the samplings took place. So, the sampling frequency is inadequate since it is too low.




Question 12
(a)
Attenuation of a signal in channels of communication is usually measured on a logarithmic rather than a linear scale because the gain of amplifiers is a series found by addition (not multiplication). The logarithmic scale provides a smaller number.

(b)
For a particular channel of communication having low attenuation, input power = 6.5mW and the attenuation per unit length = 1.8dBkm-1
(i)
Name of channel of communication:  Optic fibre

(ii)
Distance over which the power of the signal is reduced to 1.5x10-15W:
P2 : input power          P1 : output power
Attenuation/dB = 10lg(P2/P1) = 10lg({6.5x10-3}/{1.5x10-15}) = 126
Length = 126/1.8 = 70km



16 comments:

  1. can you please do w13 qp 43
    question number 1c?

    ReplyDelete
    Replies
    1. the question is explained at
      http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-86.html

      Delete
  2. Salam
    Plz explain question 4 c) ii) 2 ...

    ReplyDelete
    Replies
    1. Wslm.
      The details have been updated. Check again

      Delete
  3. Asalamualaikum Sir,
    Could you please further explain how you've derived the eqns. (in the step: "Either E = p2/2m or p = (2Em)½") in question 7 part (c)?

    ReplyDelete
    Replies
    1. The derivations have been added, though this equation may be used directly.

      Delete
  4. can you please draw 9bii) in w13 qp 43?

    ReplyDelete
    Replies
    1. See solution 664 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-133.html

      Delete
    2. Thank you. can you please also do w13 qp 42 9b ?

      Delete
    3. More details have been added for question 9

      Delete
  5. HI, could you please show how to sketch Q6(c)(ii)? Or some diagrams similar to it? Thanks!

    ReplyDelete
  6. hello,can you please explain why we multiply by 2 and square in 4 aii) w13 qp 42?

    ReplyDelete
    Replies
    1. I have added a few lines at the solution for the question. See if it helps.

      Delete
  7. Could you please show me an example of the sketch for Q6(c)(ii) ? Thanks.

    ReplyDelete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | 9702 November 2013 Paper 41 42 Worked Solutions | A-Level Physics